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The Biochemical Basis of Heredity Practice Test (page 3)

By — McGraw-Hill Professional
Updated on Aug 21, 2011

DNA Repair Questions

  1. Exposure to UV light results in the formation of thymine dimers between adjacent thymines in a single DNA strand. Describe how repair of this damage might lead to a mutation.
  2. A new mutant strain of E. coli has been identified that is observed to produce progeny containing spontaneous mutations in many different genes at a higher rate than wild-type E. coli. The mutant is described as having a "mutator" phenotype. Hypothesize about the wild-type function of the mutated gene and why it has this mutant phenotype.

Answers

Vocabulary

  1. semiconservative
  2. promoter
  3. codon
  4. primer
  5. complementation
  6. alpha helix
  7. transcription
  8. Okazaki fragments
  9. introns
  10. photoreactivation

Multiple-Choice

  1. c
  2. e
  3. c
  4. d
  5. d
  6. a
  7. c
  8. e
  9. d
  10. c

DNA and Protein Synthesis

  1. (a)   5'-ATGGCTCATGACTGA 3'     (b)   5'-AUGGCUCAUGACUGA 3'     (c)   met-ala-his-asp-(stop)     (d) The complementary strand shown in the answer for part (a) is the sense or untranscribed strand.
  2. (a)   The R bacteria have become "transformed" into virulent bacteria by obtaining material from the dead S strain cells.     (b)   No, the material from the dead S strain could be DNA, RNA, protein, or any other cellular material.
  3. 1.43
  4. In one population of a bacteriophage, the 32P radioisotope can be incorporated into DNA; and in a second phage population, the 35S radioisotope can be incorporated into proteins (some amino acids have sulfur in their side chains). These two populations of phage can be used to infect E. coli cells in separate experiments. If DNA is the hereditary material, then it should be injected into the cell. Thus, the 32P radioisotope label should be found within the bacterial cells, once the empty phage particles are removed from the outside of the cell. If protein is the hereditary material, the 35S radioisotope label will be found within the bacterial cells.
  5. 1 × 2 × 6 × 4 = 48
  6. Approximately 150 amino acids
  7. (a)   The protein would be slightly shorter than normal. Since region A does not seem to interact with other portions of the polypeptide chain, the mutant enzyme should still function normally (barring unpredicted interaction of the side chain of the mutant amino acid with other parts of the molecule). If a nonsense mutation had occurred in region D, however, a very small chain would have been produced that would be devoid of a catalytic site, because proteins are synthesized beginning at the NH2 end. (b) Same sense mutants produce no change in their polypeptide products from normal. (c) The polypeptide would be one amino acid shorter than normal. Since region C does not seem to be critical to the tertiary shape of the molecule, the mutant enzyme would probably function normally. (d) An incorrect amino acid would be present in region B. As long as its side chain did not alter the tertiary shape of the molecule, the mutant enzyme would be expected to function normally, (e) A frame shift mutant in region C is bound to create many missense codons (or perhaps a nonsense codon) from that point on through the carboxyl terminus, including the enzymatic site. Such a protein would be catalytically inactive.
  8. Nonfunctional DNA fragments might be produced if DNA replication occurs before all of the critical repairs have been made.
  9.  
  10. (a) Answers to Supplementary Problems

    (b) G was added, A was deleted (bold)

  11. 2800 genes
  12. 50 proteins, assuming only one strand of the DNA is transcribed into mRNA

Mutations

  1. (a), (c), (d), (e), (g), (i) = large plaque, (b), (f), (h), (j) = small plaque
  2. Seed parent contributes two sets of chromosomes to triploid endosperm; one Dt gene gives 7.2 mutations per kernel, two Dt genes increase mutations to 22.2 per kernel
  3. 20% more mutations in the individual receiving 0.5 roentgen per year.
  4. .
  5. Answers to Supplementary Problems
  6. The polypeptide chain folds into a configuration such that noncontiguous regions form portions of the catalytic or antibody-combining sites.
  7. The suppressing mutation could be in that portion of a gene specifying the anticodon region of a tRNA molecule. For example, a tyrosine suppressor gene changes the anticodon of tRNAtyr from 3'-AUG-5' to 3'-AUC–5', thereby allowing it to recognize UAG mRNA nonsense codons. If the genes for tRNAtyr exists in multiple copies and only one of the tRNAtyr genes was mutated to a suppressor form, there would still be other normal (non-suppressor) tRNAtyr genes to make some normal proteins. The efficiency of suppression must be low to be compatible with survival of the organism.
  8. (1) A change in one of the ribosomal proteins in the 30S subunit could cause misreading of the codon-anticodon alignment, resulting in substitution of an "acceptable" (although perhaps not the normal) amino acid in a manner analogous to the misreading induced by the antibiotic streptomycin. In a cell-free system with synthetic poly-U mRNA, streptomycin causes isoleucine tRNA to be substituted for that of phenylalanine tRNA. (2) A mutation in a gene coding for an amino-acid-activating enzyme (amino-acyl synthetase) causes a different amino acid to occasionally be attached to a given species of tRNA. For example, if AUU (isoleucine mRNA codon) is mutated to UUU (phenylalanine mRNA codon), its effect may be suppressed by the occasional misattachment of isoleucine to tRNAphe by a mutant amino-acyl synthetase that is less than 100% specific in its normal action of attaching phenylalanine to tRNAphe.
  9. Such proteins are normally homopolymers (quaternary complexes consisting of two or more identical polypeptide chains). If two mutant polypeptide chains contain compensating amino acid substitutions, they may aggregate into a heterodimer that exhibits at least partial enzymatic activity.
  10. Wild-type alleles usually code for complete, functional enzymes or other proteins. One active wild-type allele can often cause enough enzyme to be produced so that normal or nearly normal phenotypes result (dominance). Mutations of normally functioning genes are more likely to destroy the biological activities of proteins. Only in the complete absence of the wild-type gene product would the mutant phenotype be expressed (recessiveness).
  11. Loss of function can potentially occur by point mutations at a number of sites within a cistron coding for a given fermentation enzyme or in any gene coding for one of the multiple enzymes in a common biosynthetic pathway such as those in histidine synthesis. The loss of such an indispensable function is lethal. Streptomycin distorts ribosomes, causing misreading of the genetic code. Only a limited number of changes in the ribosomal proteins or rRNA could render the ribosome immune from interference by streptomycin and still preserve the way these components normally interact with mRNA, tRNA, initiation factors, etc., during protein synthesis.

DNA Repair

  1. Repair of thymine dimers occurs using excision repair. This type of repair involves using the remaining template (after excision) to synthesize a new, replacement strand. This synthesis, especially when there is a lot of DNA damage, can be error-prone, thus leading to mutations.
  2. The wild-type function of this gene may be in a DNA repair function. When mutated, the cell cannot repair its DNA efficiently, leading to a higher rate of spontaneous mutations across the genome.
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