Print
Hits: 521
52

Gamow (1954) proposed that the structure of DNA deduced by Watson and Crick (1953) could be interpreted as a way of forming roughly 20 "words" of the common amino acids from the four "letters" A, T, C, and G that represent DNA nucleotides.

Crick and coworkers (1961) used a method developed by Benzer to induce mutations in the DNA of a virus by the insertion of a single nucleotide. The mutant could not infect the bacterium Escherichia coli and neither could viruses with a second insertion of a second DNA nucleotide. However, a third nucleotide insertion restored the ability of the virus to infect the bacterium.

In 1961, Nirenberg and Matthaei conducted a series of experiments to better understand the flow of genetic information from gene to protein. They discovered that in solutions containing the contents of ruptured E. coli bacterial cells from which DNA had been removed, polymers containing only one repeating amino acid, phenylalanine, would be synthesized if synthetic mRNA composed of only the single nucleotide, uracil (U), was added to the solution in which phenylalanine was also present. In solutions containing mRNA with only adenine (A) or cytosine (C) and the amino acids lysine or proline, polymers containing only these amino acids would be synthesized. The researchers found that when ribosomes were removed by filtration, these polymers did not form. Nirenberg and Leder (1964) extended this work to include other nucleotides.

A. Summarize the conclusions regarding the encoding and decoding of heritable information supported by these studies. Explain how these studies provided evidence to support the Triplet Code.

Khorana (1960) developed a technique for synthesizing RNA composed of predictable distributions of repeated pairs or triplets of nucleotides. He found, for example, that RNA synthesized when A and U were present in relative concentrations of 4:1, respectively, will produce RNA sequences with these distributions determined by their relative probabilities: AAU:AAA, AUA:AAA, and UAA:AAA; 0.82 × 0.2/0.83 = 1/4 [calculated as follows: i) 4/5 of the bases are A, so the likelihood of selecting A is 0.8; ii) the selection is repeated to determine the second letter of the three-letter codon; iii) the likelihood of selecting a U is 1 in 5; iv) the probability of selecting the set AUU is the product; v) similarly, the probability of AAA is (4/5)3; and vi) the ratio of these probabilities is their relative likelihood]: AUU:AAA, UUA:AAA, and UAU:AAA; 0.8 × 0.22/0.83 = 1/16; and UUU:AAA; 0.23/0.83 = 1/64.

B. Based on Khorana’s findings, calculate the relative distributions of the following ratios of concentrations of RNA triplet sequences from mixtures in which the relative concentrations of guanine and cytosine, G:C, are 5:1.

RatioRelative Probabilities
GGC:GGG
GCG:GGG
CGG:GGG
 
GCC:GGG
CGC:GGG
CCG:GGG
 
CCC:GGG  
Table15.2
 

C. Based on the work of Nirenberg, Matthaei, Leder, and Khorana, the following table was constructed (taken from Khorana's Nobel Prize address):

Figure shows all 64 codons. Sixty-two of these code for amino acids, and three are stop codons.
Figure 15.18
 

A solution containing the amino acids shown in the table above and equal concentrations of the two nucleotides C and G is prepared. Predict the proteins that can be synthesized from this mixture in terms of each possible codon and their relative concentrations in terms of their amino acid repeat sequences.

D. Describe the effects of the codons UAA, UAG, and UGA on protein synthesis.

53

The yeast life cycle is usually dominated by haploid cells, each with a single set of unpaired chromosomes. The cell propagates asexually, and the genetic material is replicated through mitosis. Cell division occurs every 2–4 hours, leading to 60–100 generations in a single day. Yeast also reproduce sexually, particularly under adverse environmental conditions. When two haploid cells—with DNA containing complementary mating-type alleles—conjugate, a diploid zygote results. The diploid zygote can then complete the sexual segment of the life cycle through meiosis. After meiosis, four haploid spores are produced, which can germinate.

Researchers can grow yeast easily on nutrient-containing plates. Because both asexual and sexual reproduction is rapid, yeast has become an important organism for the experimental investigation of mutagenesis and evolution among eukaryotes. Environmental factors, such as chemicals or radiation, induce mutations. High-energy UV-c radiation of less than 1 minute in duration will result in many mutated yeast cells. UV-c can be used to mutate a strain of yeast in which the synthesis of adenine is blocked. This mutation is observable because the ade-2 mutant has a red color when cultured on nutrient-containing plates. Exposure to uv-c also can result in additional mutations. In particular, one mutant, ade-7, changes the color of the ade-2 mutant to white.

A. You have a uv-c lamp, culture plates, and growth chambers at 23 °C and 37 °C. You also have available known haploid strains that are (ade-2,+,+), where + denotes the wild type. Design a plan to determine the rate of uv-c-induced mutations in nutrient-containing plates inoculated with yeast.

Earth's ozone layer removes high-energy ultraviolet radiation, uv-c, from the solar radiation received at the surface. Lower-energy ultraviolet radiation, uv-b, strikes Earth’s surface. Damage to DNA induced by ultraviolet radiation occurs with the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine and cytosine, on the same strand of DNA. A repair enzyme, photolyase, which is activated by visible light, is present in plants and most animals, but not in humans. In characterizing the relationship between environmental mutagens and cell damage, a useful assumption is often made and referred to as the linear hypothesis. This assumption states that the extent of damage is proportional to the amount of radiation received.

Mutation rates for a strain (preac) that does not produce photolyase and a wild-type (+) strain were studied. Cultures of the two strains of yeast were diluted, and nutrient-containing plates were inoculated in triplicate at 23 °C. The plates were exposed to bright sunlight for varying time intervals. After exposure, the plates were incubated in the dark at 23 °C. After incubation between 1 and 8 hours, data shown in the table below were collected by counting the density of living cells relative to the control, and averaging these among replicates.

B. Using the data table below, graph the average survival fraction, relative to the wild-type control. Predict the number of mutations in a sample of 1,000 cells of the preac type that are exposed to bright sunlight for 15 seconds.

.
Figure 15.19 This is a 5 column table, showing Incubation time, in hours in the left most column, ranging from 1 to 8. A 10 second exposure has the following values for an incubation time of 1 to 8: 0.83, 1.00, 0.92, 0.75, 0.99, 0.81, 0.80, 1.05 and 0.89 with a standard deviation of 0.11. A 20 second exposure has the following values for an incubation time of 1 to 8: 0.58, 0.43, 0.38, 0.35, 0.49, 0.42, 0.32, 0.59, 0.45, with a standard deviation of 0.10. A 30 second exposure has the following values for an incubation time of 1 to 8: 0.33, 0.17, 0.12, 0.08, 0.11, 0.12, 0.09, 0.11, 0.14, with a standard deviation of 0.08. A 40 second exposure has the following values for an incubation time of 1 to 8: 0.17, 0.09, 0.03, 0.01, 0.01, 0.01, 0.01, 0.01, 0.04, with a standard deviation of .0.06. A 50 second exposure has the following values for an incubation time of 1 to 8: 0.08, 0.04, 0.01, 0.00, 0.00, 0.00, 0.00, 0.00, 0.02 with a standard deviation of 0.03
 

Yeast can also be used to study sexual reproduction, a somewhat puzzling phenomenon. Cloning of cells through mitosis is molecularly much less complex than meiosis, consumes less energy, and is less risky. Two alternative explanations for the evolution of sexual reproduction are popular. In one model, through assortment of genes, meiosis leads to an increase in the frequency of beneficial mutations. In the second model, detrimental mutations are purged from a population through sex. Studies using yeast (Gray and Goddard, Evol. Biol., 2012 and McDonald et al., Nature 2012) have provided a mechanism to study these models. As shown below, the fitness (defined as the log of the ratio of the number of cells in successive generations) of yeast is graphed as a function of number of mitotic reproductions in yeast grown in low-stress and high-stress environments, and with and without alternating induction of sexual reproduction.

The line graph is titled: Fitness of Yeast as a function of the number of mitotic reproductions in High and low stress environments with or without sexual reproduction. The key at the bottom states that red tringles are high-stress environments with sexual reproduction, purple triangles are high-stress environments without sexual reproduction, blue squares are low-stress environment with sexual reproduction and green squares are low stress environments without sexual reproduction. The x-axis is labelled Number of mitotic reproductions and has tick marks for 0, 50, 100, 150, 200, 250, and 300. The Y axis is labelled 0 approximately ¼ of the way up the graph. There is a red diagonal line that starts at 0,0 and ends near the top of the y-axis when the x-axis value is 300. There is a purple diagonal line that starts at 0, 0 and ends slightly above 0 at 300.  At 50 on the x axis, there is a blue square at 0 on the y-axis, a purple and a red triangle above zero . At 100 on the x-axis, there is a green square and a blue square under the 0 point and a red square above the 0 point. Between 100 and 150, there is a blue square and green square below and a red triangle and a purple triangle above the 0.  At the 150 point on the x axis, there is a blue square well below the 0, a green square a bit above the 0, an purple and red triangles well above the 0. Between 150 and 200, all four colors are above 0. At the 200 point, the blue and green squares are below 0 and the red and purple triangles are well above 0. Between 200 and 250, the blue square is below 0 and all other colors are above 0. At 250 the green square is well below 0, the blue square is right on 0, the purple triangle is above 0, and a red triangle near the top of the graph.  Between 250 and 300 the purple and green triangles are below zero, the blue square is above zero, and the red triangle is near the top of the graph. At 300, the purple triangle and blue square are below zero, the green square is above zero, and the red triangle is near the top of the graph.
Figure 15.20
 

C. Based on these data, evaluate the merits of the alternative theories of the adaptive advantage provided by sexual reproduction.

54

A. Describe the storage and retrieval of genetic information with the following model. Use the list to fill in the blanks with the letter corresponding to the correct term.

  1. amino acid
  2. tRNA
  3. DNA
  4. transcription
  5. mRNA
  6. translation
  7. protein
  8. RNA polymerase
  9. rRNA

Within the cytoplasm, __ is synthesized from __ bound to __ in a sequence that corresponds to information provided by __. This process is called __.

Within the nucleus, information originating in __ is encoded as a sequence of bases in __, which is synthesized by the enzyme __ that is embedded in the __. This process is called __.

B. During development, cell differentiation occurs, and the expression of genes is permanently switched off. Using the model summarized above, explain where information flow is most effectively blocked.

C. A chemical message is received by the cell regulating the timing of events controlled by gene expression. Using the model summarized above, explain where information flow is most effectively managed.

55

Structure and function in biology result from both the presence of genetic information and the expression of that information. Some genes are continually expressed, whereas the expression of most genes is regulated, commonly at the level of transcription. At the initiation of transcription, the TATA-binding protein (TBP) provides access to the DNA strand to be transcribed. The 5’TATAAA3’ sequence called the TATA box is found in prokaryotes, archaebacteria, and eukaryotes. Even among eukarya, when the TATA box is not present among eukaryotes, the initiation of transcription involves TBP. Scientists attribute this common characteristic to the relative thermostability of the A-T interaction. Hydrogen bonds hold the two strands of the DNA double helix together. This type of bond has the smallest interaction energy of all intermolecular forces; as temperature increases, these bonds are broken.

A. Explain the advantage, in terms of the energy required, which is provided by an AT-rich region in the sequence where transcription is initiated.

B. The fact that the TATA box or the associated TBP are common to all domains provides evidence of common ancestry among all life. Pose a scientific question that would need to be addressed by a valid alternative explanation of this fact.

C. A whole-genome survey of prokaryotes (Zheng and Wu, BMC Bioinformatics, 2010) showed that the relative amounts of guanine and cytosine in DNA poorly predicted the temperature range conditions that are suitable for an organism. Refine the question posed in part B, taking this result into account.

56

Only a fraction of DNA encodes proteins. The noncoding portion of a gene is referred to as the intron. The intron fraction depends upon the gene. Introns are rare in prokaryotic and mitochondrial DNA; in human nuclear DNA, this fraction is about 95%. The intron is transcribed into mRNA, but this noncoding mRNA is edited out before translation of the coding portion, or exon, of a gene. The edited exon segments are then spliced together by a spliceosome, a very large and complex collection of RNAs and proteins.

Although introns do not encode proteins, they have functions. In particular, they amplify expression of the exon, although the mechanism is unknown. When introns are very long, which is common among mammalian genes with roles in development, they can significantly extend the time required to complete transcription. Analysis of genes common to different plant and animal species shows many shared intronic positions and base sequences, although in some organisms, such as yeast, many introns have been deleted. Because introns do not encode proteins, mutations can remain silent and accumulate.

A. As described above, introns are ancestral remnants that are replicated because they do not disadvantage the organism. Consider the claim that introns are “junk DNA.” Evaluate the claim with supporting evidence.

B. Introns may be retained during transcription. Explain how the retention of a transcribed intron between two transcribed exons within a gene could do the following:

  • block expression of one polypeptide sequence
  • increase expression of a polypeptide
  • alter the polypeptide expressed