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! How could this result be explained? Perhaps it was impossible for a white-eyed female fly to exist; such individuals might not be viable for some unknown reason. To test this idea, Morgan testcrossed the female F1 progeny with the original white-eyed male. He obtained both white-eyed and red-eyed males and females in... |
, while the male produces both X and Y gametes. When fertilization involves an X sperm, the result is an XX zygote, which develops into a female; when fertilization involves a Y sperm, the result is an XY zygote, which develops into a male. The solution to Morgan’s puzzle is that the gene causing the white-eye trait in... |
monitored crossing over between two genes, the recessive carnation eye color (car) and the dominant bar-shaped eye (B), on chromosomes with physical peculiarities visible under a microscope. Whenever these genes recombined through crossing over, the chromosomes recombined as well. Therefore, the recombination of genes... |
expected to occur in an average of 1% of gametes. A map unit is now called a centimorgan, after Thomas Hunt Morgan. In recent times new technologies have allowed geneticists to create gene maps based on the relative positions of specific gene sequences called restriction sites because they are recognized by DNA-cleavi... |
ature wing r Rudimentary wing Recombination frequencies y and w v and m v and r v and w v and y w and m y and m w and r 0.010 0.030 0.269 0.300 0.322 0.327 0.355 0.450 Genetic map.58 r.34.31.01 0 m v w y Chapter 13 Patterns of Inheritance 265 Analyzing a Three-Point Cross. The first genetic map was constructed by A. H.... |
linked with w+.) y w m y w+ m+ → _______ y+ w+ m+ y+ w m In order to see all the recombinant types that might be present among the gametes of these heterozygous flies, Sturtevant conducted a testcross. He crossed female heterozygous flies to males recessive for all three traits and examined the progeny. Because males ... |
608%, or 32.6 centimorgans. Similarly, body ( y) and wing (m) are separated by a recombination distance of 33.832%, or 33.8 centimorgans. From this, then, we can construct our genetic map. The biggest distance, 33.8 centimorgans, separates the two outside genes, which are evidently y and m. The gene w is between them, ... |
thousands of small fragments of DNA whose relative positions are known. Investigators wishing to study a particular gene will first use techniques described in chapter 19 to screen this library and determine which fragment carries the gene of interest. They will then be able to analyze that fragment in detail. In para... |
ifuss muscular dystrophy Diabetes insipidus, renal Myotubular myopathy, X-linked FIGURE 13.34 The human X chromosome gene map. Over 59 diseases have been traced to specific segments of the X chromosome. Many of these disorders are also influenced by genes on other chromosomes. Chapter 13 Patterns of Inheritance 267 Hum... |
females fail to undergo these changes. Among fishes and in some species of reptiles, environmental changes can cause changes in the expression of this sex-determining gene, and thus of the sex of the adult individual. 268 Part IV Reproduction and Heredity FIGURE 13.35 A human karyotype. This karyotype shows the colore... |
called Y, which has few other transcribed genes. FIGURE 13.37 A calico cat. The coat coloration of this cat is due to the random inactivation of her X chromosome during early development. The female is heterozygous for orange coat color, but because only one coat color allele is expressed, she exhibits patches of oran... |
with trisomy 21 or trisomy 22 are always mentally retarded. Down Syndrome. The developmental defect produced by trisomy 21 (figure 13.38) was first described in 1866 by J. Langdon Down; for this reason, it is called Down syndrome (formerly “Down’s syndrome”). About 1 in every 750 children exhibits Down syndrome, and t... |
of the eggs a woman will ever produce have developed to the point of prophase in meiosis I by the time she is born. By the time she has children, her eggs are as old as she is. In contrast, men produce new sperm daily. Therefore, there is a much greater chance for problems of various kinds, including those that cause ... |
the XYY zygotes develop into fertile males of normal appearance. The frequency of the XYY genotype (Jacob’s syndrome) is about 1 per 1000 newborn males, but it is approximately 20 times higher among males in penal and mental institutions. This observation has led to the highly controversial suggestion that XYY males a... |
s pedigree, it is sometimes possible to estimate the likelihood that the person is a carrier for certain disorders. For example, if one of your relatives has been afflicted with a recessive genetic disorder such as cystic fibrosis, it is possible that you are a heterozygous carrier of the recessive allele for that diso... |
to test directly for the proper functioning of enzymes involved in genetic disorders. The lack of normal enzymatic activity signals the presence of the disorder. Thus, the lack of the enzyme responsible for breaking down phenylalanine signals PKU (phenylke- Amniotic fluid Uterus Hypodermic syringe Fetal cells FIGURE 1... |
when they cut strands of DNA at particular places (see chapter 18). Therefore, these mutations produce what are called restriction fragment length polymorphisms, or RFLPs (figure 13.42). Many gene defects can be detected early in pregnancy, allowing for appropriate planning by the prospective parents. Chapter 13 Patte... |
.com/raven6e http://www.biocourse.com Questions Media Resources 1. Why weren’t the implications of Koelreuter’s results recognized for a century? 2. What characteristics of the garden pea made this organism a good choice for Mendel’s experiments on heredity? 3. To determine whether a purple-flowered pea plant of unknow... |
= 1⁄4, or 0.25), plus the probability that they will both be on chromosome 2 (also 1⁄2 × 1⁄2 = 1⁄4, or 0.25), for an overall probability of 1⁄2, or 0.5. In general, the probability that two randomly selected characters will be on the same chromosome is equal to 1⁄n where n is the number of chromosome pairs. Humans hav... |
You inherit a racehorse and decide to put him out to stud. In looking over the stud book, however, you discover that the horse’s grandfather exhibited a rare disorder that causes brittle bones. The disorder is hereditary and results from homozygosity for a recessive allele. If your horse is heterozygous for the allele... |
and 50% black-eyed flies. What were the genotypes of your original two flies? 12. Hemophilia is a recessive sex-linked human blood disease that leads to failure of blood to clot normally. One form of hemophilia has been traced to the royal family of England, from which it spread throughout the royal families of Europe... |
many chromosomes would you expect to find in the karyotype of a person with Turner syndrome? 17. A woman is married for the second time. Her first husband has blood type A and her child by that marriage has type O. Her new husband has type B blood, and when they have a child its blood type is AB. What is the woman’s b... |
2 3 Muscle Muscle tissue tissue B lo d e o v el s s How primary tumors kill off the competition. Tumors require an ample blood supply to fuel their growth. The growth of new blood vessels is called angiogenesis. Inhibiting angiogenesis offers a possible way to block tumor growth. tumors, allowing the primary tumor to ... |
ahead and read the ending—in the real world of research, you never know how things are going to turn out. This story starts when a Harvard University researcher, Dr. Judah Folkman, followed up on a familiar observation made by many oncologists (cancer specialists), that removal of a primary tumor often leads to more r... |
transcriptional factor (HIF-1) in angiogenesis, Johnson and his co-workers were faced with the problem that HIF-1 has many other effects on cell growth. To get a clear look at its role in angiogenesis, the researchers turned to embryonic stem cells. Embryonic stem cells are cells harvested from early embryos, before t... |
the VEGF protein analysis (see graph b above). Levels of the protein VEGF rise in wild-type cells under conditions of hypoxia, increasing the immediate availability of oxygen to the tumor by promoting capillary formation. The researchers found levels of VEGF protein were lower in null cell tumors, and responded to hyp... |
encodes. FIGURE 14.1 DNA. The hereditary blueprint in each cell of all living organisms is a very long, slender molecule called deoxyribonucleic acid (DNA). The realization that patterns of heredity can be ex- plained by the segregation of chromosomes in meiosis raised a question that occupied biologists for over 50 y... |
Cells To test his hypothesis, Hammerling selected individuals from two species of the genus Acetabularia in which the caps look very different from one another: A. mediterranea has a disk-shaped cap, and A. crenulata has a branched, flower-like cap. Hammerling grafted a stalk from A. crenulata to a foot from A. medite... |
�s experiments suggested that the nucleus is the repository of hereditary information in a cell. A direct test of this hypothesis was carried out in 1952 by American embryologists Robert Briggs and Thomas King. Using a glass pipette drawn to a fine tip and working with a microscope, Briggs and King removed the nucleus ... |
instructions. Egg (two nucleoli) UV light destroys nucleus, or it is removed with micropipette. Tadpole (one nucleolus) Epithelial cells are isolated from tadpole intestine. Nucleus is removed in micropipette. No growth 1 2 Embryo Abnormal embryo 3 Epithelial cell nucleus is inserted into enucleate egg. Embryo Tadpole... |
a mutant strain of S. pneumoniae that lacked the virulent strain’s polysaccharide coat, the mice showed no ill effects. The coat was apparently necessary for virulence. The normal pathogenic form of this bacterium is referred to as the S form because it forms smooth colonies on a culture dish. The mutant form, which l... |
were unharmed. (4) But when Griffith injected a mixture of dead bacteria with polysaccharide coats and live bacteria without such coats, many of the mice died, and virulent bacteria with coats were recovered. Griffith concluded that the live cells had been “transformed” by the dead ones; that is, genetic information s... |
bacteriophage infects a bacterial cell, it first binds to the cell’s outer surface and then injects its hereditary information into the cell. There, the hereditary information directs the production of thousands of new viruses within the bacterium. The bacterial cell eventually ruptures, or lyses, releasing the newly ... |
? The Chemical Nature of Nucleic Acids A German chemist, Friedrich Miescher, discovered DNA in 1869, only four years after Mendel’s work was published. Miescher extracted a white substance from the nuclei of human cells and fish sperm. The proportion of nitrogen and phosphorus in the substance was different from that i... |
carbon atom. The 5′ phosphate and 3′ hydroxyl groups allow DNA and RNA to form long chains of nucleotides, because these two groups can react chemically with each other. The reaction between the phosphate group of one nucleotide and the hydroxyl group of another is a dehydration synthesis, eliminating a water molecule... |
Not a Simple Repeating Polymer the phosphate group is now linked to the two sugars by means of a pair of ester (P— O—C) bonds. The two-unit polymer resulting from this reaction still has a free 5′ phosphate group at one end and a free 3′ hydroxyl group at the other, so it can link to other nucleotides. In this way, ma... |
the DNA (table 14.1). This strongly suggested that DNA was not a simple repeating polymer and might have the information-encoding properties genetic material must have. Despite DNA’s complexity, however, Chargaff observed an important underlying regularity in doublestranded DNA: the amount of adenine present in DNA al... |
that the DNA molecule had the shape of a helix, or corkscrew, with a diameter of about 2 nanometers and a complete helical turn every 3.4 nanometers (figure 14.9c). 286 Part V Molecular Genetics 5 2 nm 3 TA T A G•••C C•••G T A 3.4 nm G•••C T A G•••C T A G•••C 0.34 nm TA T A G•••C C•••G Minor groove Major groove Major ... |
in a double helix, adenine forms two hydrogen bonds with thymine, but it will not form hydrogen bonds properly with cytosine. Similarly, guanine forms three hydrogen bonds with cytosine, but it will not form hydrogen bonds properly with thymine. Consequently, adenine and thymine will always occur in the same proportio... |
each strand of the duplex becomes part of another duplex. Two other hypotheses of gene replication were also proposed. The conservative model stated that the parental double helix would remain intact and generate DNA copies consisting of entirely new molecules. The dispersive model predicted that parental DNA would be... |
.11). Meselson and Stahl interpreted their results as follows: after the first round of replication, each daughter DNA duplex was a hybrid possessing one of the heavy strands of the parent molecule and one light strand; when this hybrid duplex replicated, it contributed one heavy strand to form another hybrid duplex an... |
of 14N-DNA and 15N-DNA, indicating that only one strand of each duplex contained 15N. After two generations in 14N medium, two bands were obtained; one of intermediate density (in which one of the strands contained 15N) and one of low density (in which neither strand contained 15N). Meselson and Stahl concluded that r... |
polymerase III than any other organism’s DNA polymerase, and so will describe it in detail here. Other DNA polymerases are thought to be broadly similar. DNA polymerase III contains 10 different kinds of polypeptide chains, as illustrated in figure 14.15. The enzyme is a dimer, with two similar multisubunit complexes.... |
polymerase III Helicase Primase FIGURE 14.16 A DNA replication fork. Helicase enzymes separate the strands of the double helix, and single-strand binding proteins stabilize the single-stranded regions. Replication occurs by two mechanisms. (1) Continuous synthesis: After primase adds a short RNA primer, DNA polymerase... |
are about 100 to 200 nucleotides long in eukaryotes and 1000 to 2000 nucleotides long in prokaryotes. Each Okazaki fragment is synthesized by DNA polymerase III in the 5′ → 3′ direction, beginning at the replication fork and moving away from it. When the polymerase reaches the 5′ end of the lagging strand, another enz... |
torque generated by unwinding. For replication to proceed at 1000 nucleotides per second, the parental helix ahead of the replication fork must rotate 100 revolutions per second! To relieve the resulting twisting, called torque, enzymes known as topisomerases—or, more informally, gyrases—cleave a strand of the helix, ... |
000 base-pairs long. Each replication unit has its own origin of replication, and multiple units may be undergoing replication at any given time, as can be seen in electron micrographs of replicating chromosomes (figure 14.19). Each unit replicates in a way fundamentally similar to prokaryotic DNA replication, using si... |
Mendelian traits and that they had resulted from changes in the hereditary information in an ancestor of the affected families. Beadle and Tatum: Genes Specify Enzymes From Garrod’s finding, it took but a short leap of intuition to surmise that the information encoded within the DNA of chromosomes acts to specify part... |
Enzyme H Chromosome Encoded enzyme Substrate in biochemical pathway Glutamate Ornithine Citruline Arginosuccinate Arginine FIGURE 14.21 Evidence for the “one-gene/one-polypeptide” hypothesis. The chromosomal locations of the many arginine mutants isolated by Beadle and Tatum cluster around three locations. These locat... |
omal positions were located, the arg mutations were found to cluster in three areas (figure 14.21). One-Gene/One-Polypeptide For each enzyme in the arginine biosynthetic pathway, Beadle and Tatum were able to isolate a mutant strain with a defective form of that enzyme, and the mutation was always located at one of a f... |
certain definite order. The information needed to specify a protein such as an enzyme, therefore, is an ordered list of amino acids. Ingram: Single Amino Acid Changes in a Protein Can Have Profound Effects Following Sanger’s pioneering work, Vernon Ingram in 1956 discovered the molecular basis of sickle cell anemia, a... |
hhe.com/raven6e http://www.biocourse.com Questions Media Resources Chapter 14 Summary 14.1 What is the genetic material? • Eukaryotic cells store hereditary information within the nucleus. • In viruses, bacteria, and eukaryotes, the hereditary information resides in nucleic acids. The transfer of nucleic acids can lead... |
enzyme is encoded by a specific region of the DNA called a gene. 298 Part V Molecular Genetics 6. What hypothesis did Beadle and Tatum test in their experiments on Neurospora? What did they do to change the DNA in individuals of this organism? How did they determine whether any of these changes affected enzymes in bio... |
feathers, and two eyes rather than one. The color of your eyes, the texture of your fingernails, and all of the other traits you receive from your parents are recorded in the cells of your body. As we have seen, this information is contained in long molecules of DNA (figure 15.1). The essence of heredity is the abilit... |
subunit Large ribosomal subunit E site P site A site Small subunit E P A mRNA binding site Small ribosomal subunit FIGURE 15.2 A ribosome is composed of two subunits. The smaller subunit fits into a depression on the surface of the larger one. The A, P, and E sites on the ribosome, discussed later in this chapter, pla... |
the strand into the gene. As it encounters each DNA nucleotide, it adds the corresponding complementary RNA nucleotide to a growing mRNA strand. Thus, guanine (G), cytosine (C), thymine (T), and adenine (A) in the DNA would signal the addition of C, G, A, and uracil (U), respectively, to the mRNA. When the RNA polymer... |
loop contains the anticodon sequence, which is complementary to a three-base sequence on messenger RNA. Amino acids attach to the free, single-stranded —OH end. (b) In the three-dimensional structure, the loops of tRNA are folded. DNA Transcription mRNA Translation Protein FIGURE 15.5 The Central Dogma of gene express... |
genetic message shifted, and the downstream gene was transcribed as nonsense. However, when they made three deletions, the correct reading frame was restored, and the sequences downstream were transcribed correctly. They obtained the same results when they made additions to the DNA consisting of one, two, or three nuc... |
production of the polypeptide polyphenylalanine (a string of phenylalanine amino acids). Therefore, one of the three-nucleotide sequences specifying phenylalanine is UUU. In 1964, Nirenberg and Philip Leder developed a powerful triplet binding assay in which a specific triplet was tested to see which radioactive amino... |
the third nucleotide (ACU, ACC, ACA, and ACG). The Code Is Practically Universal The genetic code is the same in almost all organisms. For example, the codon AGA specifies the amino acid arginine in bacteria, in humans, and in all other organisms whose genetic code has been studied. The universality of the genetic cod... |
the mechanism behind the transcription process, it is useful to focus first on RNA polymerase, the remarkable enzyme responsible for carrying it out (figure 15.7). RNA Polymerase RNA polymerase is best understood in bacteria. Bacterial RNA polymerase is very large and complex, consisting of five subunits: two α subuni... |
circles are RNA polymerase molecules bound to several promoter sites on bacterial virus DNA. scribe only once every 10 minutes. Most strong promoters have unaltered –35 and –10 sequences, while weak promoters often have substitutions within these sites. Initiation The binding of RNA polymerase to the promoter is the f... |
bubble moves down the DNA at a constant rate, about 50 nucleotides per second, leaving the growing RNA strand protruding from the bubble. After the transcription bubble passes, the now transcribed DNA is rewound as it leaves the bubble. Unlike DNA polymerase, RNA polymerase has no proofreading capability. Transcriptio... |
are rapidly degraded. 3′ poly-A tails. The 3′ end of eukaryotic transcript is cleaved off at a specific site, often containing the sequence AAUAAA. A special poly-A polymerase enzyme then adds about 250 A ribonucleotides to the 3′ end of the transcript. Called a 3′ poly-A tail, this long string of As protects the tran... |
odon and thus only one tRNA molecule. Others recognize two, three, four, or six different tRNA molecules, each with a different anticodon but coding for the same amino acid (see table 15.1). If one considers the nucleotide sequence of mRNA a coded message, then the 20 activating enzymes are responsible for decoding tha... |
the reading frame—that is, which groups of three nucleotides will be read as codons. Moreover, the complex must bind to the beginning of the mRNA molecule, so that all of the transcribed gene will be translated. In bacteria, the beginning of each mRNA molecule is marked by a leader sequence complementary to one of the... |
which releases the initial methionine from its tRNA and attaches it instead by a peptide bond to the second amino acid. fMet Leader sequence Initiation factor fMet tRNAfMet A U G C A U Large ribosomal subunit fMet E site fMet P site A site mRNA U A C G U A mRNA Small ribosomal subunit (containing ribosomal RNA) Initia... |
on at the A site, placing its amino acid adjacent to the growing chain. The chain then transfers to the new amino acid, and the entire process is repeated. Elongation continues in this fashion until a chain-terminating nonsense codon is exposed (for example, UAA in figure 15.14). Nonsense codons do not bind to tRNA, bu... |
. In particular, they found that eukaryotic proteins are encoded by RNA segments that are excised from several locations along what is called the primary RNA transcript (or primary transcript) and then spliced together to form the mRNA that is eventually translated in the cytoplasm. The experiment that revealed this un... |
zymes cut these segments (introns) out and splice together the remaining segments (exons). (b) The seven loops are the seven introns represented in the schematic drawing (c) of the mature mRNA transcript hybridized to DNA. Chapter 15 Genes and How They Work 309 1 DNA 3 Nuclear membrane 2 Primary RNA transcript 3 Intron... |
are cut out of the primary transcript before it is used in polypeptide synthesis; therefore, those sequences are not translated. The remaining sequences, which correspond to the exons, are spliced together to form the final, “processed” mRNA molecule that 310 Part V Molecular Genetics Differences between Bacterial and... |
of DNA nucleotides corresponds exactly to the sequence of amino acids in the encoded polypeptide. (b) Eukaryotic genes are typically different, containing long stretches of nucleotides called introns that do not correspond to amino acids within the encoded polypeptide. Introns are removed from the primary RNA transcri... |
How does it specify where the polypeptide should end? 6. What roles do elongation factors play in translation? 15.3 Genes are first transcribed, then translated. • During transcription, the enzyme RNA polymerase manufactures mRNA molecules with nucleotide sequences complementary to particular segments of the DNA. • Du... |
in eukaryotes operates at a distance. Designing a Complex Gene Control System. Eukaryotic genes use a complex collection of transcription factors and enhancers to aid the polymerase in transcription. The Effect of Chromosome Structure on Gene Regulation. The tight packaging of eukaryotic DNA into nucleosomes does not ... |
Binding the protein to the regulatory sequence either blocks transcription by getting in the way of RNA polymerase, or stimulates transcription by facilitating the binding of RNA polymerase to the promoter. Transcriptional Control in Prokaryotes Control of gene expression is accomplished very differently in bacteria t... |
the reversible metabolic adjustments bacterial cells make to the environment. In all multicellular organisms, changes in gene expression within particular cells serve the needs of the whole organism, rather than the survival of individual cells. Posttranscriptional Control Gene expression can be regulated at many leve... |
proteins employ one of a small set of structural, or DNA-binding, motifs, particular bends of the protein chain that permit it to interlock with the major groove of the DNA helix. Regulatory proteins identify specific sequences on the DNA double helix, without unwinding it, by inserting DNA-binding motifs into the maj... |
doubles the zone of contact between protein and DNA and so greatly strengthens the bond that forms between them. Recognition helix FIGURE 16.3 The helix-turn-helix motif. One helical region, called the recognition helix, actually fits into the major groove of DNA. There it contacts the edges of base-pairs, enabling it... |
the cluster, the stronger the protein binds to the DNA. In other forms of the zinc finger motif, the β sheet’s place is taken by another helical segment. Zn The Leucine Zipper Motif In yet another DNA-binding motif, two different protein subunits cooperate to create a single DNA-binding site. This motif is created whe... |
ia coli uses proteins encoded by a cluster of five genes to manufacture the amino acid tryptophan. All five genes are transcribed together as a unit called an operon, producing a single, long piece of mRNA. RNA polymerase binds to a promoter located at the beginning of the first gene, and then proceeds down the DNA, tr... |
binding of tryptophan to the repressor increases the distance between the two recognition helices in the repressor, allowing the repressor to fit snugly into two adjacent portions of the major groove in DNA. RNA polymerase from binding to the trp promoter. The trp genes are transcribed, and the cell proceeds to manufa... |
actose. Gene for repressor protein Promoter for I gene CAP binding site Operator Gene for permease Promoter for lac operon Gene for -galactosidase Gene for transacetylase I CAP Plac O PI Z Y A Regulatory region Coding region lac control system FIGURE 16.9 The lac region of the Escherichia coli chromosome. The lac opero... |
operon is said to have been “induced” by lactose. This two-switch control mechanism thus causes the cell to produce lactose-utilizing proteins whenever lactose is present but glucose is not, enabling it to make a metabolic decision to produce only what the cell needs, conserving its resources (figure 16.12). Bacteria ... |
�) when CAP is bound and when lactose binding to the repressor changes its shape so that it can no longer sit on the operator site and block RNA polymerase activity. CAP binding site RNA-polymerase binding site (promoter) Operator lacZ gene + + + Operon OFF because CAP is not bound Operon OFF both because lac repressor... |
or decrease its rate. They include the TATA-binding protein, the first of the basal factors to bind to the core promoter sequence. Coactivators (the tan shapes that form the bulk of the transcription complex, named according to their molecular weights) are transcription factors that link the basal factors with regulat... |
. When the bacterial activator NtrC binds to an enhancer, it causes the DNA to loop over to a distant site where RNA polymerase is bound, activating transcription. While such enhancers are rare in bacteria, they are common in eukaryotes. Enhancers A key advance in the evolution of eukaryotic gene transcription was the ... |
proteins to form nucleosomes (figure 16.17) and then the strand of nucleosomes is twisted into 30-nm filaments. Promoter Blocking by Nucleosomes Intensive study of eukaryotic chromosomes has shown that histones positioned over promoters block the assembly of transcription factor complexes. Therefore, transcription fac... |
Methylation 1 6 N H N C O CH3 H C C NH2 C N H Cytosine 5-methylcytosine FIGURE 16.18 DNA methylation. Cytosine is methylated, creating 5-methylcytosine. Because the methyl group is positioned to the side, it does not interfere with the hydrogen bonds of a GC basepair. Chapter 16 Control of Gene Expression 325 Posttran... |
ome and looped intron form. 5 5 end of intron is cut and attached near 3 end of intron, forming a lariat. The 3 end of the intron is then cut. 5 Spliceosome 3 3 Exons are spliced; spliceosome disassembles. Exon Exon Excised intron Mature mRNA 5 3 FIGURE 16.19 How spliceosomes process RNA. Particles called snRNPs contai... |
the Processed Transcript Out of the Nucleus Processed mRNA transcripts exit the nucleus through the nuclear pores described in chapter 5. The passage of a transcript across the nuclear membrane is an active process that requires that the transcript be recognized by receptors lining the interior of the pores. Specific ... |
abulary of Gene Expression activator A regulatory protein that promotes gene transcription by binding to DNA sequences upstream of a promoter. Activator binding stimulates RNA polymerase activity. anticodon The three-nucleotide sequence on one end of a tRNA molecule that is complementary to and base-pairs with an amino... |
, however, are usually much less stable, with half-lives of less than 1 hour. What makes these particular transcripts so unstable? In many cases, they contain specific sequences near their 3′ ends that make them attractive targets for enzymes that degrade mRNA. A sequence of A and U nucleotides near the 3′ poly-A tail ... |
expression by speeding or slowing protein synthesis. RNA polymerase 1. Initiation of transcription. Most control of gene expression is achieved by regulating the frequency of transcription initiation. 3 Poly-A tail Introns 3 5 3 5 Primary RNA transcript 5 Cap Exons 3 Poly-A tail mRNA Exon splicing. Gene expression 2. ... |
polymerase to the promoter. • Transcription is often controlled by a combination of repressors and activators. 3. Describe the mechanism by which the transcription of trp genes is regulated in Escherichia coli when tryptophan is present in the environment. 4. Describe the mechanism by which the transcription of lac ge... |
paths. Pattern Formation. Diffusion of chemical inducers governs pattern formation in fly embryos. Expression of Homeotic Genes. Master genes determine the form body segments will take. Programmed Cell Death. Some genes, when activated, kill their cells. 17.3 Four model developmental systems have been extensively rese... |
, where the adult individuals contain a variety of specialized cells organized into tissues and organs. A hallmark of plant development is flexibility; as a plant develops, the precise array of tissues it achieves is greatly influenced by its environment. In animals, development is complex and rigidly controlled, produ... |
blastomeres, until a solid ball of cells is produced (figure 17.4). This initial period of cell division, termed cleavage, is not accompanied by any increase in the overall size of the embryo; rather, the contents of the zygote are simply partitioned into the daughter cells. The two ends of the zygote are traditionall... |
) Neurulation. (e) Cell migration. ( f ) Organogenesis. ( g) Growth. 334 Part V Molecular Genetics Formation of the Blastula The outermost blastomeres (figure 17.5a) in the ball of cells produced during cleavage are joined to one another by tight junctions, which, as you may recall from chapter 7, are belts of protein ... |
it. The thickening is produced by the elongation of certain ectodermal cells. Those cells then assume a wedge shape by contracting bundles of actin filaments at one end. This change in shape causes the neural tissue to roll up into a tube, which eventually pinches off from the rest of the ectoderm and gives rise to th... |
much genetic research. Maternal Genes The development of an insect like Drosophila begins before fertilization, with the construction of the egg. Specialized nurse cells that help the egg to grow move some of their own mRNA into the end of the egg nearest them (figure 17.7a). As a result, mRNAs produced by maternal ge... |
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