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similar counterparts to human signaling. [2] Based on the Evolution Connection, which of the following best describes the evolution of kinases? a. The tyrosine kinases evolved before yeast diverged from other eukaryotes, but the other fifty-five subfamilies of kinases evolved after yeast diverged. b. Fifty-five subfamilies of kinases evolved before yeast diverged from other eukaryotes, but the tyrosine kinases evolved after yeast diverged. c. All kinases evolved in yeast, but yeast later lost the tyrosine kinases because they do not need them. d. The evolution of tyrosine kinases involved in cellular communication occurred about 2.5 billion years ago. 2. G. Manning, G.D. Plowman, T. Hunter, S. Sudarsanam, “Evolution of Protein Kinase Signaling from Yeast to Man,” Trends in Biochemical Sciences 27, no. 10 (2002): 514–520. 392 Chapter 9 | Cell Communication Watch this collection (http://openstaxcollege.org/l/bacteria_biofilm) of interview clips with biofilm researchers in “What Are Bacterial Biofilms?” Recurrent urinary tract infections occur when the urinary tract becomes reinfected by the same bacteria. Why are recurrent urinary infections difficult to treat? a. Bacteria often form biofilms in recurrent infections and these may be more antibiotic resistant. b. Bacteria rarely form biofilms in recurrent infections, making them more resistant to antibiotics than if they were not in a biofilm. c. Bacteria produce biofilms which behave like a unicellular organism. d. Bacteria don't produce biofilms in recurrent infections but become resistant due to repeated exposure to antibiotics. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 9 | Cell Communication 393 KEY TERMS apoptosis programmed cell death autocrine signal signal that is sent and received by the same or similar nearby cells autoinducer signaling molecule secreted by bacteria to communicate with other bacteria of its kind and others cell-surface receptor cell-surface protein that transmits a signal from the exterior of the cell to the interior, even though the ligand does not enter the cell chemical synapse small space between axon terminals and dendrites of nerve cells where neurotransmitters function cyclic AMP |
(cAMP) second messenger that is derived from ATP cyclic AMP-dependent kinase (also, protein kinase A, or PKA) kinase that is activated by binding to cAMP diacylglycerol (DAG) cleavage product of PIP2 that is used for signaling within the plasma membrane dimer chemical compound formed when two molecules join together dimerization (of receptor proteins) interaction of two receptor proteins to form a functional complex called a dimer endocrine cell cell that releases ligands involved in endocrine signaling (hormones) endocrine signal long-distance signal that is delivered by ligands (hormones) traveling through an organism's circulatory system from the signaling cell to the target cell enzyme-linked receptor cell-surface receptor with intracellular domains that are associated with membrane-bound enzymes extracellular domain region of a cell-surface receptor that is located on the cell surface G-protein-linked receptor cell-surface receptor that activates membrane-bound G-proteins to transmit a signal from the receptor to nearby membrane components growth factor ligand that binds to cell-surface receptors and stimulates cell growth inhibitor molecule that binds to a protein (usually an enzyme) and keeps it from functioning inositol phospholipid lipid present at small concentrations in the plasma membrane that is converted into a second messenger; it has inositol (a carbohydrate) as its hydrophilic head group inositol triphosphate (IP3) cleavage product of PIP2 that is used for signaling within the cell intercellular signaling communication between cells internal receptor (also, intracellular receptor) receptor protein that is located in the cytosol of a cell and binds to ligands that pass through the plasma membrane intracellular mediator (also, second messenger) small molecule that transmits signals within a cell intracellular signaling communication within cells ion channel-linked receptor cell-surface receptor that forms a plasma membrane channel, which opens when a ligand binds to the extracellular domain (ligand-gated channels) kinase enzyme that catalyzes the transfer of a phosphate group from ATP to another molecule ligand molecule produced by a signaling cell that binds with a specific receptor, delivering a signal in the process mating factor signaling molecule secreted by yeast cells to communicate to nearby yeast cells that they are available to mate and communicating their mating orientation 394 Chapter 9 | Cell Communication neurotransmitter chemical ligand that carries a signal from one nerve cell to the next |
paracrine signal signal between nearby cells that is delivered by ligands traveling in the liquid medium in the space between the cells phosphatase enzyme that removes the phosphate group from a molecule that has been previously phosphorylated phosphodiesterase enzyme that degrades cAMP, producing AMP, to terminate signaling quorum sensing method of cellular communication used by bacteria that informs them of the abundance of similar (or different) bacteria in the environment receptor protein in or on a target cell that bind to ligands second messenger causes its release small, non-protein molecule that propagates a signal within the cell after activation of a receptor signal integration interaction of signals from two or more different cell-surface receptors that merge to activate the same response in the cell signal transduction propagation of the signal through the cytoplasm (and sometimes also the nucleus) of the cell signaling cell cell that releases signal molecules that allow communication with another cell signaling pathway (also signaling cascade) chain of events that occurs in the cytoplasm of the cell to propagate the signal from the plasma membrane to produce a response synaptic signal chemical signal (neurotransmitter) that travels between nerve cells target cell cell that has a receptor for a signal or ligand from a signaling cell CHAPTER SUMMARY 9.1 Signaling Molecules and Cellular Receptors Cells communicate by both inter- and intracellular signaling. Signaling cells secrete ligands that bind to target cells and initiate a chain of events within the target cell. The four categories of signaling in multicellular organisms are paracrine signaling, endocrine signaling, autocrine signaling, and direct signaling across gap junctions. Paracrine signaling takes place over short distances. Endocrine signals are carried long distances through the bloodstream by hormones, and autocrine signals are received by the same cell that sent the signal or other nearby cells of the same kind. Gap junctions allow small molecules, including signaling molecules, to flow between neighboring cells. Internal receptors are found in the cell cytoplasm. Here, they bind ligand molecules that cross the plasma membrane; these receptor-ligand complexes move to the nucleus and interact directly with cellular DNA. Cell-surface receptors transmit a signal from outside the cell to the cytoplasm. Ion channel-linked receptors, when bound to their ligands, form a pore through the plasma membrane through which certain ions can pass. G-protein-linked receptors interact with a G-protein on the cytoplasmic side of the plasma membrane, promoting |
the exchange of bound GDP for GTP and interacting with other enzymes or ion channels to transmit a signal. Enzyme-linked receptors transmit a signal from outside the cell to an intracellular domain of a membrane-bound enzyme. Ligand binding causes activation of the enzyme. Small hydrophobic ligands (like steroids) are able to penetrate the plasma membrane and bind to internal receptors. Water-soluble hydrophilic ligands are unable to pass through the membrane; instead, they bind to cell-surface receptors, which transmit the signal to the inside of the cell. 9.2 Propagation of the Signal Ligand binding to the receptor allows for signal transduction through the cell. The chain of events that conveys the signal through the cell is called a signaling pathway or cascade. Signaling pathways are often very complex because of the interplay between different proteins. A major component of cell signaling cascades is the phosphorylation of molecules by enzymes known as kinases. Phosphorylation adds a phosphate group to serine, threonine, and tyrosine residues in a protein, changing their shapes, and activating or inactivating the protein. Small molecules like nucleotides can also be phosphorylated. Second messengers are small, non-protein molecules that are used to transmit a signal within a cell. Some examples of second messengers are calcium ions (Ca2+), cyclic AMP (cAMP), diacylglycerol (DAG), and inositol This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 9 | Cell Communication 395 triphosphate (IP3). 9.3 Response to the Signal The initiation of a signaling pathway is a response to external stimuli. This response can take many different forms, including protein synthesis, a change in the cell’s metabolism, cell growth, or even cell death. Many pathways influence the cell by initiating gene expression, and the methods utilized are quite numerous. Some pathways activate enzymes that interact with DNA transcription factors. Others modify proteins and induce them to change their location in the cell. Depending on the status of the organism, cells can respond by storing energy as glycogen or fat, or making it available in the form of glucose. A signal transduction pathway allows muscle cells to respond to immediate requirements for energy in the form of glucose. Cell growth is almost always stimulated by external signals called growth factors. Uncontrolled cell growth leads |
to cancer, and mutations in the genes encoding protein components of signaling pathways are often found in tumor cells. Programmed cell death, or apoptosis, is important for removing damaged or unnecessary cells. The use of cellular signaling to organize the dismantling of a cell ensures that harmful molecules from the cytoplasm are not released into the spaces between cells, as they are in uncontrolled death, necrosis. Apoptosis also ensures the efficient recycling of the components of the dead cell. Termination of the cellular signaling cascade is very important so that the response to a signal is appropriate in both timing and intensity. Degradation of signaling molecules and dephosphorylation of phosphorylated intermediates of the pathway by phosphatases are two ways to terminate signals within the cell. 9.4 Signaling in Single-Celled Organisms Yeasts and multicellular organisms have similar signaling mechanisms. Yeasts use cell-surface receptors and signaling cascades to communicate information on mating with other yeast cells. The signaling molecule secreted by yeasts is called mating factor. Bacterial signaling is called quorum sensing. Bacteria secrete signaling molecules called autoinducers that are either small, hydrophobic molecules or peptide-based signals. The hydrophobic autoinducers, such as AHL, bind transcription factors and directly affect gene expression. The peptide-based molecules bind kinases and initiate signaling cascades in the cells. REVIEW QUESTIONS 1. Which of the following properties prevents the ligands of cell-surface receptors from entering the cell? a. The molecules bind to the extracellular domain. b. The molecules are hydrophilic and cannot penetrate the hydrophobic interior of the plasma membrane. c. The molecules are attached to transport proteins that deliver them through the bloodstream to target cells. d. The ligands are able to penetrate the membrane, directly influencing gene expression upon receptor binding. a. b. c. d. Ions are too large to diffuse through the membrane. Ions are charged particles and cannot diffuse through the hydrophobic interior of the membrane. Ions bind to hydrophobic molecules within the ion channels. Ions bind to carrier proteins in the bloodstream, which must be removed before transport into the cell. 4. Why are endocrine signals transmitted more slowly than paracrine signals? 2. The secretion of hormones by the pituitary gland is an example of which type of signaling? a. The ligands are transported through the bloodstream and travel greater distances. a |
. autocrine signaling b. The target and signaling cells are close together. b. direct signaling across gap junctions c. The ligands are degraded rapidly. c. endocrine signaling d. paracrine signaling 3. Why are ion channels necessary to transport ions into or out of a cell? d. The ligands do not bind to carrier proteins during transport. 5. Aldosterone is a steroid hormone that regulates reabsorption of sodium ions in the kidney tubular cells. What is the probable mechanism of action of aldosterone? 396 Chapter 9 | Cell Communication a. b. c. d. It binds gated ion channels and causes a flow of ions in the cell. It binds cell surface receptors and activates synthesis of cAMP. It binds to cell surface receptors and activates a phosphorylation cascade. It binds to an intracellular receptor and activates gene transcription. 6. The gas nitric oxide has been identified as a signaling molecule. Which of the following mechanisms of action would you expect from a gaseous molecule? a. A tyrosine kinase enzyme must be activated. b. GDP is exchanged for GTP. c. The receptor forms a dimer. d. The insulin molecule is internalized in the cytoplasm. 11. What is the function of a phosphatase? a. A phosphatase removes phosphorylated amino acids from proteins. b. A phosphatase removes the phosphate group from phosphorylated amino acid residues in a protein. c. A phosphatase phosphorylates serine, threonine, and tyrosine residues. d. A phosphatase degrades second messengers in It binds to a G-protein-linked receptor. It binds to a receptor tyrosine kinase. It binds to a gated ion channel. a. b. c. d. It binds to an intracellular receptor. the cell. 7. Where do DAG and IP3 originate? 12. How does NF-κB induce gene expression? a. They are formed by phosphorylation of cAMP. a. A small, hydrophobic ligand binds to NF-κB, b. They are ligands expressed by signaling cells. c. They are hormones that diffuse through the plasma membrane to stimulate protein production. d. They are the cleavage products of the inositol phospholipid, PIP2. 8. What property enables the residues of the amino acids serine, threonine, and |
tyrosine to be phosphorylated? a. They are polar. b. They are nonpolar. c. They contain a hydroxyl group. d. They occur more frequently in the amino acid sequence of signaling proteins. 9. Dopamine is a neurotransmitter in the brain that causes long-term responses in neurons and binds to a G-proteinlinked receptor. Which of the following chemicals would you expect to increase in concentration after dopamine binds its receptor? a. ATP b. cAMP c. calcium ions d. sodium ions 10. The hormone insulin binds to a receptor tyrosine kinase on the surface of target cells. Which of the following steps takes place before phosphorylation of tyrosine residues? activating it. b. NF-κB is phosphorylated and is then free to enter the nucleus to bind DNA. c. NF-κB is a kinase that phosphorylates a transcription factor that binds DNA and promotes protein production. d. Phosphorylation of the inhibitor IκB dissociates the complex between it and NF-κB, allowing NF-κB to enter the nucleus and stimulate transcription. 13. Apoptosis can occur in a cell under what conditions? a. when a cell is infected by a virus b. when a cell is damaged c. when a cell is no longer needed d. all of the above 14. Cancer cells that continue to divide when defective often show changes in what cellular function? a. apoptosis b. c. d. their mechanism of glycolysis the mechanism of protein biosynthesis replication of DNA 15. Epinephrine mediates the fight-or-fight response of the body. One of the effects is to increase the amount of glucose available to muscles. What does the signaling pathway triggered by epinephrine cause to occur in liver cells? a. activation of metabolism b. cell division c. d. inhibition of glucose metabolism by liver cells synthesis of enzymes This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 9 | Cell Communication 397 16. Which type of molecule acts as a signaling molecule in yeasts? a. autoinducer b. mating factor c. d. second messenger steroid 17. When is quorum sensing triggered to begin? a. a sufficient number of bacteria are present b. bacteria release growth hormones a. autocrine b. endocrine c. paracrine d. gap junction 19. |
The bioluminescent bacteria Vibrio fischeri produces luminescence only if the population reaches a certain density. What is the advantage of an autoinducer? a. An autoinducer allows the producer to act independently of the presence of other cells. b. An autoinducer does not diffuse away from the c. bacterial protein expression is switched on cell. d. treatment with antibiotics occurs 18. Yeast releasing mating factor can be classified as which type of signal? CRITICAL THINKING QUESTIONS 20. What is the difference between intracellular signaling and intercellular signaling? a. b. c. d. Intracellular signaling occurs between cells of two different species. Intercellular signaling occurs between two cells of the same species. Intracellular signaling occurs between two cells of same species. Intercellular signaling occurs between cells of two different species. Intracellular signaling occurs within a cell. Intercellular signaling occurs between cells. Intracellular signaling occurs between cells. Intercellular signaling occurs within cell. 21. What are the differences between internal receptors and cell-surface receptors? c. An autoinducer allows a positive feedback loop, which increases the response in proportion to the population size. d. An autoinducer presents no advantage for the cell. a. b. c. d. Internal receptors bind to ligands that are hydrophobic and the ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Cell surface receptors bind to hydrophilic ligands and initiate a signaling cascade that indirectly influences the making of a functional protein. Internal receptors bind to ligands that are hydrophilic and ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Cell-surface receptors bind to hydrophobic ligands and initiate a signaling cascade that indirectly influences the making of a functional protein. Internal receptors bind to ligands that are hydrophobic and initiate the signaling cascade that indirectly influences the making of a functional protein. Cell-surface receptors bind to hydrophilic ligands and a ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Internal receptors are integral membrane proteins that bind to hydrophobic ligands, initiating a signaling cascade, which indirectly influences the making of a functional protein. Cell-surface receptors bind to hydrophilic ligands and the ligand-receptor complex directly enters the nucleus, initiating transcription and translation. 22. Cells grown in the laboratory are mixed with a dye molecule that is unable |
to pass through the plasma membrane. If a ligand is added to the cells, the dye is observed entering the cells. What type of receptor did the ligand bind to on the cell surface? 398 Chapter 9 | Cell Communication a. G-protein-linked R receptor b. ligand-gated ion channel c. voltage-gated ion channel d. receptor tyrosine kinase 23. The same second messengers are used in many different cells, but the response to second messengers is different in each cell. How is this possible? a. Different cells produce the same receptor, which bind to the same ligands, but have a different response in each cell type. b. Cells produce variants of a particular receptor for a particular ligand through alternative splicing, resulting in different response in each cell c. Cells contain different genes, which produce different receptors that bind to same ligand, activating different responses in each cell. d. Cells produce different receptors that bind to the same ligand or the same receptor that binds to the same ligand with different signaling components, activating different responses in each cell. 24. What would happen if the intracellular domain of a cell-surface receptor was switched with the domain from another receptor? a. b. It would activate the pathway normally triggered by the receptor that contributed the intracellular domain. It would activate the same pathway even after the intracellular domain is changed with the domain from another receptor. c. The receptor will be mutated and become non- functional, not activating any pathway. d. The receptor will become mutated and lead to continuous cell signaling, even in the absence of a ligand. 25. Explain how a chemical that blocks the binding of EGF to the EGFR would interfere with the replication of cancerous cells that overexpress EGFR. a. b. c. d. It will activate the EGFR pathway. It will block the EGFR pathway. It will have no effect and the EGFR pathway will continue normally It will lead to overexpression of the EGFR pathway 26. How does the extracellular matrix control the growth of cells? a. Contact of receptors with the extracellular matrix maintains equilibrium of the cell and provides optimal pH for the growth of the cells. b. Contact of the receptor with the extracellular matrix helps maintain concentration gradients across membrane, resulting in the flow of ions. c. The extracellular matrix provides nutrients for the cell. d. The extracellular matrix connects the |
cell to the external environment and ensures correct positioning of the cell to prevent metastasis. 27. Give an example for each one of the following effects of a cell signal: on protein expression, cellular metabolism, and cell division. a. protein expression: binding of epinephrine (adrenaline) to a G-protein-linked receptor; cellular metabolism: the MAP-kinase cascade; cell division: promoted by the binding of the EGF to its receptor tyrosine kinase b. protein expression: the MAP-kinase cascade; cellular metabolism- binding of epinephrine (adrenaline) to a G-protein-linked receptor; cell division promoted by the binding of the EGF to its receptor tyrosine kinase c. protein expression: binding of the EGF to its receptor tyrosine kinase; cellular metabolism: the MAP-kinase cascade; cell division: FASRAS signaling. d. protein expression: RAS signaling; cellular metabolism: binding of the EGF to its receptor tyrosine kinase promotes an increase; cell division: binding of epinephrine (adrenaline) to a G-protein-linked receptor. 28. The mitogen-activated protein (MAP) kinase cascade triggered by RTKs results in cell division. Create a few possible scenarios of abnormalities in the MAPK pathway leading to uncontrolled cell proliferation. a. gain of function mutation in RAS protein, mutation in I κ -B, loss of function mutation in genes for MAPK kinase pathway, regulated phosphorylation cascade b. loss of function mutation in RAS protein and gain of function mutation in RAF protein, I κ -B permanently bound to NF- κ B, regulated phosphorylation cascade c. RAS protein unable to hydrolyze its bound GTP, loss of function mutation in I κ -B, gain of function mutation in genes for MAPK kinase pathway, unregulated phosphorylation cascade d. unregulated phosphorylation cascade, loss of function mutation in RAS and RAF protein, mutation in genes for MAPK kinase pathway, regulated phosphorylation cascade This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 9 | Cell Communication 399 29. What characteristics make yeast a good model for learning about signaling in humans? a. Yeasts are prokaryotes. They have a short life cycle, easy to grow, and share similarities with humans in certain |
regulating mechanisms. b. Yeasts are eukaryotes. They have a short life cycle, easy to grow, and share similarities with humans in certain regulating mechanisms. c. Yeasts are multicellular organisms. They have a short life cycle, easy to grow, and share similarities with humans in certain regulating mechanisms. d. Yeasts are single celled organisms. They have a complex life cycle like that of humans and share similarities in regulating mechanisms. 30. Why is signaling in multicellular organisms more complicated than signaling in single-celled organisms? a. Multicellular organisms coordinate between distantly located cells; single-celled organisms communicate only with nearby cells. b. Multicellular organisms involve receptors for signaling; single-celled organisms communicate by fusion of plasma membrane with the nearby cells. c. Multicellular organisms require more time for signal transduction than single-celled organisms, as they show compartmentalization. d. Multicellular organisms require more time for signal transduction than single-celled organisms, as they lack compartmentalization. TEST PREP FOR AP® COURSES 33. Upon ingestion of bacteria, white blood cells release a chemical messenger into the blood stream that causes the synthesis of inflammation response proteins by liver cells. What is this is an example of? a. autocrine signaling b. endocrine signaling c. paracrine signaling d. synaptic signaling 34. Molecules do not flow between the endothelial cells in the brain capillaries. The membranes of the cells must be joined by what? a. gap junctions b. c. d. ligand-gated channels synapses tight junctions 35. Analyze the possible benefits of having autocrine signaling. 31. Biofilms are a prominent danger in infectious disease treatment today because it is difficult to find drugs that can penetrate the biofilm. What characteristics would a drug have if it aimed to prevent bacteria from forming biofilms in the first place? Explain your answer. 32. Support the hypothesis that signaling pathways appeared early in evolution and are well-conserved using the yeast mating factor as an example. a. Signaling in yeast uses the RTK pathway and is evolutionarily conserved, like insulin signaling in humans. b. Signaling in yeast uses G-protein coupled receptors for signaling and is evolutionarily conserved, like insulin signaling in humans. c. Signaling in yeast uses an endocrine pathway and is evolutionarily conserved, like insulin signaling in humans. d. |
Mating factor in yeast uses an autocrine signaling pathway and is evolutionarily conserved. a. Autocrine signaling helps to communicate with distantly located cells. b. Autocrine signaling connects nearby located cells. c. Autocrine signaling helps to amplify the signal by inducing more signaling production from the cell itself. d. Autocrine signaling is specific only for the cell that produced it. 36. If a chemical is an inhibitor of the enzyme adenylyl cyclase, which of the following steps in the G-protein signaling pathway would be blocked? a. activation of gene transcription b. exchange of GTP for GDP c. d. ligand bound receptor activation of G-protein synthesis of cAMP 37. Thyroid hormone is a lipid-soluble signal molecule that crosses the membrane of all cells. Why would a cell 400 Chapter 9 | Cell Communication fail to respond to the thyroid hormone? a. The MAPK cascade leading to cell activation is defective in the target cells. b. The DNA sequence it binds to underwent a mutation. c. There is no intracellular receptor for thyroid hormone in the cell. d. The second messenger does not recognize the signal from the receptor. 38. The poison form the krait snake’s bungarotoxin binds irreversibly to acetylcholine receptors interfering with acetylcholine binding at the synapse. What is the effect of bungarotoxin binding on the post synaptic cell? a. cAMP production is inhibited. b. Bungarotoxin G-proteins are not activated. c. Ion movement in the cell is inhibited. d. Phosphorylation cascade is inhibited. 39. In autoimmune lymphoproliferative syndrome (ALPS), lymphocytes which multiplied during an infection persist in the body and damage tissue. The syndrome is caused by a mutation in the FAS gene which encodes a cell surface receptor. Which signaling pathway does the receptor initiate? a. activated metabolism b. apoptosis c. cell division d. cell differentiation 40. Place the following events in their sequential order: 1. protein kinase A is activated 2. glycogen breakdown 3. epinephrine binds to G-protein-linked receptor 4. G-protein activates adenylyl cyclase 5. GTP is exchanged for GDP on the G-protein 6. ATP is converted to cAMP a. 1, 3, 5, 4, 6, 2 b. 3, 5, 4, 1, 6, 2 c. 3, 4 |
, 5, 1, 6, 2 d. 3, 5, 4, 6, 1, 2 41. The RAS protein is a G-protein connected with the response to RTKs that initiates the MAPK kinase cascade when GDP is released and GTP uploaded. Mutations in the RAS protein which interfere with its GTPase activity are common in cancer. Evaluate the connection between the inability of RAS to hydrolyze GTP and uncontrolled cell proliferation. a. RAS, when bound to GTP, becomes permanently inactive even in the presence of the ligand, and no longer regulates cell division. b. RAS, when bound to GTP, becomes permanently active even in the absence of the ligand, and no longer regulates cell division. c. RAS, when bound to GTP, forms a dimer after binding to the ligand, and causes uncontrolled division, but it remains inactive when the ligand is absent. d. RAS, when bound to GTP, does not form a dimer after binding to the ligand but stimulates downstream signaling to occur and causes uncontrolled cell division. 42. Common medications called β-blockers bind to Gprotein-linked receptors in heart muscles, blocking adrenaline. They are prescribed to patients with high blood pressure. Can you formulate a hypothesis on their mechanism of action? a. Adrenaline has a stimulatory effect on heart rate and blood pressure. β-blockers are antagonistic to adrenaline and produces inhibitory effect. b. Adrenaline has both a stimulatory and an inhibitory effect on heart rate and blood pressure. β-blockers bind to G-protein and stimulate the inhibitory effect of adrenaline. c. Adrenaline has an inhibitory effect on heart rate and blood pressure. β-blockers have a synergistic effect along with adrenaline producing an inhibitory effect. d. Adrenaline has both a stimulatory and an inhibitory effect on heart rate and blood pressure. β-blockers bind to G-protein and intervene with the inhibitory effect of adrenaline. SCIENCE PRACTICE CHALLENGE QUESTIONS 43. The figure below shows a series of states for typical G protein signal transduction. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 9 | Cell Communication 401 binding is positively cooperative. The affinity of oxygen for heme increases as the number of bound oxygen molecules increases. A. Describe |
the features in the graph above for hemoglobin that demonstrate positive cooperativity. B. The insulin receptor (IR) is a tyrosine kinase receptor that has two sites to which insulin can attach. IR is negatively cooperative. In the diagram above, the dependence of the bound fraction on available insulin is similar to the curve for k = 1 with negative cooperativity. Describe the features of this curve in the graph above that demonstrate negative cooperativity. C. When viewed from above the cell-surface, the representation shows receptors with one and two bound insulin molecules. Explain the negative cooperation for this receptor based on the free energy of conformational changes in the receptor-peptide chains. D. E. Explain the advantages in terms of selection of two-site binding with negative cooperation relative to one-site binding. F. Three binding curves with negative cooperativity and different values of k are shown on the graph. Describe conditions in which there is an advantage in having a low value of k with negative cooperativity. 45. Organisms, including plants, have evolved chemical signaling pathways to direct physiological responses to environmental changes. Stomata are pores, typically on the underside of leaves that regulate CO2, O2, and H2O exchange between plants and the external environment. This interaction controls photosynthetic rate and transpiration rate. The opening and closing of stomata are controlled by specialized guard cells that surround the stomatal pore. The osmotic state within the guard cells determines their turgor; when the guard cells are flaccid, stomata close. Turgor in the guard cells is regulated by the active transport of several ions, including K+ and H+, across the plasma membrane. Several environmental factors can cause stomatal closing: water deficit, darkness, microbes, ozone, and sulfur dioxide and other pollutants. Intracellular carbon dioxide concentration and light can trigger stomata to open. The system is regulated by a phytohormone (plant hormone) called abscisic acid (ABA) and the amino acid precursor of the synthesis of a second phytohormone Figure 9.19 Use this representation to describe the following stages in this signaling process: A. between A and B B. between B and C C. between C and D D. between D and E E. between E and A 44. Tyrosine kinase receptors are pairs of proteins that span the plasma membrane. On the extracellular side of the membrane, one or |
more sites are present that bind to signaling ligands such as insulin or growth factors. On the intracellular side, the ends of peptide chains on each protein phosphorylate the other member of the pair, providing active docking sites that initiate cellular responses. The signal is switched off by dissociation of the ligand. For each ligand-receptor system, the equilibrium constant, k, controls the distribution of receptor-bound and unbound ligands. In systems with large values of k, a site is likely to be occupied, even at low concentrations of ligand. When k is small, the likelihood of binding is low, even when the concentration of ligand is high. To initiate a new stimulus response cycle for the receptor, the ligand must dissociate. Larger values of k mean that the receptor is more likely to be occupied and thus unavailable to bind another ligand. Some ligand-binding systems have multiple binding sites. For example, hemoglobin binds four oxygen molecules, whereas myoglobin has only a single binding site. When multiple binding sites are present, the presence of an already-bound ligand can cooperatively affect the binding of other ligands on the same protein. For hemoglobin, the 402 Chapter 9 | Cell Communication called ethylene (ACC). The second messengers NO and Ca2+ in the signal response to changes in the concentrations of these hormones activate transcription factors that affect ion transport across guard cell membranes. High CO2 levels and light also alter phytohormone concentrations. are used in many different signaling pathways. Construct an explanation by analogy to other phenomena in which combining a small set of events (for example, 0 and 1 in a computer, the musical scale, or the R, G, and B components of a color) can lead to a vast assortment of outcomes. A. Explain why plants must regulate the opening and closing of stomata. Explain how this response relates to the capture of free energy for cellular processes. 46. Construct a graphical representation of information as a function of time during the transduction of a signal along a signaling pathway. A. Use your graph to describe trends in the amount of information rather than the actual magnitude. In sketching your graph, consider how the shape of the curve would change during these events: i. extracellular first messenger ii. receptor binding and conformational changes iii. release of second messengers iv. cellular responses v. halt signal and degrade intermediates B. Annotate your representation for a specific signaling system, such |
as the effect of epinephrine on the free energy released from glucose. 47. Bacteria and fungi produce several extracellular chemicals, including antibiotics that affect other organisms in the environment. Antibiotics also are produced industrially in large bacteria-containing fermentation tanks. However, antibiotics that have been used by humans to control microbes are now found at subinhibitory concentrations in the environment. Low levels of antibiotics in the environment are mutagenic for bacteria and promote the development of antibiotic resistance. Bacteria produce chemical signals that detect population density and regulate gene expression, a phenomenon called quorum sensing. Density is signaled by the extracellular concentration of small amino acid derivatives. To combat antibiotic resistance, an emerging strategy for the control of bacterial disease is quorum quenching. A. Describe the advantage of antibiotics to the organisms that produce them. B. Based on the name of the emerging strategy for controlling bacterial infections, describe a possible mechanism by which bacteria determine their population density. Justify the claim that quorum quenching may provide a more sustainable approach to disease control than the use of antibiotics. B. Construct an explanation in terms of the water potential, Y, for the efflux (outward flow) of H+ during water stress (drought). C. Consider a scenario involving environmental factors, such as water stress and daylight, which have opposing effects on the opening and closing of stomata; stomata would be signaled to close under drought conditions and to open during photosynthesis. Pose two scientific questions regarding the response of the system, one involving the phytohormones ABA and ACC, and the second involving the concentration of second messengers. D. The data shown in the table below were obtained by treating rockcress (Arabidopsis) with doses of ABA, ACC, and ABA plus ACC. Using the terms and and or, describe the expected and unexpected responses of the system just after 10 minutes and around 45 minutes, as displayed by these data. Figure 9.20 E. Researchers are investigating the interactions among multiple signaling pathways, a phenomenon referred to as “crosstalk.” The same second messengers, NO and Ca2+, This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 403 10 | CELL REPRODUCTION Figure 10.1 A sea urchin begins life as a single cell that (a) divides |
to form two cells, visible by scanning electron microscopy. After four rounds of cell division, (b) there are 16 cells, as seen in this SEM image. After many rounds of cell division, the individual develops into a complex, multicellular organism, as seen in this (c) mature sea urchin. (credit a: modification of work by Evelyn Spiegel, Louisa Howard; credit b: modification of work by Evelyn Spiegel, Louisa Howard; credit c: modification of work by Marco Busdraghi; scale-bar data from Matt Russell) Chapter Outline 10.1: Cell Division 10.2: The Cell Cycle 10.3: Control of the Cell Cycle 10.4: Cancer and the Cell Cycle 10.5: Prokaryotic Cell Division Introduction A human, as well as every sexually reproducing organism, begins life as a fertilized egg (embryo) or zygote. Trillions of cell divisions subsequently occur in a controlled manner to produce a complex, multicellular human. In other words, that original single cell is the ancestor of every other cell in the body. Once a being is fully grown, cell reproduction is still necessary to repair or regenerate tissues. For example, new blood and skin cells are constantly being produced. All multicellular organisms use cell division for growth, maintenance, and repair of tissues. Cell division is tightly regulated, and the occasional failure of regulation can have life-threatening consequences. Single-celled organisms use cell division as their method of reproduction. Not all cells in the body reproduce to repair tissues. Most nerve tissues, for example, are not capable of regeneration. This means people who have damaged their nerves or nervous system are often left paralyzed. However, this may change in the future; scientists have discovered a new drug called intracellular signal peptide (ISP), which helps nerve cells regenerate in rats. It works by blocking an enzyme that causes scar tissue in damaged nerve cells allowing the nervous system a chance to repair itself. The full research study is located here (http://openstaxcollege.org/l/ 32scar). 10.1 | Cell Division In this section, you will explore the following question: • What is the relationship between chromosomes, genes, and traits in prokaryotes and eukaryotes? 404 Chapter 10 | Cell Reproduction Connection for AP® Courses All organisms, from bacteria to complex animals, must be able to store, retrieve, and transmit genetic information to |
continue life. In later chapters, we will explore how a cell’s genetic information encoded in DNA, its genome, is replicated and passed to the next generation to direct the production of proteins, determining an organism’s traits. Prokaryotes have single circular chromosome of DNA, whereas eukaryotes have multiple, linear chromosomes composed of chromatin (DNA wrapped around a histone protein) surrounded by a nuclear membrane. Cell division involves both mitosis, the division of the chromosomes, and cytokinesis, the division of the cytoplasm. Human somatic cells consist of 46 chromosomes—22 pairs of autosomal chromosomes and a pair of sex chromosomes. Prior to mitosis, each chromosome is duplicated to ensure that daughter cells receive the full amount of hereditary material contributed by both parents. The total number of autosomal chromosomes is referred to as the diploid (2n) number. (In the next chapter, we will study meiosis, the second type of cell division in sexually reproducing organisms.) Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework, as shown in the table. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA, and in some cases, RNA are the primary sources of heritable information. The continuity of life from one cell to another has its foundation in the reproduction of cells by way of the cell cycle. The cell cycle is an orderly sequence of events that describes the stages of a cell’s life from the division of a single parent cell to the production of two new daughter cells. The mechanisms involved in the cell cycle are highly regulated. Genomic DNA Before discussing the steps a |
cell must undertake to replicate, a deeper understanding of the structure and function of a cell’s genetic information is necessary. A cell’s DNA, packaged as a double-stranded DNA molecule, is called its genome. In prokaryotes, the genome is composed of a single, double-stranded DNA molecule in the form of a loop or circle (Figure 10.2). The region in the cell containing this genetic material is called a nucleoid. Some prokaryotes also have smaller loops of DNA called plasmids that are not essential for normal growth. Bacteria can exchange these plasmids with other bacteria, sometimes receiving beneficial new genes that the recipient can add to their chromosomal DNA. Antibiotic resistance is one trait that often spreads through a bacterial colony through plasmid exchange. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 405 Figure 10.2 Prokaryotes, including bacteria and archaea, have a single, circular chromosome located in a central region called the nucleoid. In eukaryotes, the genome consists of several double-stranded linear DNA molecules (Figure 10.3). Each species of eukaryotes has a characteristic number of chromosomes in the nuclei of its cells. Human body cells have 46 chromosomes, while human gametes (sperm or eggs) have 23 chromosomes each. A typical body cell, or somatic cell, contains two matched sets of chromosomes, a configuration known as diploid. The letter n is used to represent a single set of chromosomes; therefore, a diploid organism is designated 2n. Human cells that contain one set of chromosomes are called gametes, or sex cells; these are eggs and sperm, and are designated 1n, or haploid. Figure 10.3 There are 23 pairs of homologous chromosomes in a female human somatic cell. The condensed chromosomes are viewed within the nucleus (top), removed from a cell in mitosis and spread out on a slide (right), and artificially arranged according to length (left); an arrangement like this is called a karyotype. In this image, the chromosomes were exposed to fluorescent stains for differentiation of the different chromosomes. A method of staining called “chromosome painting” employs fluorescent dyes that highlight chromosomes in different colors. (credit: National Human Genome Project/NIH) Matched pairs of |
chromosomes in a diploid organism are called homologous (“same knowledge”) chromosomes. Homologous chromosomes are the same length and have specific nucleotide segments called genes in exactly the same location, or locus. Genes, the functional units of chromosomes, determine specific characteristics by coding for specific proteins. Traits are the variations of those characteristics. For example, hair color is a characteristic with traits that are blonde, brown, or black. Each copy of a homologous pair of chromosomes originates from a different parent; therefore, the genes themselves are not identical. The variation of individuals within a species is due to the specific combination of the genes inherited from both parents. Even a slightly altered sequence of nucleotides within a gene can result in an alternative trait. For example, there are three possible gene sequences on the human chromosome that code for blood type: sequence A, sequence B, and sequence O. Because all diploid human cells have two copies of the chromosome that determines blood type, the blood type 406 Chapter 10 | Cell Reproduction (the trait) is determined by which two versions of the marker gene are inherited. It is possible to have two copies of the same gene sequence on both homologous chromosomes, with one on each (for example, AA, BB, or OO), or two different sequences, such as AB. Minor variations of traits, such as blood type, eye color, and handedness, contribute to the natural variation found within a species. However, if the entire DNA sequence from any pair of human homologous chromosomes is compared, the difference is less than one percent. The sex chromosomes, X and Y, are the single exception to the rule of homologous chromosome uniformity: Other than a small amount of homology that is necessary to accurately produce gametes, the genes found on the X and Y chromosomes are different. Eukaryotic Chromosomal Structure and Compaction If the DNA from all 46 chromosomes in a human cell nucleus was laid out end to end, it would measure approximately two meters; however, its diameter would be only 2 nm. Considering that the size of a typical human cell is about 10 µm (100,000 cells lined up to equal one meter), DNA must be tightly packaged to fit in the cell’s nucleus. At the same time, it must also be readily accessible for the genes to be expressed. During some stages of the cell cycle, the long strands of DNA are condensed into compact chromosomes. There |
are a number of ways that chromosomes are compacted. In the first level of compaction, short stretches of the DNA double helix wrap around a core of eight histone proteins at regular intervals along the entire length of the chromosome (Figure 10.4). The DNA-histone complex is part of the chromatin. Each beadlike histone-DNA complex is called a nucleosome, and DNA connecting the nucleosomes is called linker DNA. A DNA molecule in this form is about seven times shorter than the double helix without the histones, and the beads are about 10 nm in diameter, in contrast with the 2-nm diameter of a DNA double helix. The next level of compaction occurs as the nucleosomes and the linker DNA between them are coiled into a 30-nm chromatin fiber. This coiling further shortens the chromosome so that it is now about 50 times shorter than the extended form. In the third level of packing, a variety of fibrous proteins is used to pack the chromatin. These fibrous proteins also ensure that each chromosome in a non-dividing cell occupies a particular area of the nucleus that does not overlap with that of any other chromosome (see the top image in Figure 10.3). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 407 Figure 10.4 Double-stranded DNA wraps around histone proteins to form nucleosomes that have the appearance of “beads on a string.” The nucleosomes are coiled into a 30-nm chromatin fiber. When a cell undergoes mitosis, the chromosomes condense even further. DNA replicates in the S phase of interphase. After replication, the chromosomes are composed of two linked sister chromatids. When fully compact, the pairs of identically packed chromosomes are bound to each other by cohesin proteins. The connection between the sister chromatids is closest in a region called the centromere. The conjoined sister chromatids, with a diameter of about 1 µm, are visible under a light microscope. The centromeric region is highly condensed and thus will appear as a constricted area. 408 Chapter 10 | Cell Reproduction This animation (http://openstaxcollege.org/l/Packaged_DNA) illustrates the different levels of chromosome packing. Why is nucleosome formation required for |
the packaging of DNA? a. Nucleosome formation results in compaction of the DNA to form chromatin. b. Nucleosome formation results in DNA synthesis. c. Nucleosome formation decreases the number of introns in DNA. d. Nucleosome formation increases the number of introns in the DNA. Think About It What is the relationship between a genome and chromosomes? 10.2 | The Cell Cycle In this section, you will explore the following questions: • What processes occur during the three stages of interphase? • How do the chromosomes behave during the mitotic phase? Connection for AP® Courses The cell cycle describes an orderly sequence of events that are highly regulated. In eukaryotes, the cell cycle consists of a long preparatory period (interphase) followed by mitosis and cytokinesis. Interphase is divided into three phases: Gap 1 (G1), DNA synthesis (S), and Gap 2 (G2). Interphase represents the portion of the cell cycle between nuclear divisions. During this phase, preparations are made for division that include growth, duplication of most cellular contents, and replication of DNA. The cell’s DNA is replicated during the S stage. (We will study the details of DNA replication in the chapter on DNA structure and function.) Following the G2 stage of interphase, the cell begins mitosis, the process of active division by which duplicated chromosomes (chromatids) attach to spindle fibers, align themselves along the equator of the cell, and then separate from each other. Following mitosis, the cell undergoes cytokinesis, the splitting of the parent cell into two daughter cells, complete with a full complement of genetic material. In animal cells, daughter cells are separated by an actin ring, whereas plant cells are separated by the cell plate, which will grow into a new cell wall. Sometimes cells enter a Gap zero (G0) phase, during which they do not actively prepare to divide; the G0 phase can be temporary until triggered by an external signal to enter G1, or permanent, such as mature cardiac muscle cells and nerve cells. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework, as shown in the tables. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory This OpenStax book is available for free at |
http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 409 experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 3.7 The student can make predictions about natural phenomena occurring during the cell cycle. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. 3.8 The student can describe the events that occur in the cell cycle. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 5.3 The student can evaluate the evidence provided by data sets in relation to a particular scientific question. 3.11 The student is able to evaluate evidence provided by data sets to support the claim that heritable information is passed from one generation to another generation through mitosis. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 2.35][APLO 2.15][APLO 2.19][APLO 3.11][APLO 2.33][APLO 2.36][APLO 2.37][APLO 2.31] The cell cycle is an ordered series of events involving cell growth and cell division that produces two new daughter cells. Cells on the path to cell division proceed through a series of precisely timed and carefully regulated stages of growth, DNA replication, and division that produces two identical (clone) cells. The cell cycle has two major phases: interphase and the mitotic phase (Figure 10.5). During interphase, the cell grows and DNA |
is replicated. During the mitotic phase, the replicated DNA and cytoplasmic contents are separated, and the cell divides. 410 Chapter 10 | Cell Reproduction Figure 10.5 The cell cycle consists of interphase and the mitotic phase. During interphase, the cell grows and the nuclear DNA is duplicated. Interphase is followed by the mitotic phase. During the mitotic phase, the duplicated chromosomes are segregated and distributed into daughter nuclei. The cytoplasm is usually divided as well, resulting in two daughter cells. Interphase During interphase, the cell undergoes normal growth processes while also preparing for cell division. In order for a cell to move from interphase into the mitotic phase, many internal and external conditions must be met. The three stages of interphase are called G1, S, and G2. G1 Phase (First Gap) The first stage of interphase is called the G1 phase (first gap) because, from a microscopic aspect, little change is visible. However, during the G1 stage, the cell is quite active at the biochemical level. The cell is accumulating the building blocks of chromosomal DNA and the associated proteins as well as accumulating sufficient energy reserves to complete the task of replicating each chromosome in the nucleus. S Phase (Synthesis of DNA) Throughout interphase, nuclear DNA remains in a semi-condensed chromatin configuration. In the S phase, DNA replication can proceed through the mechanisms that result in the formation of identical pairs of DNA molecules—sister chromatids—that are firmly attached to the centromeric region. The centrosome is duplicated during the S phase. The two centrosomes will give rise to the mitotic spindle, the apparatus that orchestrates the movement of chromosomes during mitosis. At the center of each animal cell, the centrosomes of animal cells are associated with a pair of rod-like objects, the centrioles, which are at right angles to each other. Centrioles help organize cell division. Centrioles are not present in the centrosomes of other eukaryotic species, such as plants and most fungi. G2 Phase (Second Gap) In the G2 phase, the cell replenishes its energy stores and synthesizes proteins necessary for chromosome manipulation. Some cell organelles are duplicated, and the cytoskeleton is dismantled to provide resources for the mitotic phase. There may be additional cell growth during G2. The final preparations for |
the mitotic phase must be completed before the cell is able to enter the first stage of mitosis. The Mitotic Phase The mitotic phase is a multistep process during which the duplicated chromosomes are aligned, separated, and move into two new, identical daughter cells. The first portion of the mitotic phase is called karyokinesis, or nuclear division. The This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 411 second portion of the mitotic phase, called cytokinesis, is the physical separation of the cytoplasmic components into the two daughter cells. Revisit the stages of mitosis at this site (http://openstaxcollege.org/l/Cell_cycle_mito). Gout is a form of arthritis that causes a painful inflammation of joints. One treatment for gout is colchicine, a medication that inhibits mitosis. Explain why this medication is beneficial for people with gout and why it can cause undesirable side effects, such as low white blood cell counts. a. Colchicine increases inflammation by inhibiting mitosis. Inhibition of mitosis results in decreased white blood count. b. Colchicine decreases inflammation by inhibiting mitosis. Inhibition of mitosis results in decreased white blood count. c. Colchicine increases inflammation by inhibiting mitosis. Inhibition of mitosis results in increased white blood count. d. Colchicine decreases inflammation by inhibiting mitosis. Inhibition of mitosis results in increased white blood count. Karyokinesis (Mitosis) Karyokinesis, also known as mitosis, is divided into a series of phases—prophase, prometaphase, metaphase, anaphase, and telophase—that result in the division of the cell nucleus (Figure 10.7). 412 Chapter 10 | Cell Reproduction These budding plants demonstrate asexual reproduction, one of the main purposes of mitosis. The other two purposes are growth and repair. Figure 10.6 This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 413 Which of the following statements best describes the relationship between mitosis and asexual reproduction? a. Mitosis is a process that can result in asexual reproduction. b. Mitosis is a process |
that always results in asexual reproduction. c. Asexual reproduction is a process that always results in mitosis. d. Asexual reproduction is a process that can result in mitosis. Figure 10.7 Karyokinesis (or mitosis) is divided into five stages—prophase, prometaphase, metaphase, anaphase, and telophase. The pictures at the bottom were taken by fluorescence microscopy (hence, the black background) of cells artificially stained by fluorescent dyes: blue fluorescence indicates DNA (chromosomes) and green fluorescence indicates microtubules (spindle apparatus). (credit “mitosis drawings”: modification of work by Mariana Ruiz Villareal; credit “micrographs”: modification of work by Roy van Heesbeen; credit “cytokinesis micrograph”: Wadsworth Center/New York State Department of Health; scale-bar data from Matt Russell) Which of the following is the correct order of events in mitosis? a. Sister chromatids line up at the metaphase plate. The kinetochore becomes attached to the mitotic spindle. The nucleus reforms and the cell divide. Cohesin proteins break down and the sister chromatids separate. b. The kinetochore becomes attached to the mitotic spindle. Cohesin proteins break down and the sister chromatids separate. Sister chromatids line up at the metaphase plate. The nucleus reforms and the cell divides. c. The kinetochore becomes attached to the cohesin proteins. Sister chromatids line up at the metaphase plate. The kinetochore breaks down and the sister chromatids separate. The nucleus reforms and the cell divides. d. The kinetochore becomes attached to the mitotic spindle. Sister chromatids line up at the metaphase plate. Cohesin proteins break down and the sister chromatids separate. The nucleus reforms and the cell divide. 414 Chapter 10 | Cell Reproduction During prophase, the “first phase,” the nuclear envelope starts to dissociate into small vesicles, and the membranous organelles (such as the Golgi complex or Golgi apparatus, and endoplasmic reticulum), fragment and disperse toward the periphery of the cell. The nucleolus disappears (disperses). The centrosomes begin to move to opposite poles of |
the cell. Microtubules that will form the mitotic spindle extend between the centrosomes, pushing them farther apart as the microtubule fibers lengthen. The sister chromatids begin to coil more tightly with the aid of condensin proteins and become visible under a light microscope. During prometaphase, the “first change phase,” many processes that were begun in prophase continue to advance. The remnants of the nuclear envelope fragment. The mitotic spindle continues to develop as more microtubules assemble and stretch across the length of the former nuclear area. Chromosomes become more condensed and discrete. Each sister chromatid develops a protein structure called a kinetochore in the centromeric region (Figure 10.8). The proteins of the kinetochore attract and bind mitotic spindle microtubules. As the spindle microtubules extend from the centrosomes, some of these microtubules come into contact with and firmly bind to the kinetochores. Once a mitotic fiber attaches to a chromosome, the chromosome will be oriented until the kinetochores of sister chromatids face the opposite poles. Eventually, all the sister chromatids will be attached via their kinetochores to microtubules from opposing poles. Spindle microtubules that do not engage the chromosomes are called polar microtubules. These microtubules overlap each other midway between the two poles and contribute to cell elongation. Astral microtubules are located near the poles, aid in spindle orientation, and are required for the regulation of mitosis. Figure 10.8 During prometaphase, mitotic spindle microtubules from opposite poles attach to each sister chromatid at the kinetochore. In anaphase, the connection between the sister chromatids breaks down, and the microtubules pull the chromosomes toward opposite poles. During metaphase, the “change phase,” all the chromosomes are aligned in a plane called the metaphase plate, or the equatorial plane, midway between the two poles of the cell. The sister chromatids are still tightly attached to each other by cohesin proteins. At this time, the chromosomes are maximally condensed. During anaphase, the “upward phase,” the cohesin proteins degrade, and the sister chromatids separate at the centromere. Each chromatid, now called a chromosome, is pulled rapidly toward the centros |
ome to which its microtubule is attached. The cell becomes visibly elongated (oval shaped) as the polar microtubules slide against each other at the metaphase plate where they overlap. During telophase, the “distance phase,” the chromosomes reach the opposite poles and begin to decondense (unravel), relaxing into a chromatin configuration. The mitotic spindles are depolymerized into tubulin monomers that will be used to assemble cytoskeletal components for each daughter cell. Nuclear envelopes form around the chromosomes, and nucleosomes appear within the nuclear area. Cytokinesis Cytokinesis, or “cell motion,” is the second main stage of the mitotic phase, during which cell division is completed via the physical separation of the cytoplasmic components into two daughter cells. Division is not complete until the cell components have been apportioned and completely separated into the two daughter cells. Although the stages of mitosis are similar for most eukaryotes, the process of cytokinesis is quite different for eukaryotes that have cell walls, such as plant cells. In cells such as animal cells that lack cell walls, cytokinesis starts during late anaphase. A contractile ring composed of actin filaments forms just inside the plasma membrane at the former metaphase plate. The actin filaments pull the equator of the This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 415 cell inward, forming a fissure. This fissure, or “crack,” is called the cleavage furrow. The furrow deepens as the actin ring contracts, and eventually the membrane is cleaved in two (Figure 10.9). In plant cells, a new cell wall must form between the daughter cells. During interphase, the Golgi apparatus accumulates enzymes, structural proteins, and glucose molecules prior to breaking into vesicles and dispersing throughout the dividing cell. During telophase, these Golgi vesicles are transported on microtubules to form a phragmoplast (a vesicular structure) at the metaphase plate. There, the vesicles fuse and coalesce from the center toward the cell walls; this structure is called a cell plate. As more vesicles fuse, the cell plate enlarges until it mer |
ges with the cell walls at the periphery of the cell. Enzymes use the glucose that has accumulated between the membrane layers to build a new cell wall. The Golgi membranes become parts of the plasma membrane on either side of the new cell wall (Figure 10.9). Figure 10.9 During cytokinesis in animal cells, a ring of actin filaments forms at the metaphase plate. The ring contracts, forming a cleavage furrow, which divides the cell in two. In plant cells, Golgi vesicles coalesce at the former metaphase plate, forming a phragmoplast. A cell plate formed by the fusion of the vesicles of the phragmoplast grows from the center toward the cell walls, and the membranes of the vesicles fuse to form a plasma membrane that divides the cell in two. 416 Chapter 10 | Cell Reproduction Activity • Use a set of pipe cleaners (or other materials as directed by your teacher) that you can use to model chromosomes during mitosis and meiosis: 1. Each of the pipe cleaners represents a single, unreplicated chromosome. Each chromosome should differ in size, as they do in most organisms. Assume that your dividing cell contains 3 chromosomes: numbered chromosome 1, 2, and 3. 2. Using both members of each homologous pair for chromosomes 1–3, model how the chromosomes would appear in a cell that had just finished the S phase of the cell cycle. Once your teacher has approved your model, have one member of your group document the model by photographing or drawing it. 3. Now, repeat step 2 but show the cell at metaphase during mitosis. 4. Finally, model the two daughter cells that will result from mitosis. Again, have one member of your group document the model. 5. Repeat steps 2–5 for both meiosis I and meiosis II. Remember that you should have four daughter cells at the end of meiosis II. Also remember to ask your teacher for approval and document your model before moving on to the next phase of meiosis. 6. Exchange/ copy all of the drawings or photographs that your group took of your models. As a group or individually (as directed by your teacher) create a report to turn in that labels and explain each picture of your model. • An organism’s ploidy count is the total number of chromosome sets contained in each body cell. Most organisms have a ploidy level of 2, meaning that they have two |
sets of chromosomes due to presence of homologous pairs. However, some plants are multiploid, meaning they can have ploidy levels greater than 2. The table shows possible multiploid levels of some common crop plants. Common name Multiploid chromosome count Normal chromosome count Bananas Potatoes Wheat Sugar cane 33 48 42 80 11 12 7 10 Analyze the data with a partner of in a group as directed by your teacher. On a separate sheet of paper, answer the following questions. a. How does the multiploid count of the crop plants relate to their normal chromosome count? b. Explain the basis for the relationship you described in part a, in terms of what occurs to chromosomes during replication and meiosis. c. Give one additional example of a possible multiploid chromosome count for each species in the table above. Exercise 10.1 A. A comparison of the relative time intervals of mitotic stages can be made by completing the task described. In evaluating each time interval, the problem suggests that you assume that the length of time to complete one cell cycle is 24 hours. How can that assumption be tested? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 417 Suppose that you have a growth chamber in which roots of a newly germinated plant can be examined visually with a lens that provides a magnification from which lengths can be determined with a precision of ± 0.05 mm. The field of view can be rotated so that measurements can be made of both the length and diameter of the growing tip. A large number of growing roots can be studied. Tips can be sampled, sectioned, and examined microscopically with a 25× magnification so that estimates of the diameter and length of cells can be made. Cells in the growing tip of the root rapidly undergo mitosis, just as the whitefish blastula described in Figure 10.10. With increasing distance from the growing tip, the rate at which mitosis occurs slows until tissue is reached in which the initiation of the cell cycle is delayed. A. Describe a sequence of measurements that could be used to test the assumption that the cell cycle, once started, has a total time interval of 24 hours. Hint: Rather than counting cells, it might be useful to measure the length of the root tip and the average length of a cell. B. Using the data obtained from your measurements described in part A, how can the rate of cell division be |
calculated? An experiment that is perhaps similar to one you have proposed was conducted previously (Beemster and Baxter, 1998), and the results are shown in the table. Distance (mm) Per hour 0 0.1 0.2 0.3 0.4 0.5 0.035 ± 0.01 0.047 ± 0.005 0.044 ± 0.01 0.039 ± 0.01 0.042 ± 0.01 0.031 ± 0.005 Table 10.1 C. Using these data, estimate the length of time of the cell cycle, including an estimate of precision by calculating the standard deviation. Growth factors are signals that initiate cell division in eukaryotes. (The data in the table above show that cells in the plant root less than a mm from the root tip are showing a reduction of growth rate.) The interaction of two plant hormones, auxin and brassinosteroids, have been shown [Chaiwanon and Wang, Cell, 164(6), 1257, 2016] to regulate cell division in root tips. Auxin concentrations are higher near the root tip and decrease with distance from the tip. Brassinosteroids decrease in concentration near the root tip. Auxin is actively transported between cells, whereas brassinosteroids have limited transport between cells. D. Based on these data and the observed distribution of brassinosteroids and auxin in the growing root, predict a mechanism for their interaction and justify the claim that brassinosteroid synthesis is negatively regulated by auxin transported to the cell, and that auxin is positively regulated and amplified. Think About It Chemotherapy drugs such as vincristine and colchicines disrupt mitosis by binding to tubulin (the subunit of microtubules) and interfering with microtubule assembly and disassembly. What mitotic structure is targeted by these drugs, and what effect would this have on cell division? G0 Phase Not all cells adhere to the classic cell cycle pattern in which a newly formed daughter cell immediately enters the preparatory phases of interphase, closely followed by the mitotic phase. Cells in G0 phase are not actively preparing to divide. The cell is in a quiescent (inactive) stage that occurs when cells exit the cell cycle. Some cells enter G0 temporarily until an external signal triggers the onset of G1. Other cells that never or rarely divide, such as mature cardiac muscle and nerve cells, remain in G0 permanently. 418 Chapter 10 | Cell Reprodu |
ction Determine the Time Spent in Cell Cycle Stages Problem: How long does a cell spend in interphase compared to each stage of mitosis? Background: A prepared microscope slide of blastula cross-sections will show cells arrested in various stages of the cell cycle. It is not visually possible to separate the stages of interphase from each other, but the mitotic stages are readily identifiable. If 100 cells are examined, the number of cells in each identifiable cell cycle stage will give an estimate of the time it takes for the cell to complete that stage. Problem Statement: Given the events included in all of interphase and those that take place in each stage of mitosis, estimate the length of each stage based on a 24-hour cell cycle. Before proceeding, state your hypothesis. Test your hypothesis: Test your hypothesis by doing the following: 1. Place a fixed and stained microscope slide of whitefish blastula cross-sections under the scanning objective of a light microscope. 2. Locate and focus on one of the sections using the scanning objective of your microscope. Notice that the section is a circle composed of dozens of closely packed individual cells. 3. Switch to the low-power objective and refocus. With this objective, individual cells are visible. 4. Switch to the high-power objective and slowly move the slide left to right, and up and down to view all the cells in the section (Figure 10.10). As you scan, you will notice that most of the cells are not undergoing mitosis but are in the interphase period of the cell cycle. Figure 10.10 Slowly scan whitefish blastula cells with the high-power objective as illustrated in image (a) to identify their mitotic stage. (b) A microscopic image of the scanned cells is shown. (credit “micrograph”: modification of work by Linda Flora; scale-bar data from Matt Russell) 5. Practice identifying the various stages of the cell cycle, using the drawings of the stages as a guide (Figure 10.7). 6. Once you are confident about your identification, begin to record the stage of each cell you encounter as you scan left to right, and top to bottom across the blastula section. 7. Keep a tally of your observations and stop when you reach 100 cells identified. 8. The larger the sample size (total number of cells counted), the more accurate the results. If possible, gather and record group data prior to calculating percentages and making estimates. Record your observations: Make a |
table similar to Table 10.2 in which you record your observations. Results of Cell Stage Identification Phase or Stage Individual Totals Group Totals Percent Interphase Prophase This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 419 Results of Cell Stage Identification Phase or Stage Individual Totals Group Totals Percent Metaphase Anaphase Telophase Cytokinesis Totals Table 10.2 100 100 100 percent Analyze your data/report your results: To find the length of time whitefish blastula cells spend in each stage, multiply the percent (recorded as a decimal) by 24 hours. Make a table similar to Table 10.3 to illustrate your data. Estimate of Cell Stage Length Phase or Stage Percent (as Decimal) Time in Hours Interphase Prophase Metaphase Anaphase Telophase Cytokinesis Table 10.3 10.3 | Control of the Cell Cycle In this section, you will explore the following questions: • What are examples of internal and external mechanisms that control the cell cycle? • What molecules are involved in controlling the cell cycle through positive and negative regulation? Connection for AP® Courses Each step of the cell cycle is closely monitored by external signals and internal controls called checkpoints. There are three major checkpoints in the cell cycle: one near the end of G1, a second at the G2/M transition, and the third during metaphase. Growth factor proteins arriving at the dividing cell’s plasma membrane can trigger the cell to begin dividing. Cyclins and cyclin-dependent kinases (Cdks) are internal molecular signals that regulate cell transitions through the various checkpoints. Passage through the G1 checkpoint makes sure that the cell is ready for DNA replication in the S stage of interphase; passage through the G2 checkpoint triggers the separation of chromatids during mitosis. Positive regulator molecules like the cyclins and Cdks allow the cell cycle to advance to the next stage; negative regulator molecules, such as tumor suppressor proteins, monitor cellular conditions and can halt the cycle until specific requirements are met. Errors in the regulation of the cell cycle can cause cancer, which is characterized by uncontrolled cell division. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined 420 Chapter 10 | Cell Reproduction in Big Idea 3 of the AP® Biology Curriculum Framework, as shown in the tables. The Learning Objectives |
listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 3.7 The student can make predictions about natural phenomena occurring during the cell cycle. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. 3.8 The student can describe the events that occur in the cell cycle. The length of the cell cycle is highly variable, even within the cells of a single organism. In humans, the frequency of cell turnover ranges from a few hours in early embryonic development, to an average of two to five days for epithelial cells, and to an entire human lifetime spent in G0 by specialized cells, such as cortical neurons or cardiac muscle cells. There is also variation in the time that a cell spends in each phase of the cell cycle. When fast-dividing mammalian cells are grown in culture (outside the body under optimal growing conditions), the length of the cycle is about 24 hours. In rapidly dividing human cells with a 24-hour cell cycle, the G1 phase lasts approximately nine hours, the S phase lasts 10 hours, the G2 phase lasts about four and one-half hours, and the M phase lasts approximately one-half hour. In early embryos of fruit flies, the cell cycle is completed in about eight minutes. The timing of events in the cell cycle is controlled by mechanisms that are both internal and external to the cell. Regulation of the Cell Cycle by External Events Both the initiation and inhibition of cell division are triggered by events external to the cell when it is about to begin the replication process. An event may be as simple as the death of a nearby cell |
or as sweeping as the release of growthpromoting hormones, such as human growth hormone (HGH). A lack of HGH can inhibit cell division, resulting in dwarfism, whereas too much HGH can result in gigantism. Crowding of cells can also inhibit cell division. Another factor that can initiate cell division is the size of the cell; as a cell grows, it becomes inefficient due to its decreasing surface-tovolume ratio. The solution to this problem is to divide. Whatever the source of the message, the cell receives the signal, and a series of events within the cell allows it to proceed into interphase. Moving forward from this initiation point, every parameter required during each cell cycle phase must be met or the cycle cannot progress. Regulation at Internal Checkpoints It is essential that the daughter cells produced be exact duplicates of the parent cell. Mistakes in the duplication or distribution of the chromosomes lead to mutations that may be passed forward to every new cell produced from an abnormal cell. To prevent a compromised cell from continuing to divide, there are internal control mechanisms that operate at three main cell cycle checkpoints. A checkpoint is one of several points in the eukaryotic cell cycle at which the progression of a cell to the next stage in the cycle can be halted until conditions are favorable. These checkpoints occur near the end of G1, at the G2/M transition, and during metaphase (Figure 10.11). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 421 Figure 10.11 The cell cycle is controlled at three checkpoints. The integrity of the DNA is assessed at the G1 checkpoint. Proper chromosome duplication is assessed at the G2 checkpoint. Attachment of each kinetochore to a spindle fiber is assessed at the M checkpoint. The G1 Checkpoint The G1 checkpoint determines whether all conditions are favorable for cell division to proceed. The G1 checkpoint, also called the restriction point (in yeast), is a point at which the cell commits to the cell division process. External influences, such as growth factors, play a large role in carrying the cell past the G1 checkpoint. In addition to adequate reserves and cell size, there is a check for genomic DNA damage at the G1 checkpoint. A cell that does not meet all the requirements will not be allowed to progress into the S phase. The cell can halt the cycle and attempt to remedy the problematic condition |
, or the cell can advance into G0 and await further signals when conditions improve. The G2 Checkpoint The G2 checkpoint bars entry into the mitotic phase if certain conditions are not met. As at the G1 checkpoint, cell size and protein reserves are assessed. However, the most important role of the G2 checkpoint is to ensure that all of the chromosomes have been replicated and that the replicated DNA is not damaged. If the checkpoint mechanisms detect problems with the DNA, the cell cycle is halted, and the cell attempts to either complete DNA replication or repair the damaged DNA. The M Checkpoint The M checkpoint occurs near the end of the metaphase stage of karyokinesis. The M checkpoint is also known as the spindle checkpoint, because it determines whether all the sister chromatids are correctly attached to the spindle microtubules. Because the separation of the sister chromatids during anaphase is an irreversible step, the cycle will not proceed until the kinetochores of each pair of sister chromatids are firmly anchored to at least two spindle fibers arising from opposite poles of the cell. 422 Chapter 10 | Cell Reproduction Watch what occurs at cell_checkpnts) to see an animation of the cell cycle. the G1, G2, and M checkpoints by visiting this website (http://openstaxcollege.org/l/ Down Syndrome is a genetic, developmental condition caused by nondisjunction of chromosome 21 during meiosis. Explain how a problem with the spindle checkpoint can cause this to occur in the cell. a. Failure in spindle checkpoint results in the formation of one gamete cell with two extra chromosomes and another gamete cell lacking chromosomes. b. Failure in spindle checkpoint yields the same number of chromosomes in each gamete cell. c. Failure in spindle checkpoint will form two gamete cells without any chromosomes. d. Failure in spindle checkpoint results in the formation of one gamete cell with an extra chromosome and another gamete cell lacking a chromosome. Regulator Molecules of the Cell Cycle In addition to the internally controlled checkpoints, there are two groups of intracellular molecules that regulate the cell cycle. These regulatory molecules either promote progress of the cell to the next phase (positive regulation) or halt the cycle (negative regulation). Regulator molecules may act individually, or they can influence the activity or production of other regulatory proteins. Therefore, the failure of a single regulator may have almost no effect on the cell cycle, especially if more than |
one mechanism controls the same event. Conversely, the effect of a deficient or non-functioning regulator can be wide-ranging and possibly fatal to the cell if multiple processes are affected. Positive Regulation of the Cell Cycle Two groups of proteins, called cyclins and cyclin-dependent kinases (Cdks), are responsible for the progress of the cell through the various checkpoints. The levels of the four cyclin proteins fluctuate throughout the cell cycle in a predictable pattern (Figure 10.12). Increases in the concentration of cyclin proteins are triggered by both external and internal signals. After the cell moves to the next stage of the cell cycle, the cyclins that were active in the previous stage are degraded. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 423 Figure 10.12 The concentrations of cyclin proteins change throughout the cell cycle. There is a direct correlation between cyclin accumulation and the three major cell cycle checkpoints. Also note the sharp decline of cyclin levels following each checkpoint (the transition between phases of the cell cycle), as cyclin is degraded by cytoplasmic enzymes. (credit: modification of work by "WikiMiMa"/Wikimedia Commons) Cyclins regulate the cell cycle only when they are tightly bound to Cdks. To be fully active, the Cdk/cyclin complex must also be phosphorylated in specific locations. Like all kinases, Cdks are enzymes (kinases) that phosphorylate other proteins. Phosphorylation activates the protein by changing its shape. The proteins phosphorylated by Cdks are involved in advancing the cell to the next phase. (Figure 10.13). The levels of Cdk proteins are relatively stable throughout the cell cycle; however, the concentrations of cyclin fluctuate and determine when Cdk/cyclin complexes form. The different cyclins and Cdks bind at specific points in the cell cycle and thus regulate different checkpoints. 424 Chapter 10 | Cell Reproduction Figure 10.13 Cyclin-dependent kinases (Cdks) are protein kinases that, when fully activated, can phosphorylate and thus activate other proteins that advance the cell cycle past a checkpoint. To become fully activated, a Cdk must bind to a cyclin protein and then be phosphorylated by another kinase. Since the cyclic fluctuations of cyclin levels are based on the timing of the cell cycle and |
not on specific events, regulation of the cell cycle usually occurs by either the Cdk molecules alone or the Cdk/cyclin complexes. Without a specific concentration of fully activated cyclin/Cdk complexes, the cell cycle cannot proceed through the checkpoints. Although the cyclins are the main regulatory molecules that determine the forward momentum of the cell cycle, there are several other mechanisms that fine-tune the progress of the cycle with negative, rather than positive, effects. These mechanisms essentially block the progression of the cell cycle until problematic conditions are resolved. Molecules that prevent the full activation of Cdks are called Cdk inhibitors. Many of these inhibitor molecules directly or indirectly monitor a particular cell cycle event. The block placed on Cdks by inhibitor molecules will not be removed until the specific event that the inhibitor monitors is completed. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 425 Negative Regulation of the Cell Cycle The second group of cell cycle regulatory molecules are negative regulators. Negative regulators halt the cell cycle. Remember that in positive regulation, active molecules cause the cycle to progress. The best understood negative regulatory molecules are retinoblastoma protein (Rb), p53, and p21. Retinoblastoma proteins are a group of tumor-suppressor proteins common in many cells. The 53 and 21 designations refer to the functional molecular masses of the proteins (p) in kilodaltons. Much of what is known about cell cycle regulation comes from research conducted with cells that have lost regulatory control. All three of these regulatory proteins were discovered to be damaged or non-functional <|endoftext|>in cells that had begun to replicate uncontrollably (became cancerous). In each case, the main cause of the unchecked progress through the cell cycle was a faulty copy of the regulatory protein. Rb, p53, and p21 act primarily at the G1 checkpoint. p53 is a multi-functional protein that has a major impact on the commitment of a cell to division because it acts when there is damaged DNA in cells that are undergoing the preparatory processes during G1. If damaged DNA is detected, p53 halts the cell cycle and recruits enzymes to repair the DNA. If the DNA cannot be repaired, p53 can trigger apoptosis, or cell death, to prevent the duplication of damaged chromosomes. As p53 levels rise, the production of p21 is triggered. p |
21 enforces the halt in the cycle dictated by p53 by binding to and inhibiting the activity of the Cdk/cyclin complexes. As a cell is exposed to more stress, higher levels of p53 and p21 accumulate, making it less likely that the cell will move into the S phase. Rb exerts its regulatory influence on other positive regulator proteins. Chiefly, Rb monitors cell size. In the active, dephosphorylated state, Rb binds to proteins called transcription factors, most commonly, E2F (Figure 10.14). Transcription factors “turn on” specific genes, allowing the production of proteins encoded by that gene. When Rb is bound to E2F, production of proteins necessary for the G1/S transition is blocked. As the cell increases in size, Rb is slowly phosphorylated until it becomes inactivated. Rb releases E2F, which can now turn on the gene that produces the transition protein, and this particular block is removed. For the cell to move past each of the checkpoints, all positive regulators must be “turned on,” and all negative regulators must be “turned off.” 426 Chapter 10 | Cell Reproduction Figure 10.14 Rb halts the cell cycle and releases its hold in response to cell growth. Rb and other proteins that negatively regulate the cell cycle are sometimes called tumor supressors. Why do you think the name tumor suppressor might be appropriate for these proteins? a. They inhibit cell division. b. They enhance the rate of cell division. c. They start the cell cycle, thereby suppressing tumor formation. d. These proteins, when phosphorylated, allow the cell cycle to proceed. • Rb is a negative regulator that blocks the cell cycle at the G1 checkpoint until the cell achieves a requisite size. What is the most likely mechanism that Rb employs to halt the cell cycle? • A cell has a mutation that results in the production of an abnormal cyclin-dependent kinase at the G2/M checkpoint. What is a likely consequence of the mutation on the cell cycle? 10.4 | Cancer and the Cell Cycle In this section, you will explore the following question: • What causes uncontrolled cell growth, and why does it often cause cancer? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 427 Connection for AP® Cour |
ses Cancer results from unchecked cell division caused by a breakdown of the mechanisms that regulate the cell cycle. The loss of control begins with a change in the DNA sequence of a gene that codes for one of the regulatory molecules. Faulty instructions lead to a protein that does not function as it should. One culprit that has been identified is the p53 protein (coded for by the p53 gene), a major regulator at the G1 checkpoint. Normally, p53 proteins monitor DNA. If they find cells with damaged DNA, p53 will trigger repair mechanisms or destroy the cells, thus suppressing the formation of a tumor. However, mutations in p53 can result in abnormal p53 proteins that fail to stop cell division if the cell’s DNA is damaged. This results in an increased number of mutations, leading to abnormal daughter cells. Eventually, all checkpoints in the cell become nonfunctional, and the abnormal cells can crowd out normal cells. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework, as shown in the table. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 3.7 The student can make predictions about natural phenomena occurring during the cell cycle. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 2.32][APLO 2.34][APLO 3.6][APLO 3.7][APLO 3.8][APLO 4.6][APLO 4.14][APLO 4.22] Cancer comprises many different diseases caused by a common mechanism: uncontrolled cell growth. Despite the redundancy and overlapping levels of cell cycle control, errors do occur. One of the critical processes monitored by the cell cycle |
checkpoint surveillance mechanism is the proper replication of DNA during the S phase. Even when all of the cell cycle controls are fully functional, a small percentage of replication errors (mutations) will be passed on to the daughter cells. If changes to the DNA nucleotide sequence occur within a coding portion of a gene and are not corrected, a gene mutation results. All cancers start when a gene mutation gives rise to a faulty protein that plays a key role in cell reproduction. The change in the cell that results from the malformed protein may be minor: perhaps a slight delay in the binding of Cdk to cyclin or an Rb protein that detaches from its target DNA while still phosphorylated. Even minor mistakes, however, may allow subsequent mistakes to occur more readily. Over and over, small uncorrected errors are passed from the parent cell to the daughter cells and amplified as each generation produces more non-functional proteins from uncorrected DNA damage. Eventually, the pace of the cell cycle speeds up as the effectiveness of the control and repair mechanisms decreases. Uncontrolled growth of the mutated cells outpaces the growth of normal cells in the area, and a tumor (“-oma”) can result. Proto-oncogenes The genes that code for the positive cell cycle regulators are called proto-oncogenes. Proto-oncogenes are normal genes that, when mutated in certain ways, become oncogenes, genes that cause a cell to become cancerous. Consider what might happen to the cell cycle in a cell with a recently acquired oncogene. In most instances, the alteration of the DNA sequence will result in a less functional (or non-functional) protein. The result is detrimental to the cell and will likely prevent the cell from completing the cell cycle; however, the organism is not harmed because the mutation will not be carried forward. If a cell cannot reproduce, the mutation is not propagated and the damage is minimal. Occasionally, however, a gene mutation causes a change that increases the activity of a positive regulator. For example, a mutation that allows Cdk to be activated 428 Chapter 10 | Cell Reproduction without being partnered with cyclin could push the cell cycle past a checkpoint before all of the required conditions are met. If the resulting daughter cells are too damaged to undergo further cell divisions, the mutation would not be propagated and no harm would come to the organism. However, if the atypical daughter cells are able to undergo further cell divisions, subsequent generations of cells |
will probably accumulate even more mutations, some possibly in additional genes that regulate the cell cycle. The Cdk gene in the above example is only one of many genes that are considered proto-oncogenes. In addition to the cell cycle regulatory proteins, any protein that influences the cycle can be altered in such a way as to override cell cycle checkpoints. An oncogene is any gene that, when altered, leads to an increase in the rate of cell cycle progression. Tumor Suppressor Genes Like proto-oncogenes, many of the negative cell cycle regulatory proteins were discovered in cells that had become cancerous. Tumor suppressor genes are segments of DNA that code for negative regulator proteins, the type of regulators that, when activated, can prevent the cell from undergoing uncontrolled division. The collective function of the bestunderstood tumor suppressor gene proteins, Rb, p53, and p21, is to put up a roadblock to cell cycle progression until certain events are completed. A cell that carries a mutated form of a negative regulator might not be able to halt the cell cycle if there is a problem. Tumor suppressors are similar to brakes in a vehicle: Malfunctioning brakes can contribute to a car crash. Mutated p53 genes have been identified in more than one-half of all human tumor cells. This discovery is not surprising in light of the multiple roles that the p53 protein plays at the G1 checkpoint. A cell with a faulty p53 may fail to detect errors present in the genomic DNA (Figure 10.15). Even if a partially functional p53 does identify the mutations, it may no longer be able to signal the necessary DNA repair enzymes. Either way, damaged DNA will remain uncorrected. At this point, a functional p53 will deem the cell unsalvageable and trigger programmed cell death (apoptosis). The damaged version of p53 found in cancer cells, however, cannot trigger apoptosis. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 429 Figure 10.15 The role of normal p53 is to monitor DNA and the supply of oxygen (hypoxia is a condition of reduced oxygen supply). If damage is detected, p53 triggers repair mechanisms. If repairs are unsuccessful, p53 signals apoptosis. A cell with an abnormal p53 protein cannot repair damaged DNA and thus cannot signal apoptosis. Cells with |
abnormal p53 can become cancerous. (credit: modification of work by Thierry Soussi) Human papillomavirus can cause cervical cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity? a. E6 activates p53 b. E6 inactivates p53 c. E6 mutates p53 d. E6 binding marks p53 for degradation The loss of p53 function has other repercussions for the cell cycle. Mutated p53 might lose its ability to trigger p21 production. Without adequate levels of p21, there is no effective block on Cdk activation. Essentially, without a fully functional p53, the G1 checkpoint is severely compromised and the cell proceeds directly from G1 to S regardless of internal and external conditions. At the completion of this shortened cell cycle, two daughter cells are produced that have inherited the mutated p53 gene. Given the non-optimal conditions under which the parent cell reproduced, it is likely that the daughter cells will have acquired other mutations in addition to the faulty tumor suppressor gene. Cells such as these daughter cells quickly accumulate both oncogenes and non-functional tumor suppressor genes. Again, the result is tumor growth. 430 Chapter 10 | Cell Reproduction Go to this website (http://openstaxcollege.org/l/cancer) to watch an animation of how cancer results from errors in the cell cycle. Treating cancer can be described as a fight against natural biologic processes. Explain what this means in terms of tumor formation. a. Cancer forms when natural defenses are inhibited and cells divide uncontrollably. b. Mutated cells undergo apoptosis, causing cells to divide uncontrollably. c. Cancer treatment would require inhibiting apoptosis which is a natural defense. d. In cancerous cells, apoptosis occurs. Think About It Human papillomavirus (HPV) can cause cervical cancer. The virus encodes E6, a protein that binds p53. Predict the most likely effect of E6 binding on p53 activity, and explain the basis for your prediction. 10.5 | Prokaryotic Cell Division In this section, you will explore the following question: • How does the process of binary fission in prokaryotes differ from cell division in eukaryotes? Prokaryotes, such as bacteria, propagate by binary fission. |
For unicellular organisms, cell division is the only method to produce new individuals. In both prokaryotic and eukaryotic cells, the outcome of cell reproduction is a pair of daughter cells that are genetically identical to the parent cell. In unicellular organisms, daughter cells are individuals. To achieve the outcome of cloned offspring, certain steps are essential. The genomic DNA must be replicated and then allocated into the daughter cells; the cytoplasmic contents must also be divided to give both new cells the machinery to sustain life. In bacterial cells, the genome consists of a single, circular DNA chromosome; therefore, the process of cell division is simplified. Karyokinesis is unnecessary because there is no nucleus and thus no need to direct one copy of the multiple chromosomes into each daughter cell. This type of cell division is called binary (prokaryotic) fission. Binary Fission Due to the relative simplicity of the prokaryotes, the cell division process, called binary fission, is a less complicated and much more rapid process than cell division in eukaryotes. The single, circular DNA chromosome of bacteria is not enclosed in a nucleus, but instead occupies a specific location, the nucleoid, within the cell (Figure 10.2). Although the DNA of the nucleoid is associated with proteins that aid in packaging the molecule into a compact size, there are no histone proteins and thus no nucleosomes in prokaryotes. The packing proteins of bacteria are, however, related to the cohesin and condensin proteins involved in the chromosome compaction of eukaryotes. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 431 The bacterial chromosome is attached to the plasma membrane at about the midpoint of the cell. The starting point of replication, the origin, is close to the binding site of the chromosome to the plasma membrane (Figure 10.16). Replication of the DNA is bidirectional, moving away from the origin on both strands of the loop simultaneously. As the new double strands are formed, each origin point moves away from the cell wall attachment toward the opposite ends of the cell. As the cell elongates, the growing membrane aids in the transport of the chromosomes. After the chromosomes have cleared the midpoint of the elongated cell, cytoplasmic separation begins. The formation of a ring composed of repeating units of |
a protein called FtsZ directs the partition between the nucleoids. Formation of the FtsZ ring triggers the accumulation of other proteins that work together to recruit new membrane and cell wall materials to the site. A septum is formed between the nucleoids, extending gradually from the periphery toward the center of the cell. When the new cell walls are in place, the daughter cells separate. Figure 10.16 These images show the steps of binary fission in prokaryotes. (credit: modification of work by “Mcstrother”/Wikimedia Commons) 432 Chapter 10 | Cell Reproduction The precise timing and formation of the mitotic spindle is critical to the success of eukaryotic cell division. Prokaryotic cells, on the other hand, do not undergo karyokinesis and therefore have no need for a mitotic spindle. However, the FtsZ protein that plays such a vital role in prokaryotic cytokinesis is structurally and functionally very similar to tubulin, the building block of the microtubules that make up the mitotic spindle fibers that are necessary for eukaryotes. FtsZ proteins can form filaments, rings, and other threedimensional structures that resemble the way tubulin forms microtubules, centrioles, and various cytoskeletal components. In addition, both FtsZ and tubulin employ the same energy source, GTP (guanosine triphosphate), to rapidly assemble and disassemble complex structures. FtsZ and tubulin are homologous structures derived from common evolutionary origins. In this example, FtsZ is the ancestor protein to tubulin (a modern protein). While both proteins are found in extant organisms, tubulin function has evolved and diversified tremendously since evolving from its FtsZ prokaryotic origin. A survey of mitotic assembly components found in present-day unicellular eukaryotes reveals crucial intermediary steps to the complex membrane-enclosed genomes of multicellular eukaryotes (Table 10.4). Cell Division Apparatus among Various Organisms Structure of genetic material There is no nucleus. The single, circular chromosome exists in a region of cytoplasm called the nucleoid. Prokaryotes Division of nuclear material Occurs through binary fission. As the chromosome is replicated, the two copies move to opposite ends of the cell by an unknown mechanism. Some protists Linear chromosomes exist in the nucleus. Chromos |
omes attach to the nuclear envelope, which remains intact. The mitotic spindle passes through the envelope and elongates the cell. No centrioles exist. Other protists Linear chromosomes exist in the nucleus. A mitotic spindle forms from the centrioles and passes through the nuclear membrane, which remains intact. Chromosomes attach to the mitotic spindle, which separates the chromosomes and elongates the cell. Animal cells Linear chromosomes exist in the nucleus. A mitotic spindle forms from the centrosomes. The nuclear envelope dissolves. Chromosomes attach to the mitotic spindle, which separates the chromosomes and elongates the cell. Separation of daughter cells FtsZ proteins assemble into a ring that pinches the cell in two. Microfilaments form a cleavage furrow that pinches the cell in two. Microfilaments form a cleavage furrow that pinches the cell in two. Microfilaments form a cleavage furrow that pinches the cell in two. Table 10.4 FtsZ is a prokaryotic protein and tubulin is a eukaryotic protein. These two proteins share many structural and functional similarities and are believed to have evolved from the same ancestral protein. However, there are also some important differences between these proteins. In what way are these proteins different? a. Tubulin proteins can rapidly disassemble, but FtsZ proteins cannot. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 433 b. Tubulin proteins can form long filaments, but FtsZ proteins cannot. c. Tubulin uses GTP as an energy source, but FtsZ does not. d. Tubulin pulls chromosomes apart, but FtsZ does not. 434 Chapter 10 | Cell Reproduction KEY TERMS anaphase stage of mitosis during which sister chromatids are separated from each other binary fission prokaryotic cell division process cell cycle ordered series of events involving cell growth and cell division that produces two new daughter cells cell cycle checkpoint cycle stages mechanism that monitors the preparedness of a eukaryotic cell to advance through the various cell cell plate structure formed during plant cell cytokinesis by Golgi vesicles, forming a temporary structure (phragmoplast) and fusing at the metaphase plate; ultimately leads to the formation of cell walls that separate the two daughter cells centriole rod- |
like structure constructed of microtubules at the center of each animal cell centrosome centromere region at which sister chromatids are bound together; a constricted area in condensed chromosomes chromatid single DNA molecule of two strands of duplicated DNA and associated proteins held together at the centromere cleavage furrow constriction formed by an actin ring during cytokinesis in animal cells that leads to cytoplasmic division condensin proteins that help sister chromatids coil during prophase cyclin one of a group of proteins that act in conjunction with cyclin-dependent kinases to help regulate the cell cycle by phosphorylating key proteins; the concentrations of cyclins fluctuate throughout the cell cycle cyclin-dependent kinase one of a group of protein kinases that helps to regulate the cell cycle when bound to cyclin; it functions to phosphorylate other proteins that are either activated or inactivated by phosphorylation cytokinesis division of the cytoplasm following mitosis that forms two daughter cells. diploid cell, nucleus, or organism containing two sets of chromosomes (2n) FtsZ tubulin-like protein component of the prokaryotic cytoskeleton that is important in prokaryotic cytokinesis (name origin: Filamenting temperature-sensitive mutant Z) G0 phase distinct from the G1 phase of interphase; a cell in G0 is not preparing to divide G1 phase G2 phase (also, first gap) first phase of interphase centered on cell growth during mitosis (also, second gap) third phase of interphase during which the cell undergoes final preparations for mitosis gamete haploid reproductive cell or sex cell (sperm, pollen grain, or egg) gene physical and functional unit of heredity, a sequence of DNA that codes for a protein. genome total genetic information of a cell or organism haploid cell, nucleus, or organism containing one set of chromosomes (n) histone one of several similar, highly conserved, low molecular weight, basic proteins found in the chromatin of all eukaryotic cells; associates with DNA to form nucleosomes homologous chromosomes chromosomes of the same morphology with genes in the same location; diploid organisms have pairs of homologous chromosomes (homologs), with each homolog derived from a different parent interphase period of the cell cycle leading up to mitosis; includes G1, S, and G2 phases (the interim |
period between two consecutive cell divisions karyokinesis mitotic nuclear division This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 435 kinetochore protein structure associated with the centromere of each sister chromatid that attracts and binds spindle microtubules during prometaphase locus position of a gene on a chromosome metaphase stage of mitosis during which chromosomes are aligned at the metaphase plate metaphase plate equatorial plane midway between the two poles of a cell where the chromosomes align during metaphase mitosis (also, karyokinesis) period of the cell cycle during which the duplicated chromosomes are separated into identical nuclei; includes prophase, prometaphase, metaphase, anaphase, and telophase mitotic phase period of the cell cycle during which duplicated chromosomes are distributed into two nuclei and cytoplasmic contents are divided; includes karyokinesis (mitosis) and cytokinesis mitotic spindle apparatus composed of microtubules that orchestrates the movement of chromosomes during mitosis nucleosome subunit of chromatin composed of a short length of DNA wrapped around a core of histone proteins oncogene mutated version of a normal gene involved in the positive regulation of the cell cycle origin (also, ORI) region of the prokaryotic chromosome where replication begins (origin of replication) p21 cell cycle regulatory protein that inhibits the cell cycle; its levels are controlled by p53 p53 cell cycle regulatory protein that regulates cell growth and monitors DNA damage; it halts the progression of the cell cycle in cases of DNA damage and may induce apoptosis prometaphase stage of mitosis during which the nuclear membrane breaks down and mitotic spindle fibers attach to kinetochores prophase stage of mitosis during which chromosomes condense and the mitotic spindle begins to form proto-oncogene normal gene that when mutated becomes an oncogene quiescent refers to a cell that is performing normal cell functions and has not initiated preparations for cell division retinoblastoma protein (Rb) regulatory molecule that exhibits negative effects on the cell cycle by interacting with a transcription factor (E2F) S phase second, or synthesis, stage of interphase during which DNA replication occurs septum structure formed in a bacterial cell as a precursor to the separation of the cell into two daughter cells telophase |
stage of mitosis during which chromosomes arrive at opposite poles, decondense, and are surrounded by a new nuclear envelope tumor suppressor gene uncontrolled division segment of DNA that codes for regulator proteins that prevent the cell from undergoing CHAPTER SUMMARY 10.1 Cell Division Prokaryotes have a single circular chromosome composed of double-stranded DNA, whereas eukaryotes have multiple, linear chromosomes composed of chromatin, all surrounded by a nuclear membrane. The 46 chromosomes of human somatic cells are composed of 22 pairs of autosomes (matched pairs) and a pair of sex chromosomes, which may or may not be matched. This is the 2n or diploid state. Human gametes have 23 chromosomes representing one complete set of chromosomes; a set of chromosomes is complete with either one of the sex chromosomes. This is the n or haploid state. Genes are segments of DNA that code for a specific protein. An organism’s traits are determined by the genes inherited from each parent. Duplicated chromosomes are composed of two sister chromatids. Chromosomes are compacted using a 436 Chapter 10 | Cell Reproduction variety of mechanisms during certain stages of the cell cycle. Several classes of protein are involved in the organization and packing of the chromosomal DNA into a highly condensed structure. The condensing complex compacts chromosomes, and the resulting condensed structure is necessary for chromosomal segregation during mitosis. 10.2 The Cell Cycle The cell cycle is an orderly sequence of events. Cells on the path to cell division proceed through a series of precisely timed and carefully regulated stages. In eukaryotes, the cell cycle consists of a long preparatory period, called interphase. Interphase is divided into G1, S, and G2 phases. The mitotic phase begins with karyokinesis (mitosis), which consists of five stages: prophase, prometaphase, metaphase, anaphase, and telophase. The final stage of the mitotic phase is cytokinesis, during which the cytoplasmic components of the daughter cells are separated either by an actin ring (animal cells) or by cell plate formation (plant cells). 10.3 Control of the Cell Cycle Each step of the cell cycle is monitored by internal controls called checkpoints. There are three major checkpoints in the cell cycle: one near the end of G1, a second at the G2/M transition, and the third during metaphase. Positive regulator molecules allow the cell cycle |
to advance to the next stage. Negative regulator molecules monitor cellular conditions and can halt the cycle until specific requirements are met. 10.4 Cancer and the Cell Cycle Cancer is the result of unchecked cell division caused by a breakdown of the mechanisms that regulate the cell cycle. The loss of control begins with a change in the DNA sequence of a gene that codes for one of the regulatory molecules. Faulty instructions lead to a protein that does not function as it should. Any disruption of the monitoring system can allow other mistakes to be passed on to the daughter cells. Each successive cell division will give rise to daughter cells with even more accumulated damage. Eventually, all checkpoints become nonfunctional, and rapidly reproducing cells crowd out normal cells, resulting in a tumor or leukemia (blood cancer). 10.5 Prokaryotic Cell Division In both prokaryotic and eukaryotic cell division, the genomic DNA is replicated and then each copy is allocated into a daughter cell. In addition, the cytoplasmic contents are divided evenly and distributed to the new cells. However, there are many differences between prokaryotic and eukaryotic cell division. Bacteria have a single, circular DNA chromosome but no nucleus. Therefore, mitosis is not necessary in bacterial cell division. Bacterial cytokinesis is directed by a ring composed of a protein called FtsZ. Ingrowth of membrane and cell wall material from the periphery of the cells results in the formation of a septum that eventually constructs the separate cell walls of the daughter cells. REVIEW QUESTIONS 1. A diploid cell has how many times the number of chromosomes as a haploid cell? a. four times b. half c. one-fourth d. twice 2. The first level of DNA organization in a eukaryotic cell is maintained by which molecule? a. cohesion b. condensin c. chromatin d. histone 3. What inherited feature, in specific combinations, determines an organism’s traits? a. cell membranes b. genes c. proteins d. RNA 4. What are identical copies of chromatin held together by cohesin at the centromere called? a. histones b. nucleosomes c. chromatin d. sister chromatids 5. Chromosomes are duplicated during what stage of the cell cycle? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 4 |
37 a. G1 phase b. prophase c. pro-metaphase d. S-phase a. p53 b. Retinoblastoma protein (Rb) c. cyclin d. Cyclin-dependent kinase (Cdk) 6. Which of the following events does not occur during some stages of interphase? 13. Which negative regulatory molecule can trigger apoptosis if vital cell cycle events do not occur? a. DNA duplication b. increase in cell size c. organelle duplication d. separation of sister chromatids a. p53 b. p21 c. Retinoblastoma protein (Rb) d. Cyclin-dependent kinase (Cdk) 7. Attachment of the mitotic spindle fibers to the kinetochores is a characteristic of which stage of mitosis? 14. What is the main prerequisite for clearance at the G2 checkpoint? a. anaphase b. prophase c. prometaphase d. metaphase 8. The fusing of Golgi vesicles at the metaphase plate of dividing plant cells forms what structure? a. actin ring b. cell plate c. cleavage furrow d. mitotic spindle 9. What would be the outcome of blocking S-phase of interphase? a. The cell would enter karyokinesis. b. DNA replication would not occur. c. Centrosomes would be duplicated. d. The cytoskeleton would be dismantled. 10. At which of the cell cycle checkpoints do external forces have the greatest influence? a. G1 b. G2 checkpoint checkpoint c. M checkpoint d. G0 checkpoint 11. If the M checkpoint is not cleared, what stage of mitosis will be blocked? a. prophase b. prometaphase c. metaphase d. anaphase 12. Which protein is a positive regulator that phosphorylates other proteins when activated? a. The cell has a reached a sufficient size. b. The cell has an adequate stockpile of nucleotides. c. An accurate and complete DNA replication has occurred. d. Proper attachment of mitotic spindle fibers to kinetochores has occurred. 15. What do you call changes to the order of nucleotides in a segment of DNA that codes for a protein? a. proto-oncogenes b. tumor suppressor genes c. gene mutations d. negative regulators 16. Human papillomavirus can cause cervical |
cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity? a. E6 activates p53. b. E6 protects p53 from degradation. c. E6 mutates p53. d. E6 binding marks p53 for degradation. 17. What is a gene that codes for a positive cell cycle regulator called? a. kinase inhibitor b. oncogenes c. proto-oncogenes d. tumor suppressor genes 18. Which molecule is a Cdk inhibitor or is controlled by p53? a. anti-kinase b. cyclin c. p21 d. Rb 438 Chapter 10 | Cell Reproduction a. In both prokaryotic and eukaryotic cells, the outcome of cell reproduction is a pair of daughter cells, which are genetically identical to the parent cell. b. Karyokinesis is unnecessary in prokaryotes because there is no nucleus. c. Replication of the prokaryotic chromosome begins at the origin of replication and continues in both directions at once. d. The mitotic spindle draws the duplicated chromosomes to the opposite ends of the cell followed by formation of a septum and two daughter cells. 21. The formation of what structure, that will eventually form the new cell walls of the daughter cells, is directed by FtsZ? a. contractile ring b. cell plate c. cytoskeleton d. septum a. G1 - assessment for DNA damage, S - duplication of genetic material, G2 duplication and dismantling organelles - b. G1 - duplication of organelles, S - duplication of DNA, G2 - assessment of DNA damage c. G1 - synthesis of DNA, S - synthesis of organelle genetic material, G2 DNA damage - assessment of d. G1 - preparation for DNA synthesis, S - assessment of DNA damage, M - Division of cell 25. Chemotherapy drugs such as vincristine and colchicines disrupt mitosis by binding to tubulin (the subunit of microtubules) and interfering with microtubule assembly and disassembly. Exactly what mitotic structure do these drugs target, and what effect would that have on cell division? 19. Which eukaryotic cell cycle events are missing in binary fission? a. cell growth b. DNA duplication c. |
karyokinesis d. cytokinesis 20. Which of the following statements about binary fission is false? CRITICAL THINKING QUESTIONS 22. Compare and contrast a human somatic cell to a human gamete. a. Somatic cells have 46 chromosomes and are diploid, whereas gametes have half as many chromosomes as found in somatic cells. b. Somatic cells have 23 chromosomes and are diploid, whereas gametes have half as many chromosomes are are present in somatic cells. c. Somatic cells have 46 chromosomes and are haploid, whereas gametes have 23 chromosomes and are diploid. d. Somatic cells have 46 chromosomes with one sex chromosome. In gametes, 23 chromosomes are present with two sex chromosomes. 23. Eukaryotic chromosomes are thousands of times longer than a typical cell. Explain how chromosomes can fit inside a eukaryotic nucleus. a. The genetic material remains distributed in the nucleus, mitochondria, and chloroplast. b. The genome is present in a looped structure, thus it fits the size of the nucleus. c. The DNA remains coiled around proteins to form nucleosomes. d. The genetic material remains bound to the nuclear envelope, forming invaginations. 24. Briefly describe the events that occur in each phase of interphase. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 439 a. The drugs bind tubulin and inhibit the binding of spindle to the chromosome. This can arrest the cell cycle. b. The drugs bind the tubulin, which leads to an error in the chromosome separation. This could lead to apoptosis. c. The drugs bind the tubulin, thereby inhibiting their division in S-phase. This inhibits cell division. d. The drugs bind the spindle fiber and hinder the separation of chromatins. This promotes the division spontaneously. 26. List some reasons why a cell that has just completed cytokinesis might enter the G0 phase instead of the G1 phase. a. Some cells are physiologically inhibited from undergoing any division and remain in the G0 phase to provide assistance to their neighboring cells. b. Some cells reproduce only under certain conditions and, until then, they remain in the G0 phase. c. Suspected DNA damage can lead the cell to phase. undergo the G0 d. The |
lack of important components of cell division makes cells stay in the G0 phase. 27. Describe the general conditions that must be met at each of the three main cell cycle checkpoints. checkpoint - assessment of DNA damage, a. G1 G2 segregation of sister chromatids in anaphase. - assessment of new DNA, M checkpoint - checkpoint - assessment of new DNA, M checkpoint - Energy reserves for s phase, b. G1 G2 checkpoint- attachment of spindle to kinetochore. checkpoint - assessment of DNA damage, checkpoint - energy reserves for c. G1 G2 duplication, M checkpoint - attachment of spindle to kinetochore d. G1 checkpoint - Energy reserves for S-phase, S checkpoint - synthesis of DNA, G2 - assessment of new DNA checkpoint 28. Explain the roles of the positive cell cycle regulators compared to the negative regulators. a. Positive regulators promote the cell cycle but negative regulators block the cell cycle. b. Positive regulators block the cell division in cancerous cells but negative regulators promote in such cells. c. Positive regulators promote the cell cycle but negative regulators arrest the cell cycle until certain events have occurred. d. Positive regulators show positive feedback mechanisms but negative regulators show negative feedback in the cell cycle. 29. Describe what occurs at the M checkpoint and predict what would happen if the M checkpoint failed. a. The M checkpoint checks for proper separation of sister chromatids and if it fails, then cells may undergo nondisjunction of chromosomes. b. The M checkpoint checks if the DNA is damaged and promotes its repair. If it fails, then the daughters end up with damaged DNA. c. The M checkpoint ensures the proper duplication of DNA and if it fails, the cells may undergo nondisjunction of chromosomes. d. The M checkpoint ensures that all the components required for cell division are available and if it fails, the cell cycle will be inhibited. 30. List the regulatory mechanisms that might be lost in a cell producing faulty p53. a. assessment of damaged DNA, recruiting repair enzymes, and binding of spindle to kinetochore b. quality of DNA, triggering apoptosis, and recruiting repair enzymes c. quality of DNA, binding of spindle to kinetochore, and assessment of DNA repair d. triggering apoptosis, recruiting repair enzymes, and proper binding of spindle to kinetochore 31. p53 can trigger apoptosis if certain cell cycle events fail. How does this regulatory outcome benefit a multice |
llular organism? a. The apoptosis helps in controlling the consumption of energy by the extra cells. b. The apoptosis inhibits the production of faulty proteins, which could be produced due to the DNA damage. c. The process of apoptosis stops the invasion of viruses in the other cells. d. The cells are killed due to the production of reactive oxygen species produced, which could harm the organism. 32. Name the processes that eukaryotic cell division and binary fission have in common. 440 Chapter 10 | Cell Reproduction a. DNA duplication, division of cell organelles, division of the cytoplasmic contents b. DNA duplication, segregation of duplicated chromosomes, and division of the cytoplasmic contents c. d. formation of a septum, DNA duplication, division of the cytoplasmic contents segregation of duplicated chromosomes, formation of a septum, division of cell organelles TEST PREP FOR AP® COURSES 34. 33. The formation of what structure, that will eventually form the new cell walls of the daughter cells, is directed by FtsZ? a. contractile ring b. cell plate c. cytoskeleton d. septum a. The DNA is found wrapped around histones to form nucleosomes, which further compact and ultimately form linear chromosomes. The prokaryotic genome is found as a loop and in eukaryotes as a double-stranded linear structure. b. The DNA is wrapped around the nucleosomes to show a compact structure. The eukaryotes show a loop structure and prokaryotes show a doublestranded linear genome. c. The genetic material shows ringed heterochromatin structure. The prokaryotes show multiple loops, and eukaryotes show a condensed chromatin. d. The genetic material is wrapped around histones. The prokaryotes have a condensed structure in nucleoids, but eukaryotes have double-stranded linear structure. Which of the following statements cannot be inferred from the karyotype shown? 36. a. The cell contains DNA. b. The cell contains 46 chromosomes. c. The cell is diploid. d. The cell is prokaryotic. 35. Explain how DNA, which in humans measures approximately two meters, can fit inside a human cell that is about 10 µm. Discuss how the organization of the genetic material in eukaryotes differs from prok |
aryotes. Which of the following statements about structure 1 on the karyotype is not true? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 10 | Cell Reproduction 441 a. Structure 1 consists of homologous chromosomes. b. The two parts of structure 1 will have genes in different loci. c. The two parts of structure 1 originate from different parents. d. The two parts of structure 1 will have slightly different sequences of nucleotides. 37. boundary. The cells are In a study on cell division, researchers culture synchronously dividing human cells with thymidine. This causes the cells to arrest at the G1 then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A-D) are shown in the graph. Which sample presents the expected results for cells in S-phase? a. b. c. d. sample A sample B sample C sample D 40. In a study on cell division, researchers culture synchronously dividing human cells with thymidine. This Based on the karyotype provided, the nondisjuction of which chromosome causes Down Syndrome? a. chromosome 21 b. chromosome 22 c. X chromosome d. Y chromosome 38. Describe the sequence of mitotic cell cycle for one pair of chromosome that is undergoing normal mitotic division. a. anaphase - metaphase - prophase - cytokinesis b. anaphase - prophase - metaphase - cytokinesis c. prophase - anaphase - metaphase - cytokinesis d. prophase - metaphase - anaphase - cytokinesis 39. 442 Chapter 10 | Cell Reproduction boundary. The cells are causes the cells to arrest at the G1 then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A-D) are shown in the graph. Explain what is happening in terms of the cell cycle and DNA content in sample B. a. All the contents of the cell have been doubled. b. The DNA content of the cell |
has doubled. c. Two cells have been fused. d. The cells are showing the semiconservative mechanism of cell division. checkpoint. If damaged DNA is detected, 41. Li-Fraumeni syndrome (LFS1) is a rare hereditary disorder that leads to a predisposition to cancer. This hereditary disorder is linked to mutations in the tumor suppressor gene encoding the transcription factor p53. p53 acts at the G1 p53 halts the cell cycle. As p53 levels rise, the production of p21 is triggered. p21 enforces the halt in the cell cycle. A variant of Li-Fraumeni, called LFS2, is thought to occur due to a mutation of the CHK2 gene, which is also a tumor suppressor gene. CHK2 regulates the action of p53. Which of the following cascades is most likely to occur in a normal cell that does not contain the LFS mutation? a. 1. cell cycle progression 2. p53 3. p21 4. CHK2 b. 1. p53 2. p21 3. CHK2 4. cell cycle progression c. 1. p21 2. p53 3. CHK2 4. cell cycle progression d. 1. CHK2 2. p53 3. p21 4. cell cycle progression 42. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 The insulin growth factor (IGF-1) promotes cell proliferation as shown in the diagram. The expression of which protein in the diagram is controlled through negative feedback? a. active Cdk4 b. Cyclin D1 c. Cyclin D1/Cdk4 complex d. IGF-1 43. Explain why p53, p21, and CHK2 are considered tumor suppressor genes, not proto-oncogenes. Give an example of a proto-oncogene. a. p53, p21, and CHK2 suppress the proteins that regulate the cell cycle, whereas protooncogenes, like phosphorylated Rb, help in cell cycle progression. b. p53, p21, and CHK2 are negative cell cycle regulators, whereas Cdks are proto-oncogenes, which could cause cancer when mutated. c. p53, p21, and CHK2 suppress the proteins that regulate the cell cycle, whereas Rb is considered a proto-on |
cogene, as it is the most primitive d. The three proteins help stop the formation of tumors, whereas Cdk’s are called protooncogenes because they are the most primitive of all. Chapter 10 | Cell Reproduction 443 SCIENCE PRACTICE CHALLENGE QUESTIONS 44. Many biological processes are synchronized with the 24-hour rotational period of Earth. Circadian (24-hour) periodicity is common across phyla. One of these processes is the cell cycle. The currently accepted explanation is that the low-oxygen atmosphere of early Earth had no ozone layer to filter out the solar ultraviolet radiation that damages DNA. Completing the S phase of the cell cycle at night provided a selective advantage. The internal clock controlling the cell cycle and the circadian clock became synchronized. Research has demonstrated that changes in one clock, either the circadian clock or the cell cycle clock, disrupt timing in the other. The question was, which clock controls the other? Researchers have found that the circadian clock, which can be observed by fluorescent markers on proteins that carry the circadian signal, can be disrupted by changes in light, nutrition, or exposure to the steroid dexamethasone. Nutrition can also disrupt the cell cycle clock. Rat fibroblasts (cells constantly undergoing mitosis) were cultured on medium containing different levels of fetal bovine serum (FBS) with and without the addition of dexamethasone. Confluence is a phenomenon that occurs in tissue culture when the surface of the growth medium becomes covered with cells, and the cells stop dividing. The circadian and cell cycle periods were measured. the experimental conditions. B. Based on these data, justify the claim that in cells that are actively dividing, the circadian period is set by the cell cycle period rather than the reverse. 45. Cells in different tissues of a fully developed human show significant variations in the length of time that they remain in the G0 phase of the cell cycle: muscle (lifetime), nerve (lifetime), adipose (years), liver (year), erythrocyte (months), bone osteoclasts (weeks), leukocyte (days), and epidermal (hours). For each of these types of tissues, propose a reason based on internal and external factors and function that might account for the differences among their longevities. 46. Describe the essential components and results of mitosis and the activities that occur during interphase to prepare the cell for mitosis. 47. Cancer comprises many |
different diseases with a common cause: uncontrolled cell growth. Cancer is a complex response to a host of environmental mutagens as well as the accumulation of random mutations. Since the “war on cancer” began in 1971, the death rate due to cancer has changed very little despite the discovery of several tumor suppressor genes, including p53. Dexamethasone Confluence FBS 0% 10% 15% no no no 10% yes *20% yes 20% 10% yes yes a b c d e f g Circadian Period (hr) 24± 0.5 21.9± 1.1 19.4± 0.5 24.2± 0.5 21.25± 0.36 29± 1.05 24± 0.5 Cell Cycle Period (hr) 24± 0.5 21.3± 1.3 18.6± 0.6 20.1± 0.94 19.5± 0.42 16.05± 0.48 na no no no no no no yes Table 10.5 * Subsets of samples with 20% FBS and dexamethasone were clustered around two means for each measured period. A. Based on these data, describe the connections between the circadian period and the cell cycle period for each of 444 Chapter 10 | Cell Reproduction A. Briefly describe the multiple functions of p53, C. Mutational signatures of p53 are shown in the including the role of p53 in apoptosis. B. A principle of biology is that “form follows function.” The protein p53, which has multiple functions, is named for its molecular mass—approximately 53 kDa. This is not a large polymer by comparison with other proteins; for example, ATP synthase, which has only one function, has a molecular mass of approximately 550 kDa. Based on analogies to processes involved in cellular signaling, create a model(s) to explain how so many functions can be supported by a single, relatively simple structure. figure above [G.P. Pfeifer et al., Nature, 21(48), 2002] for the three types of cancer with the highest death rates in the U.S.: lung (~225,000 deaths in 2016), breast (246,000), and colorectal (381,000). These data can be obtained by sequencing the gene that encodes p53. Approximately 85% of lung cancers occur in smokers. Based on these data, calculate how many deaths due to lung |
cancer among nonsmokers were reported in 2016. How much does smoking increase the likelihood of death due to lung cancer? D. As shown under each graph, particular transversions (replacement of a pyrimidine by a purine of vice versa) or transitions (replacement of a purine or pyrimidine by the alternative purine or pyrimidine) are features of specific mutational signatures. Based on these data, identify the transversion or transition that seems to be induced by cigarette smoke. E. Using your answer to B above, predict possible mechanisms, that is, transversion or transition, for the different mutational signatures among lung cancers of smokers and those of other cancers, and for the very similar mutational signatures of lung cancers of nonsmokers and of breast and colorectal cancers. The partitioning of function along the length of the protein can lead to functional and nonfunctional segments. It is believed that the transversions due to smoking are caused by polyaromatic hydrocarbons. The hotspots for these mutations lie in the segment that binds to DNA. The transition hotspots are in segments that regulate apoptosis. Figure 10.17 This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 445 11 | MEIOSIS AND SEXUAL REPRODUCTION Figure 11.1 Each of us, like these other large multicellular organisms, begins life as a fertilized egg. After trillions of cell divisions, each of us develops into a complex, multicellular organism. (credit a: modification of work by Frank Wouters; credit b: modification of work by Ken Cole, USGS; credit c: modification of work by Martin Pettitt) Chapter Outline 11.1: The Process of Meiosis 11.2: Sexual Reproduction Introduction The ability to reproduce “in kind” is a basic characteristic of all living things. “In kind” means that the offspring of an organism closely resembles its parent or parents. Hippopotamuses give birth to hippopotamus calves, Joshua trees produce Joshua tree seedlings, and flamingos lay eggs that hatch into flamingo chicks. In kind can mean exactly the same. Many unicellular organisms, such as yeast, and a few multicellular organisms, such as sponges, can produce genetically identical clones of themselves through cell division. However, many single-ce |
lled organisms and most multicellular organisms reproduce regularly using a method requiring two parents. Sexual reproduction occurs through the production by each parent of a haploid cell (containing one half of an offspring’s required genetic material) and the fusion of these two haploid cells to form a single, unique diploid cell with a complete set of genetic information. In most plants and animals, through multiple rounds of mitotic cell division, this diploid cell will develop into an adult organism. Haploid cells that are necessary for sexual reproduction are produced by a type of cell division called meiosis. Sexual reproduction, specifically meiosis and fertilization, introduces variation into offspring. Variation is an important component of a species evolutionary success. The vast majority of eukaryotic organisms employs some form of meiosis and fertilization to reproduce. Not all sexually reproducing eukaryotes reproduce solely by sexual reproduction. For example, an Asian termite species, Reticulitermes speratus, can reproduce sexually or asexually. In a young colony, a single termite pair—the king and queen—produce worker offspring sexually by the union of haploid cells. However, after several years, as the queen begins to age, she produces some offspring asexually in a process called parthenogenesis. These offspring, which are destined to become new queens, are not fertilized by the king. They are genetic clones of the queen. More information about parthenogenesis in these termites can be found at this article (http://openstaxcollege.org/l/32termitequeen). 446 Chapter 11 | Meiosis and Sexual Reproduction 11.1 | The Process of Meiosis In this section, you will explore the following questions: • How do chromosomes behave during meiosis? • What cellular events occur during meiosis? • What are the similarities and differences between meiosis and mitosis? • How can the process of meiosis generate genetic variation? Connection for AP® Courses As we explored the cell cycle and mitosis in a previous chapter, we learned that cells divide to grow, replace other cells, and reproduce asexually. Without mutation, or changes in the DNA, the daughter cells produced by mitosis receive a set of genetic instructions that is identical to that of the parent cell. Because changes in genes drive both the unity and diversity of life, organisms without genetic variation cannot evolve through natural selection. Evolution occurs only because organisms have developed ways to vary their genetic material. |
This occurs through mutations in DNA, recombination of genes during meiosis, and meiosis followed by fertilization in sexually reproducing organisms. Sexual reproduction requires that diploid (2n) organisms produce haploid (1n) cells through meiosis and that these haploid cells fuse to form new, diploid offspring. The union of these two haploid cells, one from each parent, is fertilization. Although the processes of meiosis and mitosis share similarities, their end products are different. Recall that eukaryotic DNA is contained in chromosomes, and that chromosomes occur in homologous pairs (homologues). At fertilization, the male parent contributes one member of each homologous pair to the offspring, and the female parent contributes the other. With the exception of the sex chromosomes, homologous chromosomes contain the same genes, but these genes can have different variations, called alleles. (For example, you might have inherited an allele for brown eyes from your father and an allele for blue eyes from your mother.) As in mitosis, homologous chromosomes are duplicated during the S-stage (synthesis) of interphase. However, unlike mitosis, in which there is just one nuclear division, meiosis has two complete rounds of nuclear division—meiosis I and meiosis II. These result in four nuclei and (usually) four daughter cells, each with half the number of chromosomes as the parent cell (1n). The first division, meiosis I, separates homologous chromosomes, and the second division, meiosis II, separates chromatids. (Remember: during meiosis, DNA replicates ONCE but divides TWICE, whereas in mitosis, DNA replicates ONCE but divides only ONCE.). Although mitosis and meiosis are similar in many ways, they have different outcomes. The main difference is in the type of cell produced: mitosis produces identical cells, allowing growth or repair of tissues; meiosis generates reproductive cells, or gametes. Gametes, often called sex cells, unite with other sex cells to produce new, unique organisms. Genetic variation occurs during meiosis I, in which homologous chromosomes pair and exchange non-sister chromatid segments (crossover). Here the homologous chromosomes separate into different nuclei, causing a reduction in “ploidy.” During meiosis II—which is more similar to a mitotic division—the chromatids separate and segregate into |
four haploid sex cells. However, because of crossover, the resultant daughter cells do not contain identical genomes. As in mitosis, external factors and internal signals regulate the meiotic cell cycle. As we will explore in more detail in a later chapter, errors in meiosis can cause genetic disorders, such as Down syndrome. Information presented and the examples highlighted in the section support concepts and learning objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The learning objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A learning objective merges required content with one or more of the seven science practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 447 Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 3.9 The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by fertilization. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 7.1 The student can connect phenomena and models across spatial and temporal scales. 3.10 The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 1.9][APLO 2.15][APLO 2.39][APLO 3.11][APLO 3.9] You read that fertilization is the union of two sex cells from two individual organisms. If these two cells each contain one set of chromosomes, the resulting fertilized cell contains two sets of chromosomes. Haploid cells contain one set of |
chromosomes. Cells containing two sets of chromosomes are called diploid. The number of sets of chromosomes in a cell is called its ploidy level. If the reproductive cycle is to continue, a diploid cell must reduce the number of its chromosome sets before fertilization can occur again. Otherwise, the number of chromosome sets would double, and continue to double in every generation. So, in addition to fertilization, sexual reproduction includes a nuclear division that reduces the number of chromosome sets. Most animals and plants are diploid, containing two sets of chromosomes. In an organism’s somatic cells, sometimes referred to as “body” cells (all cells of a multicellular organism except the reproductive cells), the nucleus contains two copies of each chromosome, called homologous chromosomes. Homologous chromosomes are matched pairs containing the same genes in identical locations along their length. Diploid organisms inherit one copy of each homologous chromosome from each parent; all together, they are considered a full set of chromosomes. Haploid cells, containing a single copy of each homologous chromosome, are found only within an organism’’s reproductive structures, such as the ovaries and testes. Haploid cells can be either gametes or spores. Male gametes are sperm and female gametes are eggs. All animals and most plants produce gametes. Spores are haploid cells that can produce a haploid organism or can fuse with another spore to form a diploid cell. Some plants and all fungi produce spores. As you have learned, the nuclear division that forms haploid cells— meiosis—is closely related to mitosis. Mitosis is the part of a cell reproduction cycle that results in identical daughter nuclei that are also genetically identical to the original parent nucleus. In mitosis, both the parent and the daughter nuclei are at the same ploidy level—diploid for most plants and animals. Meiosis employs many of the same mechanisms as mitosis. However, the starting nucleus is always diploid and the nuclei that result at the end of a meiotic cell division are haploid. To achieve this reduction in chromosome number, meiosis consists of one round of chromosome duplication and two rounds of nuclear division. Because the events that occur during each of the division stages are analogous to the events of mitosis, the same stage names are assigned. However, because there are two rounds of division, the major process and |
the stages are designated with a “I” or a “II.” Thus, meiosis I is the first round of meiotic division and consists of prophase I, prometaphase I, and so on. Meiosis II, in which the second round of meiotic division takes place, includes prophase II, prometaphase II, and so on. Meiosis I Meiosis is preceded by an interphase consisting of the G1, S, and G2 phases, which are nearly identical to the phases preceding mitosis. The G1 phase, which is also called the first gap phase, is the first phase of the interphase and is focused on cell growth. The S phase is the second phase of interphase, during which the DNA of the chromosomes is replicated. Finally, the G2 phase, also called the second gap phase, is the third and final phase of interphase; in this phase, the cell undergoes the final preparations for meiosis. During DNA duplication in the S phase, each chromosome is replicated to produce two identical copies, called sister chromatids, that are held together at the centromere by cohesin proteins. Cohesin holds the chromatids together until 448 Chapter 11 | Meiosis and Sexual Reproduction anaphase II. The centrosomes, which are the structures that organize the microtubules of the meiotic spindle, also replicate. This prepares the cell to enter prophase I, the first meiotic phase. Prophase I Early in prophase I, before the chromosomes can be seen clearly microscopically, the homologous chromosomes are attached at their tips to the nuclear envelope by proteins. As the nuclear envelope begins to break down, the proteins associated with homologous chromosomes bring the pair close to each other. Recall that, in mitosis, homologous chromosomes do not pair together. In mitosis, homologous chromosomes line up end-to-end so that when they divide, each daughter cell receives a sister chromatid from both members of the homologous pair. The synaptonemal complex, a lattice of proteins between the homologous chromosomes, first forms at specific locations and then spreads to cover the entire length of the chromosomes. The tight pairing of the homologous chromosomes is called synapsis. In synapsis, the genes on the chromatids of the homologous chromosomes are aligned precisely with each other. The synaptonemal complex |
supports the exchange of chromosomal segments between non-sister homologous chromatids, a process called crossing over. Crossing over can be observed visually after the exchange as chiasmata (singular = chiasma) (Figure 11.2). In species such as humans, even though the X and Y sex chromosomes are not homologous (most of their genes differ), they have a small region of homology that allows the X and Y chromosomes to pair up during prophase I. A partial synaptonemal complex develops only between the regions of homology. Figure 11.2 Early in prophase I, homologous chromosomes come together to form a synapse. The chromosomes are bound tightly together and in perfect alignment by a protein lattice called a synaptonemal complex and by cohesin proteins at the centromere. Located at intervals along the synaptonemal complex are large protein assemblies called recombination nodules. These assemblies mark the points of later chiasmata and mediate the multistep process of crossover—or genetic recombination—between the non-sister chromatids. Near the recombination nodule on each chromatid, the double-stranded DNA is cleaved, the cut ends are modified, and a new connection is made between the non-sister chromatids. As prophase I progresses, the synaptonemal complex begins to break down and the chromosomes begin to condense. When the synaptonemal complex is gone, the homologous chromosomes remain attached to each other at the centromere and at chiasmata. The chiasmata remain until anaphase I. The number of chiasmata varies according to the species and the length of the chromosome. There must be at least one chiasma per chromosome for proper separation of homologous chromosomes during meiosis I, but there may be as many as 25. Following crossover, the synaptonemal complex breaks down and the cohesin connection between homologous pairs is also removed. At the end of prophase I, the pairs are held together only at the chiasmata (Figure 11.3) and are called tetrads because the four sister chromatids of each pair of homologous chromosomes are now visible. The crossover events are the first source of genetic variation in the nuclei produced by meiosis. A single crossover event between homologous non-sister chromatids leads to a reciprocal exchange |
of equivalent DNA between a maternal chromosome and a paternal chromosome. Now, when that sister chromatid is moved into a gamete cell it will carry some DNA from one parent of the individual and some DNA from the other parent. The sister recombinant chromatid has a combination of maternal and paternal genes that did not exist before the crossover. Multiple crossovers in an arm of the chromosome have the same effect, exchanging segments of DNA to create recombinant chromosomes. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 449 Figure 11.3 Crossover occurs between non-sister chromatids of homologous chromosomes. The result is an exchange of genetic material between homologous chromosomes. Prometaphase I The key event in prometaphase I is the attachment of the spindle fiber microtubules to the kinetochore proteins at the centromeres. Kinetochore proteins are multiprotein complexes that bind the centromeres of a chromosome to the microtubules of the mitotic spindle. Microtubules grow from centrosomes placed at opposite poles of the cell. The microtubules move toward the middle of the cell and attach to one of the two fused homologous chromosomes. The microtubules attach at each chromosomes' kinetochores. With each member of the homologous pair attached to opposite poles of the cell, in the next phase, the microtubules can pull the homologous pair apart. A spindle fiber that has attached to a kinetochore is called a kinetochore microtubule. At the end of prometaphase I, each tetrad is attached to microtubules from both poles, with one homologous chromosome facing each pole. The homologous chromosomes are still held together at chiasmata. In addition, the nuclear membrane has broken down entirely. Metaphase I During metaphase I, the homologous chromosomes are arranged in the center of the cell with the kinetochores facing opposite poles. The homologous pairs orient themselves randomly at the equator. For example, if the two homologous members of chromosome 1 are labeled a and b, then the chromosomes could line up a-b, or b-a. This is important in determining the genes carried by a gamete, as each will only receive one of the two homologous chromosomes |
. Recall that homologous chromosomes are not identical. They contain slight differences in their genetic information, causing each gamete to have a unique genetic makeup. This randomness is the physical basis for the creation of the second form of genetic variation in offspring. Consider that the homologous chromosomes of a sexually reproducing organism are originally inherited as two separate sets, one from each parent. Using humans as an example, one set of 23 chromosomes is present in the egg donated by the mother. The father provides the other set of 23 chromosomes in the sperm that fertilizes the egg. Every cell of the multicellular offspring has copies of the original two sets of homologous chromosomes. In prophase I of meiosis, the homologous 450 Chapter 11 | Meiosis and Sexual Reproduction chromosomes form the tetrads. In metaphase I, these pairs line up at the midway point between the two poles of the cell to form the metaphase plate. Because there is an equal chance that a microtubule fiber will encounter a maternally or paternally inherited chromosome, the arrangement of the tetrads at the metaphase plate is random. Any maternally inherited chromosome may face either pole. Any paternally inherited chromosome may also face either pole. The orientation of each tetrad is independent of the orientation of the other 22 tetrads. This event—the random (or independent) assortment of homologous chromosomes at the metaphase plate—is the second mechanism that introduces variation into the gametes or spores. In each cell that undergoes meiosis, the arrangement of the tetrads is different. The number of variations is dependent on the number of chromosomes making up a set. There are two possibilities for orientation at the metaphase plate; the possible number of alignments therefore equals 2n, where n is the number of chromosomes per set. Humans have 23 chromosome pairs, which results in over eight million (223) possible genetically-distinct gametes. This number does not include the variability that was previously created in the sister chromatids by crossover. Given these two mechanisms, it is highly unlikely that any two haploid cells resulting from meiosis will have the same genetic composition (Figure 11.4). To summarize the genetic consequences of meiosis I, the maternal and paternal genes are recombined by crossover events that occur between each homologous pair during prophase I. In addition, the random assortment of tetrads on the metaphase plate produces a unique combination of maternal and |
paternal chromosomes that will make their way into the gametes. Figure 11.4 Random, independent assortment during metaphase I can be demonstrated by considering a cell with a set of two chromosomes (n = 2). In this case, there are two possible arrangements at the equatorial plane in metaphase I. The total possible number of different gametes is 2n, where n equals the number of chromosomes in a set. In this example, there are four possible genetic combinations for the gametes. With n = 23 in human cells, there are over 8 million possible combinations of paternal and maternal chromosomes. Anaphase I In anaphase I, the microtubules pull the linked chromosomes apart. The sister chromatids remain tightly bound together at the centromere. The chiasmata are broken in anaphase I as the microtubules attached to the fused kinetochores pull the homologous chromosomes apart (Figure 11.5). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 451 Telophase I and Cytokinesis In telophase, the separated chromosomes arrive at opposite poles. The remainder of the typical telophase events may or may not occur, depending on the species. In some organisms, the chromosomes decondense and nuclear envelopes form around the chromatids in telophase I. In other organisms, cytokinesis—the physical separation of the cytoplasmic components into two daughter cells—occurs without reformation of the nuclei. In nearly all species of animals and some fungi, cytokinesis separates the cell contents via a cleavage furrow (constriction of the actin ring that leads to cytoplasmic division). In plants, a cell plate is formed during cell cytokinesis by Golgi vesicles fusing at the metaphase plate. This cell plate will ultimately lead to the formation of cell walls that separate the two daughter cells. Two haploid cells are the end result of the first meiotic division. The cells are haploid because at each pole, there is just one of each pair of the homologous chromosomes. Therefore, only one full set of the chromosomes is present. This is why the cells are considered haploid—there is only one chromosome set, even though each homolog still consists of two sister chromatids. Recall that sister chromatids are merely |
duplicates of one of the two homologous chromosomes (except for changes that occurred during crossing over). In meiosis II, these two sister chromatids will separate, creating four haploid daughter cells. Review the process of meiosis, observing how chromosomes align and migrate, at Meiosis: An Interactive Animation (http://openstaxcollege.org/l/animal_meiosis). Human males typically have XY chromosomes and females have XX chromosomes, but there are rare instances in which a male can inherit an XXY or an XYY, or a female can have three X chromosomes. Explain how an error in meiosis can cause these aberrations. a. Errors can arise only during the recombination process which may result in deletions, duplications or translocations causing such abnormalities. b. Aberrations caused when a pair of homologous chromosomes fails to separate during anaphase I or when sister chromatids fail to separate during anaphase II, the daughter cells will inherit unequal numbers of chromosomes. c. Errors during anaphase I of meiosis only cause such aberrations resulting in unequal numbers of chromosomes. d. Errors during meiosis introduce variations in the DNA sequence, which depends specifically on the size of the variant only. Meiosis II In some species, cells enter a brief interphase, or interkinesis, before entering meiosis II. Interkinesis lacks an S phase, so chromosomes are not duplicated. The two cells produced in meiosis I go through the events of meiosis II in synchrony. During meiosis II, the sister chromatids within the two daughter cells separate, forming four new haploid gametes. The mechanics of meiosis II is similar to mitosis, except that each dividing cell has only one set of homologous chromosomes. Therefore, each cell has half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. Prophase II If the chromosomes decondensed in telophase I, they condense again. If nuclear envelopes were formed, they fragment into vesicles. The centrosomes that were duplicated during interkinesis move away from each other toward opposite poles, and new spindles are formed. Prometaphase II The nuclear envelopes are completely broken down, and the spindle is fully formed. Each sister chromatid forms an 452 Chapter 11 | Meiosis and Sexual Reproduction individual kinetochore that attaches to microtub |
ules from opposite poles. Metaphase II The sister chromatids are maximally condensed and aligned at the equator of the cell. Anaphase II The sister chromatids are pulled apart by the kinetochore microtubules and move toward opposite poles. Non-kinetochore microtubules elongate the cell. Figure 11.5 The process of chromosome alignment differs between meiosis I and meiosis II. In prometaphase I, microtubules attach to the fused kinetochores of homologous chromosomes, and the homologous chromosomes are arranged at the midpoint of the cell in metaphase I. In anaphase I, the homologous chromosomes are separated. In prometaphase II, microtubules attach to the kinetochores of sister chromatids, and the sister chromatids are arranged at the midpoint of the cells in metaphase II. In anaphase II, the sister chromatids are separated. Telophase II and Cytokinesis The chromosomes arrive at opposite poles and begin to decondense. Nuclear envelopes form around the chromosomes. Cytokinesis separates the two cells into four unique haploid cells. At this point, the newly formed nuclei are both haploid. The cells produced are genetically unique because of the random assortment of paternal and maternal homologs and because of the recombining of maternal and paternal segments of chromosomes (with their sets of genes) that occurs during crossover. The entire process of meiosis is outlined in Figure 11.6. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 453 Figure 11.6 An animal cell with a diploid number of four (2n = 4) proceeds through the stages of meiosis to form four haploid daughter cells. Comparing Meiosis and Mitosis Mitosis and meiosis are both forms of division of the nucleus in eukaryotic cells. They share some similarities, but also exhibit distinct differences that lead to very different outcomes (Figure 11.7). Mitosis is a single nuclear division that results in two nuclei that are usually partitioned into two new cells. The nuclei resulting from a mitotic division are genetically identical to the original nucleus. They have the same number of sets of chromosomes, one set in the case of haploid cells and two sets in the case of di |
ploid cells. In most plants and all animal species, it is typically diploid cells that undergo mitosis to form new diploid cells. In contrast, meiosis consists of two nuclear divisions resulting in four nuclei that are usually partitioned into four new cells. The nuclei resulting from meiosis are not genetically identical and they contain one chromosome set only. This is half the number of chromosome sets in the original cell, which is diploid. 454 Chapter 11 | Meiosis and Sexual Reproduction The main differences between mitosis and meiosis occur in meiosis I, which is a very different nuclear division than mitosis. In meiosis I, the homologous chromosome pairs become associated with each other, are bound together with the synaptonemal complex, develop chiasmata and undergo crossover between sister chromatids, and line up along the metaphase plate in tetrads with kinetochore fibers from opposite spindle poles attached to each kinetochore of a homolog in a tetrad. All of these events occur only in meiosis I. When the chiasmata resolve and the tetrad is broken up with the homologs moving to one pole or another, the ploidy level—the number of sets of chromosomes in each future nucleus—has been reduced from two to one. For this reason, meiosis I is referred to as a reduction division. There is no such reduction in ploidy level during mitosis. Meiosis II is much more analogous to a mitotic division. In this case, the duplicated chromosomes (only one set of them) line up on the metaphase plate with divided kinetochores attached to kinetochore fibers from opposite poles. During anaphase II, as in mitotic anaphase, the kinetochores divide and one sister chromatid—now referred to as a chromosome—is pulled to one pole while the other sister chromatid is pulled to the other pole. If it were not for the fact that there had been crossover, the two products of each individual meiosis II division would be identical (like in mitosis). Instead, they are different because there has always been at least one crossover per chromosome. Meiosis II is not a reduction division because although there are fewer copies of the genome in the resulting cells, there is still one set of chromosomes, as there was at the end of meiosis I. Figure 11.7 Meiosis and mitosis are both preceded by one |
round of DNA replication; however, meiosis includes two nuclear divisions. The four daughter cells resulting from meiosis are haploid and genetically distinct. The daughter cells resulting from mitosis are diploid and identical to the parent cell. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 455 The Mystery of the Evolution of Meiosis Some characteristics of organisms are so widespread and fundamental that it is sometimes difficult to remember that they evolved like other simpler traits. Meiosis is such an extraordinarily complex series of cellular events that biologists have had trouble hypothesizing and testing how it may have evolved. Although meiosis is inextricably entwined with sexual reproduction and its advantages and disadvantages, it is important to separate the questions of the evolution of meiosis and the evolution of sex, because early meiosis may have been advantageous for different reasons than it is now. Thinking outside the box and imagining what the early benefits from meiosis might have been is one approach to uncovering how it may have evolved. Meiosis and mitosis share obvious cellular processes and it makes sense that meiosis evolved from mitosis. The difficulty lies in the clear differences between meiosis I and mitosis. Adam Wilkins and Robin Holliday[1] summarized the unique events that needed to occur for the evolution of meiosis from mitosis. These steps are homologous chromosome pairing, crossover exchanges, sister chromatids remaining attached during anaphase, and suppression of DNA replication in interphase. They argue that the first step is the hardest and most important, and that understanding how it evolved would make the evolutionary process clearer. They suggest genetic experiments that might shed light on the evolution of synapsis. There are other approaches to understanding the evolution of meiosis in progress. Different forms of meiosis exist in single-celled protists. Some appear to be simpler or more “primitive” forms of meiosis. Comparing the meiotic divisions of different protists may shed light on the evolution of meiosis. Marilee Ramesh and colleagues[2] compared the genes involved in meiosis in protists to understand when and where meiosis might have evolved. Although research is still ongoing, recent scholarship into meiosis in protists suggests that some aspects of meiosis may have evolved later than others. This kind of genetic comparison can tell us what aspects of meiosis are the oldest and what cellular processes they may |
have borrowed from in earlier cells. Which of the following events occurs in both mitosis and meiosis I? a. Homologous chromosomes pair together. b. Crossover occurs between chromosomes. c. Chromosomes line up at the metaphase plate. d. Sister chromatids remain attached during anaphase. 1. Adam S. Wilkins and Robin Holliday, “The Evolution of Meiosis from Mitosis,” Genetics 181 (2009): 3–12. 2. Marilee A. Ramesh, Shehre-Banoo Malik and John M. Logsdon, Jr, “A Phylogenetic Inventory of Meiotic Genes: Evidence for Sex in Giardia and an Early Eukaryotic Origin of Meiosis,” Current Biology 15 (2005):185–91. 456 Chapter 11 | Meiosis and Sexual Reproduction Click through the steps of this interactive animation to compare the meiotic process of cell division to that of mitosis: How Cells Divide (http://openstaxcollege.org/l/how_cells_dvide). Single-celled organisms, like amoebas, reproduce by mitosis. Explain how the genetic makeup of these organisms differs from organisms that undergo meiosis. a. Organisms reproducing through mitosis produce genetically different daughter cells whereas those producing through meiosis have genetically identical daughter cells. b. Crossing over or mixing of chromosomes does not occur in meiosis whereas it is prevalent in mitosis. c. Mitosis is a process of asexual reproduction in which the number of chromosomes are reduced by half producing two haploid cells whereas in meiosis two diploid cells are produced by cell division. d. Organisms producing through mitosis create genetically identical offspring as only a single parent copies its entire genetic material to the offspring. In meiosis, two parents produces gametes and the offspring have only half the number of chromosomes of each parent and hence genetic variation is introduced. Activity Create a series of diagrams with annotations to compare and contrast the processes of mitosis and meiosis in an organism with a haploid number of six. Then, using specific examples, explain how meiosis followed by fertilization increases genetic variation in a family of organisms. 11.2 | Sexual Reproduction In this section, you will explore the following questions: • Why are meiosis and sexual reproduction considered evolved traits? • Why is variation among offspring a potential evolutionary advantage to sexual reproduction? • What are the three different |
life-cycles among sexual multicellular organisms and their commonalities? Connection for AP® Courses Nearly all eukaryotes undergo sexual reproduction. The variation introduced into the reproductive cells (gametes or spores) by meiosis is advantageous for evolution via natural selection. Meiosis and fertilization alternate as the organisms pass through the haploid and diploid stages of their life cycles. In most animals, the diploid stage dominates, whereas in fungi, the haploid stage dominates. Identifying the haploid and diploid stages within the life cycles of different organisms is vital in understanding how organisms reproduce and in determining when mitosis and meiosis occur. Information presented and the examples highlighted in the section support concepts and learning objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The learning objectives listed in the Curriculum Framework provide This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 457 a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven science practices. Big Idea 3 Enduring Understanding 3.C Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. The processing of genetic information is imperfect and is a source of genetic variation. 3.C.2 Biological systems have multiple processes that increase genetic variation. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. 3.27 The student is able to compare and contrast processes by which genetic variation is produced and maintained in organisms from multiple domains. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.7][APLO 3.9][APLO 3.24][APLO 3.28] Sexual reproduction was an early evolutionary innovation after the appearance of eukaryotic cells. It appears to have been very successful because most eukaryotes are able to reproduce sexually, and in many animals, it is the only mode of reproduction. And yet, scientists recognize some real disadvantages to sexual reproduction. On the surface, creating offspring that are genetic clones of the parent appears to be a better system |
. If the parent organism is successfully occupying a habitat, offspring with the same traits would be similarly successful. There is also the obvious benefit to an organism that can produce offspring whenever circumstances are favorable by asexual budding, fragmentation, or asexual eggs. These methods of reproduction do not require another organism of the opposite sex. Indeed, some organisms that lead a solitary lifestyle have retained the ability to reproduce asexually. In addition, in asexual populations, every individual is capable of reproduction. In sexual populations, the males are not producing the offspring themselves, so in theory an asexual population could grow twice as fast. However, multicellular organisms that exclusively depend on asexual reproduction are exceedingly rare. Why is sexuality (and meiosis) so common? This is one of the important unanswered questions in biology and has been the focus of much research beginning in the latter half of the twentieth century. There are several possible explanations, one of which is that the variation that sexual reproduction creates among offspring is very important to the survival and reproduction of the population. Thus, on average, a sexually reproducing population will leave more descendants than an otherwise similar asexually reproducing population. The only source of variation in asexual organisms is mutation. This is the ultimate source of variation in sexual organisms, but in addition, those different mutations are continually reshuffled from one generation to the next when different parents combine their unique genomes and the genes are mixed into different combinations by crossovers during prophase I and random assortment at metaphase I. 458 Chapter 11 | Meiosis and Sexual Reproduction The Red Queen Hypothesis It is not in dispute that sexual reproduction provides evolutionary advantages to organisms that employ this mechanism to produce offspring. But why, even in the face of fairly stable conditions, does sexual reproduction persist when it is more difficult and costly for individual organisms? Variation is the outcome of sexual reproduction, but why are ongoing variations necessary? Enter the Red Queen hypothesis, first proposed by Leigh Van Valen in 1973.[3] The concept was named in reference to the Red Queen's race in Lewis Carroll's book, Through the Looking-Glass. All species co-evolve with other organisms; for example predators evolve with their prey, and parasites evolve with their hosts. Each tiny advantage gained by favorable variation gives a species an edge over close competitors, predators, parasites, or even prey. The only method that will allow a co-evolving species to maintain its own share of the resources is to also continually improve its fitness. As one species gains |
an advantage, this increases selection pressure on the other species; they must also develop an advantage or they will be outcompeted. No single species progresses too far ahead because genetic variation among the progeny of sexual reproduction provides all species with a mechanism to improve rapidly. Species that cannot keep up become extinct. The Red Queen’s catchphrase was, “It takes all the running you can do to stay in the same place.” This is an apt description of co-evolution between competing species. Which of the following scenarios provides the best support for the Red Queen Hypothesis? a. An asexually reproducing plant rapidly populates a hillside left barren by a fire. b. Individuals of a snail population that reproduce asexually die out after a parasite invades its territory. c. A widely dispersed population of ruffed grouse disappears because individuals have difficulty finding mates. d. A sexually-reproducing species of gophers goes extinct after a new predator is introduced. Life Cycles of Sexually Reproducing Organisms Fertilization and meiosis alternate in sexual life cycles. What happens between these two events depends on the organism. The process of meiosis reduces the chromosome number by half. Fertilization, the joining of two haploid gametes, restores the diploid condition. There are three main categories of life cycles in multicellular organisms: diploid-dominant, in which the multicellular diploid stage is the most obvious life stage, such as with most animals including humans; haploiddominant, in which the multicellular haploid stage is the most obvious life stage, such as with all fungi and some algae; and alternation of generations, in which the two stages are apparent to different degrees depending on the group, as with plants and some algae. Diploid-Dominant Life Cycle Nearly all animals employ a diploid-dominant life-cycle strategy in which the only haploid cells produced by the organism are the gametes. Early in the development of the embryo, specialized diploid cells, called germ cells, are produced within the gonads, such as the testes and ovaries. Germ cells are capable of mitosis to perpetuate the cell line and meiosis to produce gametes. Once the haploid gametes are formed, they lose the ability to divide again. There is no multicellular haploid life stage. Fertilization occurs with the fusion |
of two gametes, usually from different individuals, restoring the diploid state (Figure 11.8). 3. Leigh Van Valen, “A New Evolutionary Law,” Evolutionary Theory 1 (1973): 1–30 This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 459 Figure 11.8 In animals, sexually reproducing adults form haploid gametes from diploid germ cells. Fusion of the gametes gives rise to a fertilized egg cell, or zygote. The zygote will undergo multiple rounds of mitosis to produce a multicellular offspring. The germ cells are generated early in the development of the zygote. Haploid-Dominant Life Cycle Most fungi and algae employ a life-cycle type in which the “body” of the organism—the ecologically important part of the life cycle—is haploid. The haploid cells that make up the tissues of the dominant multicellular stage are formed by mitosis. During sexual reproduction, specialized haploid cells from two individuals, designated the (+) and (−) mating types, join to form a diploid zygote. The zygote immediately undergoes meiosis to form four haploid cells called spores. Although haploid like the “parents,” these spores contain a new genetic combination from two parents. The spores can remain dormant for various time periods. Eventually, when conditions are conducive, the spores form multicellular haploid structures by many rounds of mitosis (Figure 11.9). 460 Chapter 11 | Meiosis and Sexual Reproduction Figure 11.9 Fungi, such as black bread mold (Rhizopus nigricans), have haploid-dominant life cycles. The haploid multicellular stage produces specialized haploid cells by mitosis that fuse to form a diploid zygote. The zygote undergoes meiosis to produce haploid spores. Each spore gives rise to a multicellular haploid organism by mitosis. (credit “zygomycota” micrograph: modification of work by “Fanaberka”/Wikimedia Commons) If a mutation occurs so that a fungus is no longer able to produce a minus mating type, will it still be able to reproduce? a. No, sexual mode of reproduction is the only mode of |
reproduction in fungi. b. No, absence of minus mating types will disrupt functions in fungi. c. Yes, it will be able to reproduce asexually by the mitotic divisions of spores. d. Yes, by action of some enzymes, it will be able to reproduce asexually. Alternation of Generations The third life-cycle type, employed by some algae and all plants, is a blend of the haploid-dominant and diploid-dominant extremes. Species with alternation of generations have both haploid and diploid multicellular organisms as part of their life cycle. The haploid multicellular plants are called gametophytes, because they produce gametes from specialized cells. Meiosis is not directly involved in the production of gametes in this case, because the organism that produces the gametes is already a haploid. Fertilization between the gametes forms a diploid zygote. The zygote will undergo many rounds of mitosis and give rise to a diploid multicellular plant called a sporophyte. Specialized cells of the sporophyte will undergo meiosis and produce haploid spores. The spores will subsequently develop into the gametophytes (Figure 11.10). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 461 Figure 11.10 Plants have a life cycle that alternates between a multicellular haploid organism and a multicellular diploid organism. In some plants, such as ferns, both the haploid and diploid plant stages are free-living. The diploid plant is called a sporophyte because it produces haploid spores by meiosis. The spores develop into multicellular, haploid plants called gametophytes because they produce gametes. The gametes of two individuals will fuse to form a diploid zygote that becomes the sporophyte. (credit “fern”: modification of work by Cory Zanker; credit “sporangia”: modification of work by "Obsidian Soul"/Wikimedia Commons; credit “gametophyte and sporophyte”: modification of work by “Vlmastra”/Wikimedia Commons) Although all plants utilize some version of the alternation of generations, the relative |
size of the sporophyte and the gametophyte and the relationship between them vary greatly. In plants such as moss, the gametophyte organism is the free-living plant, and the sporophyte is physically dependent on the gametophyte. In other plants, such as ferns, both the gametophyte and sporophyte plants are free-living; however, the sporophyte is much larger. In seed plants, such as magnolia trees and daisies, the gametophyte is composed of only a few cells and, in the case of the female gametophyte, is completely retained within the sporophyte. Sexual reproduction takes many forms in multicellular organisms. However, at some point in each type of life cycle, meiosis produces haploid cells that will fuse with the haploid cell of another organism. The mechanisms of variation—crossover, random assortment of homologous chromosomes, and random fertilization—are present in all versions of sexual reproduction. The fact that nearly every multicellular organism on Earth employs sexual reproduction is strong evidence for the benefits of producing offspring with unique gene combinations, though there are other possible benefits as well. Think About It Compare and contrast the three main types of life cycles in multicellular organisms and give an example of an organism that employs each. 462 Chapter 11 | Meiosis and Sexual Reproduction KEY TERMS alternation of generations life-cycle type in which the diploid and haploid stages alternate chiasmata (singular, chiasma) the structure that forms at the crossover points after genetic material is exchanged cohesin proteins that form a complex that seals sister chromatids together at their centromeres until anaphase II of meiosis crossover exchange of genetic material between non-sister chromatids resulting in chromosomes that incorporate genes from both parents of the organism diploid-dominant life-cycle type in which the multicellular diploid stage is prevalent fertilization union of two haploid cells from two individual organisms gametophyte a multicellular haploid life-cycle stage that produces gametes germ cells specialized cell line that produces gametes, such as eggs or sperm haploid-dominant life-cycle type in which the multicellular haploid stage is prevalent interkinesis (also, interphase II) brief period of rest between meiosis I and meiosis II life cycle the sequence of events in the development of an organism and the |
production of cells that produce offspring meiosis a nuclear division process that results in four haploid cells meiosis I first round of meiotic cell division; referred to as reduction division because the ploidy level is reduced from diploid to haploid meiosis II second round of meiotic cell division following meiosis I; sister chromatids are separated into individual chromosomes, and the result is four unique haploid cells recombination nodules protein assemblies formed on the synaptonemal complex that mark the points of crossover events and mediate the multistep process of genetic recombination between non-sister chromatids reduction division nuclear division that produces daughter nuclei each having one-half as many chromosome sets as the parental nucleus; meiosis I is a reduction division somatic cell all the cells of a multicellular organism except the gametes or reproductive cells spore haploid cell that can produce a haploid multicellular organism or can fuse with another spore to form a diploid cell sporophyte a multicellular diploid life-cycle stage that produces haploid spores by meiosis synapsis formation of a close association between homologous chromosomes during prophase I synaptonemal complex protein lattice that forms between homologous chromosomes during prophase I, supporting crossover tetrad two duplicated homologous chromosomes (four chromatids) bound together by chiasmata during prophase I CHAPTER SUMMARY 11.1 The Process of Meiosis Sexual reproduction requires that diploid organisms produce haploid cells that can fuse during fertilization to form diploid offspring. As with mitosis, DNA replication occurs prior to meiosis during the S-phase of the cell cycle. Meiosis is a series of events that arrange and separate chromosomes and chromatids into daughter cells. During the interphases of meiosis, each chromosome is duplicated. In meiosis, there are two rounds of nuclear division resulting in four nuclei and usually four daughter cells, each with half the number of chromosomes as the parent cell. The first separates homologs, and the second—like mitosis—separates chromatids into individual chromosomes. During meiosis, variation in the daughter This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 463 nuclei is introduced because of crossover in prophase I and random alignment of tetrads at metaph |
ase I. The cells that are produced by meiosis are genetically unique. Meiosis and mitosis share similarities, but have distinct outcomes. Mitotic divisions are single nuclear divisions that produce daughter nuclei that are genetically identical and have the same number of chromosome sets as the original cell. Meiotic divisions include two nuclear divisions that produce four daughter nuclei that are genetically different and have one chromosome set instead of the two sets of chromosomes in the parent cell. The main differences between the processes occur in the first division of meiosis, in which homologous chromosomes are paired and exchange non-sister chromatid segments. The homologous chromosomes separate into different nuclei during meiosis I, causing a reduction of ploidy level in the first division. The second division of meiosis is more similar to a mitotic division, except that the daughter cells do not contain identical genomes because of crossover. 11.2 Sexual Reproduction Nearly all eukaryotes undergo sexual reproduction. The variation introduced into the reproductive cells by meiosis appears to be one of the advantages of sexual reproduction that has made it so successful. Meiosis and fertilization alternate in sexual life cycles. The process of meiosis produces unique reproductive cells called gametes, which have half the number of chromosomes as the parent cell. Fertilization, the fusion of haploid gametes from two individuals, restores the diploid condition. Thus, sexually reproducing organisms alternate between haploid and diploid stages. However, the ways in which reproductive cells are produced and the timing between meiosis and fertilization vary greatly. There are three main categories of life cycles: diploid-dominant, demonstrated by most animals; haploid-dominant, demonstrated by all fungi and some algae; and the alternation of generations, demonstrated by plants and some algae. REVIEW QUESTIONS 1. How many and what type of daughter cells does meiosis produce? a. b. c. d. four haploid four diploid two haploid two diploid 2. What structure is most important in forming the tetrads? a. centromere b. chiasmata c. kinetochore d. Synaptonemal complex 3. At which stage of meiosis are sister chromatids separated from each other? a. anaphase I b. anaphase II c. prophase I d. prophase II 4. At metaphase I, homologous chromosomes are connected only at what structures? a |
. chiasmata b. kinetochores c. microtubules d. recombination nodules 5. What phase(s) of mitotic interphase is missing from meiotic interkinesis? a. G0 phase b. G1 phase c. G2 phase d. S-phase 6. What part of meiosis is most similar to mitosis? a. b. reduction division interkinesis c. meiosis I d. meiosis II 7. Which of the following is not true during crossing over? a. Chiasmata are formed. b. Non-sister chromatids exchange genetic material. c. Recombination nodules mediate cross over events. d. Spindle microtubules guide the movement of chromosomal material. 8. During which phase does the second round of genetic variation occur during meiosis? a. anaphase I b. metaphase I c. prophase II d. Genetic variation only occurs during prophase I. 9. Which type of life cycle has both a haploid and a 464 Chapter 11 | Meiosis and Sexual Reproduction diploid multicellular stage? a. alternation of generations b. asexual c. diploid-dominant d. haploid-dominant 12. What is a disadvantage of sexual reproduction over asexual forms of reproduction? a. Half the population is capable of carrying offspring. b. Identical offspring are not produced. c. Adaptation to rapidly changing environments is 10. What is a source of variation in asexual reproduction? more difficult. a. crossing over of chromosomes b. mutation of DNA d. Mutation rates are slower. 13. Fungi typically display which type of life cycle? c. random assortment of chromosomes a. alternation of generations d. There is no variation in asexual reproduction. 11. What is a likely evolutionary advantage of sexual reproduction over asexual reproduction? b. asexual c. diploid-dominant d. haploid-dominant 14. What is a haploid cell produced in a diploid-dominant organism by meiosis called? a. gamete b. gametophyte c. d. spore sporophyte 16. Which of the following distinguishes metaphase I from metaphase II? a. Sexual reproduction involves fewer steps. b. Sexual reproduction results in variation in the offspring. c. Sexual reproduction is more metabolically efficient. d. Sexual reproduction uses up fewer resources in a given environment. |
CRITICAL THINKING QUESTIONS 15. Describe what happens to the tetrads after they form. a. Prophase I of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is equal chance of a microtubule fiber to encounter a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. b. Prophase II of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is equal chance of microtubule fiber to encounter maternally or paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. c. Prophase I of mitosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is equal chance of a microtubule fiber to encounter a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. d. Prophase I of meiosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is a chance of microtubule fiber to encounter maternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 11 | Meiosis and Sexual Reproduction 465 a. Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is introduced in meiosis, but not in mitosis. b. Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is reduced in meiosis, but not in mitosis. c. Metaphase and telophase portions of meiosis and mitosis are the same. Meiosis and mitosis are also the same, except for the number of chromosomes. Anaphase I and anaphase are different. d. Prophase and telophase portions of meiosis and mitosis are the same. Meiosis II and mitosis are also the same and have the same number of chromosomes. Anaphase I and anaphase are different |
. 18. Explain how the orientation of homologous chromosomes during metaphase I of meiosis contributes to greater variation in gametes. a. The random alignment of homologous chromosomes at the metaphase plate ensures the random destination of the chromosomes in the daughter cells. b. Because homologous chromosomes dissociate from the spindle fibers during metaphase I, they move randomly to the daughter cells. c. The homologous chromosomes are paired tightly during metaphase I and undergo crossover as the synaptonemal complex forms a lattice around them. d. Recombination of maternal and paternal chromosomes occurs in metaphase I because the homologous chromosomes are not connected at their centromeres. 19. Explain how the Red Queen’s catchphrase, “It takes all the running you can do to stay in the same place,” describes co-evolution between competing species. a. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. A Pair of chromosomes is pulled apart and migrate towards pole in anaphase I, while in anaphase II sister chromatids separate. Telophase I reconstitutes the nucleus and loosen the chromosomes, while telophase II mimics telophase I. b. Prophase I condenses the chromosomes and eliminates the nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. Pairs of chromosomes are pulled apart and migrate towards the poles during anaphase I, while in anaphase II sister chromatids separate. Telophase I reconstitutes the nucleus and condenses the chromosomes, while telophase II mimics telophase I. c. Prophase I condense the chromosomes and add nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. Pair of chromosomes are pulled apart and migrate towards the poles in anaphase I, while in anaphase II sister chromatids separate. Telophase I reconstitutes the nucleus and loosens the chromosomes, while |
telophase II mimics telophase I. d. Prophase I condenses the chromosomes and eliminates the nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. During Metaphase II, the chromosomes line up in a double line across the spindle. Each pair of chromosomes is pulled apart and migrate towards the poles in anaphase I, while in anaphase II sister chromatids separate. Telophase I reconstitutes the nucleus and loosen the chromosomes, while telophase II mimics telophase I. 17. Though the stages of meiosis have the same names as the stages of mitosis, they exhibit fundamental differences. What are the main differences between the two processes? 466 Chapter 11 | Meiosis and Sexual Reproduction a. When a sexually reproducing species and an asexually reproducing species compete for the same resources, they both “run [evolve] in the same place” because the increased genetic variation in the sexually reproducing species balances the loss in energy it uses to find and attract mates. b. When one species gains an advantage with a favorable variation, selection increases on another species with which it competes. This species must also develop an advantage or it will be outcompeted. The two species “run [evolve] to stay in the same place.” c. When one species develops a mutation that decreases its ability to survive, a competing species will become better able to survive even though it has not changed in any way. In effect, this species “runs [evolves] to stay in the same place.” d. When two asexually reproducing species encounter rapid environmental change, the species that is also able to reproduce sexually will outcompete the other. This way it can “run [evolve] to stay in the same place.” 20. Which three processes lead to variation among offspring that have the same two parents? a. genetic recombination, fertilization, meiosis b. crossing over, random chromosome assortment, genetic recombination c. meiosis, crossing over, genetic recombination d. fertilization, crossing over, random chromosome assortment TEST PREP FOR AP® COURSES 22. Reproductive cells in most species are different from the cells that make up the rest of the organism. What are the “body” |
cells called and how are they different from the reproductive cells? a. Body cells are called gametes and they have half the number of chromosomes found in reproductive cells. b. Body cells are called somatic cells and have the same number of chromosomes as reproductive cells. c. Body cells are called somatic cells and have double the number of chromosomes found in reproductive cells. d. Body cells are called gametes and have double the number of chromosomes found in reproductive cells. 23. Spores are structures produced by some plants and all fungi. Which is true about them? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 21. Compare the three main types of life cycles in multicellular organisms and give an example of an organism that employs each. a. b. c. d. In a diploid dominant cycle, the multicellular diploid stage is present, as in humans. Haploid dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, both haploid dominant and diploid dominant stages alternate, as in plants. In a diploid dominant cycle, the unicellular diploid stage is present, as in humans. In a haploid dominant life cycle, a unicellular haploid stage is present, as in fungi. In alternation of generations both haploid dominant and diploid dominant stages alternate, as in plants. In a diploid dominant cycle, a multicellular haploid stage is present, as in humans. In a haploid dominant life cycle, a multicellular diploid stage is present, as in fungi. In alternation of generations, both haploid dominant and diploid dominant stages alternate, as in plants. In a diploid dominant cycle, a multicellular diploid stage is present, as in algae. In a haploid dominant life cycle, a multicellular haploid stage is present, as in plants. In alternation of generations, both haploid dominant and diploid dominant stages alternate, as in fungi. a. Spores are haploid reproductive cells that can produce haploid organisms through mitosis. b. Spores are haploid precursors to gametes that give rise to gametes when environmental conditions are favorable. c. Spores are haploid reproductive cells that can produce diploid cells without fertilization. d. Spores |
are haploid cells formed only during asexual reproduction and so are not formed by meiosis. 24. In prophase I, the homologous chromosomes are paired up and linked together. What binds the chromosomes together and maintains their alignment? a. cohesin proteins b. c. d. tetrads the centromere synaptonemal complex Chapter 11 | Meiosis and Sexual Reproduction 467 25. One of the ways that sexual reproduction enhances the diversity of offspring from the same parents is through a process called crossing over. What entities does this occur between during prophase I? a. alternation of generations b. diploid-dominant c. haploid-dominant a. b. sister chromatids tetrads c. non-homologous chromosomes d. non-sister chromatids of homologous chromosomes 26. There are three sources of genetic variation in sexual reproduction. Which is not considered random? a. All are random. b. Crossing over c. Egg and sperm fertilization d. Tetrad alignment on the meiotic spindle. 27. Which one of the three types of life cycles of sexually reproducing organisms does not have a multicellular haploid stage? d. They all have a multicellular haploid stage in their life cycles. 28. How are spores produced in haploid-dominant and alternation of generation life cycles? a. by gametophytes b. by germ cells c. d. through mitosis through meiosis 29. What is one thing that is true of haploid-dominant life cycles but not of alternation of generation life cycles? a. meiosis b. c. (+) and (−) mating types spores d. a free-living haploid stage SCIENCE PRACTICE CHALLENGE QUESTIONS 30. Meiosis involves processes that are common to all eukaryotes, involving the same or similar genes. Evaluate the support for the theory of evolution provided by this evidence and, additionally, by the absence of any alternative process. 31. Meiotic phases of yeast cells were observed microscopically with fluorescent markers (Nachman et al., Cell, 131(3), 2007) to determine the time intervals of meiosis I and meiosis II. The data are displayed in the following figure: Figure 11.11 The duration of meiosis I is measured relative to the transfer of spores to the growth medium. The duration of meiosis II is measured relative to the emergence from me |
iosis I. On the y-axis, the fraction of cells observed to enter each phase are shown, where the sampling has been made in increments of 0.5 hours. A. Qualitatively compare the mean and standard deviation for these two distributions. B. The gene Ime1 is transcribed at the start of meiosis I in response to nitrogen starvation. This activates Ime2 that interacts with Ime1. If, during meiosis I, the cells are supplied with nitrogen, meiosis is halted. Based on these data, justify the claim that this interaction provides a negative feedback loop. C. Explain the advantage provided to the population and the risk to individual cells of the timing of meiosis displayed in the graph above. 32. Construct an explanation as to how DNA is transmitted to the next generation via meiosis followed by fertilization. 33. In eukaryotes, sexual reproduction involves the recombination of heritable information from both parents via meiosis followed by fertilization. Meiosis reduces the number of chromosomes from diploid (2n) to haploid (1n) during the production of gametes. Meiosis begins with the duplication of DNA, producing four strands of DNA in two pairs of homologous chromosomes: 2(2n) becomes 4(n), that is, four haploid cells, where n is the number of strands of DNA in a chromosome. A. Construct an explanation of the importance of random, independent assortment to genetic variation by creating a diagram that represents homologous chromosomes during prophase I without crossover and the possible arrangements of these chromosomes during metaphase I: • without recombination during prophase I 468 Chapter 11 | Meiosis and Sexual Reproduction • with recombination involving two chiasmata B. An alternative would be to bypass the initial duplication of DNA: 2n becomes 2(n), that is, a diploid cell becomes two haploid cells. Predict the effect that this would have on genetic variation. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 469 12 | MENDEL'S EXPERIMENTS AND HEREDITY Figure 12.1 Experimenting with thousands of garden peas, Mendel uncovered the fundamentals of genetics. (credit: modification of work by Jerry Kirkhart) Chapter Outline 12.1: Mendel’s Experiments and the Laws of Probability 12. |
2: Characteristics and Traits 12.3: Laws of Inheritance Introduction During the 19th century, long before chromosomes or genes had been identified, Johann Gregor Mendel set the framework for genetics by studying a simple biological system, the garden pea. He conducted methodical, quantitative analyses using large sample sizes. Mendel’s work laid the foundation for the fundamental principles of heredity. We now know that genes, carried on chromosomes, are the basic functional units of heredity with the capacity to be replicated, expressed, repressed, modified and mutated. Today, the postulates put forth by Mendel form the basis of classical, or Mendelian, genetics. Genes do not all obey the tenets of Mendelian genetics, but Mendel’s experiments serve as an excellent starting point for thinking about inheritance. An understanding of genetic inheritance enables scientists to study and explain complex phenomena. For example, scientists studied the remains of 84 ancient dogs from North and South America. They found that some of the dogs had greater genetic diversity, indicating that these dogs might have interbred with American wolves. Other dogs in their sample had low diversity, indicating that ancient humans were purposely breeding dogs. The study also found that dogs migrated to the Americas with humans only about 10,000 years ago. You can read more about this fascinating story here (http://openstaxcollege.org/l/32dogs). 470 Chapter 12 | Mendel's Experiments and Heredity 12.1 | Mendel’s Experiments and the Laws of Probability In this section, you will explore the following questions: • Why was Mendel’s experimental work so successful? • How do the sum and product rules of probability predict the outcomes of monohybrid crosses involving dominant and recessive alleles? Connection for AP® Courses Genetics is the science of heredity. Austrian monk Gregor Mendel set the framework for genetics long before chromosomes or genes had been identified, at a time when meiosis was not well understood. Working with garden peas, Mendel found that crosses between true-breading parents (P) that differed in one trait (e.g., color: green peas versus yellow peas) produced first generation (F1) offspring that all expressed the trait of one parent (e.g., all green or all yellow). Mendel used the term dominant to refer to the trait that was observed, and recessive to denote that non-expressed trait, or |
the trait that had “disappeared” in this first generation. When the F1 offspring were crossed with each other, the F2 offspring exhibited both traits in a 3:1 ratio. Other crosses (e.g., height: tall plants versus short plants) generated the same 3:1 ratio (in this example, tall to short) in the F2 offspring. By mathematically examining sample sizes, Mendel showed that genetic crosses behaved according to the laws of probability, and that the traits were inherited as independent events. In other words, Mendel used statistical methods to build his model of inheritance. As you have likely noticed, the AP Biology course emphasizes the application of mathematics. Two rules of probability can be used to find the expected proportions of different traits in offspring from different crosses. To find the probability of two or more independent events (events where the outcome of one event has no influence on the outcome of the other event) occurring together, apply the product rule and multiply the probabilities of the individual events. To find the probability that one of two or more events occur, apply the sum rule and add their probabilities together. The content presented in this section supports the learning objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The AP® learning objectives merge essential knowledge content with one or more of the seven science practices. These objectives provide a transparent foundation for the AP® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP® exam questions. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.3 The chromosomal basis of inheritance proposed by Mendel provides an understanding of the pattern of passage of genes from parent to offspring. Science Practice Learning Objective 3.1 The student can pose scientific questions. 3.13 The student is able to pose questions about ethical, social, or medical issues surrounding human genetic disorders. Essential Knowledge 3.A.3 The chromosomal basis of inheritance proposed by Mendel provides an understanding of the pattern of passage of genes from parent to offspring. Science Practice Learning Objective 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter |
12 | Mendel's Experiments and Heredity 471 Figure 12.2 Johann Gregor Mendel is considered the father of genetics. Johann Gregor Mendel (1822–1884) (Figure 12.2) was a lifelong learner, teacher, scientist, and man of faith. As a young adult, he joined the Augustinian Abbey of St. Thomas in Brno in what is now the Czech Republic. Supported by the monastery, he taught physics, botany, and natural science courses at the secondary and university levels. In 1856, he began a decade-long research pursuit involving inheritance patterns in honeybees and plants, ultimately settling on pea plants as his primary model system (a system with convenient characteristics used to study a specific biological phenomenon to be applied to other systems). In 1865, Mendel presented the results of his experiments with nearly 30,000 pea plants to the local Natural History Society. He demonstrated that traits are transmitted faithfully from parents to offspring independently [1] of other traits and in dominant and recessive patterns. In 1866, he published his work, Experiments in Plant Hybridization, in the proceedings of the Natural History Society of Brünn. Mendel’s work went virtually unnoticed by the scientific community that believed, incorrectly, that the process of inheritance involved a blending of parental traits that produced an intermediate physical appearance in offspring; this hypothetical process appeared to be correct because of what we know now as continuous variation. Continuous variation results from the action of many genes to determine a characteristic like human height. Offspring appear to be a “blend” of their parents’ traits when we look at characteristics that exhibit continuous variation. The blending theory of inheritance asserted that the original parental traits were lost or absorbed by the blending in the offspring, but we now know that this is not the case. Mendel was the first researcher to see it. Instead of continuous characteristics, Mendel worked with traits that were inherited in distinct classes (specifically, violet versus white flowers); this is referred to as discontinuous variation. Mendel’s choice of these kinds of traits allowed him to see experimentally that the traits were not blended in the offspring, nor were they absorbed, but rather that they kept their distinctness and could be passed on. In 1868, Mendel became abbot of the monastery and exchanged his scientific pursuits for his pastoral duties. He was not recognized for his extraordinary scientific contributions during his lifetime. In fact, it was not until 1900 that |
his work was rediscovered, reproduced, and revitalized by scientists on the brink of discovering the chromosomal basis of heredity. Mendel’s Model System Mendel’s seminal work was accomplished using the garden pea, Pisum sativum, to study inheritance. This species naturally self-fertilizes, such that pollen encounters ova within individual flowers. The flower petals remain sealed tightly until after pollination, preventing pollination from other plants. The result is highly inbred, or “true-breeding,” pea plants. These are plants that always produce offspring that look like the parent. By experimenting with true-breeding pea plants, Mendel avoided the appearance of unexpected traits in offspring that might occur if the plants were not true breeding. The garden pea also grows to maturity within one season, meaning that several generations could be evaluated over a relatively short time. Finally, large quantities of garden peas could be cultivated simultaneously, allowing Mendel to conclude that his results did not come about simply by chance. 1. Johann Gregor Mendel, Versuche über Pflanzenhybriden Verhandlungen des naturforschenden Vereines in Brünn, Bd. IV für das Jahr, 1865 Abhandlungen, 3–47. [go here for the English translation here (http://openstaxcollege.org/l/32mendelPlain) ] 472 Chapter 12 | Mendel's Experiments and Heredity Mendelian Crosses Mendel performed hybridizations, which involve mating two true-breeding individuals that have different traits. In the pea, which is naturally self-pollinating, this is done by manually transferring pollen from the anther of a mature pea plant of one variety to the stigma of a separate mature pea plant of the second variety. In plants, pollen carries the male gametes (sperm) to the stigma, a sticky organ that traps pollen and allows the sperm to move down the pistil to the female gametes (ova) below. To prevent the pea plant that was receiving pollen from self-fertilizing and confounding his results, Mendel painstakingly removed all of the anthers from the plant’s flowers before they had a chance to mature. Plants used in first-generation crosses were called P0, or parental generation one, plants (Figure 12.3). Mendel |
collected the seeds belonging to the P0 plants that resulted from each cross and grew them the following season. These offspring were called the F1, or the first filial (filial = offspring, daughter or son), generation. Once Mendel examined the characteristics in the F1 generation of plants, he allowed them to self-fertilize naturally. He then collected and grew the seeds from the F1 plants to produce the F2, or second filial, generation. Mendel’s experiments extended beyond the F2 generation to the F3 and F4 generations, and so on, but it was the ratio of characteristics in the P0−F1−F2 generations that were the most intriguing and became the basis for Mendel’s postulates. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 473 Figure 12.3 In one of his experiments on inheritance patterns, Mendel crossed plants that were true-breeding for violet flower color with plants tru<|endoftext|>e-breeding for white flower color (the P generation). The resulting hybrids in the F1 generation all had violet flowers. In the F2 generation, approximately three quarters of the plants had violet flowers, and one quarter had white flowers. Garden Pea Characteristics Revealed the Basics of Heredity In his 1865 publication, Mendel reported the results of his crosses involving seven different characteristics, each with two contrasting traits. A trait is defined as a variation in the physical appearance of a heritable characteristic. The characteristics included plant height, seed texture, seed color, flower color, pea pod size, pea pod color, and flower position. For the characteristic of flower color, for example, the two contrasting traits were white versus violet. To fully examine each characteristic, Mendel generated large numbers of F1 and F2 plants, reporting results from 19,959 F2 plants alone. His findings were consistent. What results did Mendel find in his crosses for flower color? First, Mendel confirmed that he had plants that bred true for white or violet flower color. Regardless of how many generations Mendel examined, all self-crossed offspring of parents with white flowers had white flowers, and all self-crossed offspring of parents with violet flowers had violet flowers. In addition, Mendel confirmed that, other than flower color, the pea plants |
were physically identical. 474 Chapter 12 | Mendel's Experiments and Heredity Once these validations were complete, Mendel applied the pollen from a plant with violet flowers to the stigma of a plant with white flowers. After gathering and sowing the seeds that resulted from this cross, Mendel found that 100 percent of the F1 hybrid generation had violet flowers. Conventional wisdom at that time would have predicted the hybrid flowers to be pale violet or for hybrid plants to have equal numbers of white and violet flowers. In other words, the contrasting parental traits were expected to blend in the offspring. Instead, Mendel’s results demonstrated that the white flower trait in the F1 generation had completely disappeared. Importantly, Mendel did not stop his experimentation there. He allowed the F1 plants to self-fertilize and found that, of F2-generation plants, 705 had violet flowers and 224 had white flowers. This was a ratio of 3.15 violet flowers per one white flower, or approximately 3:1. When Mendel transferred pollen from a plant with violet flowers to the stigma of a plant with white flowers and vice versa, he obtained about the same ratio regardless of which parent, male or female, contributed which trait. This is called a reciprocal cross—a paired cross in which the respective traits of the male and female in one cross become the respective traits of the female and male in the other cross. For the other six characteristics Mendel examined, the F1 and F2 generations behaved in the same way as they had for flower color. One of the two traits would disappear completely from the F1 generation only to reappear in the F2 generation at a ratio of approximately 3:1 (Table 12.1). The Results of Mendel’s Garden Pea Hybridizations Characteristic Contrasting P0 Traits F1 Offspring Traits F2 Offspring Traits F2 Trait Ratios Flower color Violet vs. white 100 percent violet Flower position Axial vs. terminal 100 percent axial Plant height Tall vs. dwarf 100 percent tall Seed texture Round vs. wrinkled 100 percent round Seed color Yellow vs. green 100 percent yellow Pea pod texture Inflated vs. constricted 100 percent inflated Pea pod color Green vs. yellow 100 percent green Table 12.1 705 violet 224 white 651 axial 207 terminal 787 tall 277 dwarf 5,474 round 1,850 wrinkled 6,022 yellow 2,001 green 882 inflated 299 constricted |
428 green 152 yellow 3.15:1 3.14:1 2.84:1 2.96:1 3.01:1 2.95:1 2.82:1 Upon compiling his results for many thousands of plants, Mendel concluded that the characteristics could be divided into expressed and latent traits. He called these, respectively, dominant and recessive traits. Dominant traits are those that are inherited unchanged in a hybridization. Recessive traits become latent, or disappear, in the offspring of a hybridization. The recessive trait does, however, reappear in the progeny of the hybrid offspring. An example of a dominant trait is the violet-flower trait. For this same characteristic (flower color), white-colored flowers are a recessive trait. The fact that the recessive trait reappeared in the F2 generation meant that the traits remained separate (not blended) in the plants of the F1 generation. Mendel also proposed that plants possessed two copies of the trait for the flower-color characteristic, and that each parent transmitted one of its two copies to its offspring, where they came together. Moreover, the physical observation of a dominant trait could mean that the genetic composition of the organism included two dominant versions of the characteristic or that it included one dominant and one recessive version. Conversely, the observation of a recessive trait meant that the organism lacked any dominant versions of this characteristic. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 475 So why did Mendel repeatedly obtain 3:1 ratios in his crosses? To understand how Mendel deduced the basic mechanisms of inheritance that lead to such ratios, we must first review the laws of probability. Think About It Students are performing a cross involving seed color in garden pea plants. Yellow seed color is dominant to green seed color. What F1 offspring would be expected when cross true-breeding plants with green seeds with true-breading plants with yellow seeds? Express the answer(s) as percentage. Probability Basics Probabilities are mathematical measures of likelihood. The empirical probability of an event is calculated by dividing the number of times the event occurs by the total number of opportunities for the event to occur. It is also possible to calculate theoretical probabilities by dividing the number of times that an event is expected to occur by the number of times that it could occur. Empirical probabilities come from observations, like |
those of Mendel. Theoretical probabilities come from knowing how the events are produced and assuming that the probabilities of individual outcomes are equal. A probability of one for some event indicates that it is guaranteed to occur, whereas a probability of zero indicates that it is guaranteed not to occur. An example of a genetic event is a round seed produced by a pea plant. In his experiment, Mendel demonstrated that the probability of the event “round seed” occurring was one in the F1 offspring of true-breeding parents, one of which has round seeds and one of which has wrinkled seeds. When the F1 plants were subsequently self-crossed, the probability of any given F2 offspring having round seeds was now three out of four. In other words, in a large population of F2 offspring chosen at random, 75 percent were expected to have round seeds, whereas 25 percent were expected to have wrinkled seeds. Using large numbers of crosses, Mendel was able to calculate probabilities and use these to predict the outcomes of other crosses. The Product Rule and Sum Rule Mendel demonstrated that the pea-plant characteristics he studied were transmitted as discrete units from parent to offspring. As will be discussed, Mendel also determined that different characteristics, like seed color and seed texture, were transmitted independently of one another and could be considered in separate probability analyses. For instance, performing a cross between a plant with green, wrinkled seeds and a plant with yellow, round seeds still produced offspring that had a 3:1 ratio of green:yellow seeds (ignoring seed texture) and a 3:1 ratio of round:wrinkled seeds (ignoring seed color). The characteristics of color and texture did not influence each other. The product rule of probability can be applied to this phenomenon of the independent transmission of characteristics. The product rule states that the probability of two independent events occurring together can be calculated by multiplying the individual probabilities of each event occurring alone. To demonstrate the product rule, imagine that you are rolling a sixsided die (D) and flipping a penny (P) at the same time. The die may roll any number from 1–6 (D#), whereas the penny may turn up heads (PH) or tails (PT). The outcome of rolling the die has no effect on the outcome of flipping the penny and vice versa. There are 12 possible outcomes of this action (Table 12.2), and each event is expected to occur with equal probability. Twelve Equally Likely Outcomes of Rolling a Die and Fl |
ipping a Penny Rolling Die Flipping Penny D1 D1 D2 D2 D3 Table 12.2 PH PT PH PT PH 476 Chapter 12 | Mendel's Experiments and Heredity Twelve Equally Likely Outcomes of Rolling a Die and Flipping a Penny Rolling Die Flipping Penny D3 D4 D4 D5 D5 D6 D6 Table 12.2 PT PH PT PH PT PH PT Of the 12 possible outcomes, the die has a 2/12 (or 1/6) probability of rolling a two, and the penny has a 6/12 (or 1/2) probability of coming up heads. By the product rule, the probability that you will obtain the combined outcome 2 and heads is: (D2) x (PH) = (1/6) x (1/2) or 1/12 (Table 12.3). Notice the word “and” in the description of the probability. The “and” is a signal to apply the product rule. For example, consider how the product rule is applied to the dihybrid cross: the probability of having both dominant traits in the F2 progeny is the product of the probabilities of having the dominant trait for each characteristic, as shown here: 3 4 × 3 4 = 9 16 On the other hand, the sum rule of probability is applied when considering two mutually exclusive outcomes that can come about by more than one pathway. The sum rule states that the probability of the occurrence of one event or the other event, of two mutually exclusive events, is the sum of their individual probabilities. Notice the word “or” in the description of the probability. The “or” indicates that you should apply the sum rule. In this case, let’s imagine you are flipping a penny (P) and a quarter (Q). What is the probability of one coin coming up heads and one coin coming up tails? This outcome can be achieved by two cases: the penny may be heads (PH) and the quarter may be tails (QT), or the quarter may be heads (QH) and the penny may be tails (PT). Either case fulfills the outcome. By the sum rule, we calculate the probability of obtaining one head and one tail as [(PH) × (QT)] + [(QH) × (PT)] = [(1/2) × (1/2)] + [(1/2) × (1/2)] = 1/ |
2 (Table 12.3). You should also notice that we used the product rule to calculate the probability of PH and QT, and also the probability of PT and QH, before we summed them. Again, the sum rule can be applied to show the probability of having just one dominant trait in the F2 generation of a dihybrid cross: 3 16 + 3 4 = 15 16 The Product Rule and Sum Rule Product Rule Sum Rule For independent events A and B, the probability (P) of them both occurring (A and B) is (PA × PB) For mutually exclusive events A and B, the probability (P) that at least one occurs (A or B) is (PA + PB) Table 12.3 To use probability laws in practice, it is necessary to work with large sample sizes because small sample sizes are prone to deviations caused by chance. The large quantities of pea plants that Mendel examined allowed him calculate the probabilities of the traits appearing in his F2 generation. As you will learn, this discovery meant that when parental traits were known, the offspring’s traits could be predicted accurately even before fertilization. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 477 12.2 | Characteristics and Traits In this section, you will explore the following questions: • What is the relationship between genotypes and phenotypes in dominant and recessive gene systems? • How can a Punnett square be used to calculate expected proportions of genotypes and phenotypes in a monohybrid cross? • How do phenomena such as incomplete dominance, codominance, recessive lethals, multiple alleles, and sex linkage explain deviations from Mendel’s model of inheritance? Connection for AP® Courses The characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits (e.g., green peas versus yellow peas). As we will explore in more detail in later chapters, the physical expression of characteristics is accomplished through the expression of genes (sequences of DNA), carried on chromosomes. The genetic makeup of peas consists of two similar, or homologous (remember this term from Chapter 11), copies of each chromosome, one from each parent. Through meiosis, diploid organisms utilize meiosis to produce haploid (1n) gametes that participate in fertilization. |
For cases in which a single gene controls a single characteristic, such as pea color, a diploid organism has genetic copies that may or may not encode the same version of the characteristic. These gene variations (e.g., green peas versus yellow peas) are called alleles. Different alleles for a given gene in a diploid organism interact to express physical characteristics such as pea color in plants or hairline appearance in humans. The observable traits of an organism are referred to as its phenotype. The organism’s underlying genetic makeup, i.e., the combination of alleles, is called its genotype. When diploid organisms carry the same alleles for a given trait, they are said to be homozygous for the genotype; when they carry different alleles, they are said to be heterozygous. For a gene whose expression is Mendelian (Section 12.1), homozygous dominant and heterozygous organisms will look identical; that is, they will have different genotypes but the same phenotype. The recessive allele will only be observed in homozygous recessive individuals. However, alleles do not always behave in dominant and recessive patterns. In other words, there are exceptions to Mendel’s model of inheritance. For example, incomplete dominance describes situation in which the heterozygote exhibits a phenotype that is intermediate between the homozygous phenotypes (e.g., a pink-flowered offspring is produced from a cross between a red-flowered parent and a white-flowered parent). Codominance describes the simultaneous expression of both of the alleles in the heterozygote (e.g., human blood types). It is also common for more than two alleles of a gene to exist in a population (e.g., variations in sizes of pumpkins). In humans, as in many animals and some plants, females have two X chromosomes, and males have one X chromosome and one Y chromosome. Genes on the X chromosome are X-linked, and males inherit and express only one allele for the gene (e.g., hemophilia, color-blindness). Some alleles can also be lethal, so their phenotype will never be observed. Many human genetic disorders, including albinism, cystic fibrosis, and Huntington’s disease can be explained on the basis of simple Mendelian inheritance patterns created by pedigree analysis. (In later Chapters we will learn how DNA analysis can be used to diagnose genetic |
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