Chapter 8 Practice Problems Flashcards

Question

The HbB^S (sickle cell) allele of the human B-globin gene changes the sixth amino acid in the B-globin chain from glutamic acid to valine. In HbB^C, the sixth amino acid in B-globin is changed from glutamic acid to lysine. What would be the other of these two mutations within the map of the B-globin gene?

Answer 1

Glutamic acid (in the wild-type protein) can be encoded by either GAA or GAG. In sickle cell anemia, this amino acid is changed to valine by a single base change. Valine is encoded by GUN with N representing any of the four bases. Therefore, the second base of the triplet was altered from A to U in the HbβS allele. In HbβC the glutamic acid codon (GAA or GAG) is changed to a lysine (AAA or AAG). The change here is in the first base of the codon. The mutation causing the HbβC allele therefore precedes the HbβS mutation in the sequence of the β-globin gene when reading the RNA-like strand in the 5′-to-3′ direction. (Stated another way, the HbβC mutation is upstream of the HbβS mutation by one nucleotide pair.)

Answer 2

wild type mRNA: GGN GCN CCN AGA/G AAA/G mutant mRNA; GGN CAU/C CAA/C GGN AAA/G

Answer 3

a. polypeptides will be produced only at high Mg2+ concentrations. The codons in the RNA will be read as …UGU GUG UGU GUG... and so polypeptides with alternating Cys and Val amino acids will be made. b. At low Mg2+ concentrations, translation will start with AUG, and the RNA will be read as AUG CAU GCA UGC AUG …; polypeptides with repeats of Met-His-Ala-Cys will be synthesized. At high Mg2+ concentrations, because translation can initiate anywhere on the RNA, the same polypeptides will be produced except they will not all begin with Met. (RNAs not read starting with AUG are read as UGC AUG CAU GCA… or GCA UGC AUG CUA…) c. At low Mg2+ concentrations, translation will start with AUG: AUG UAU GUA UGU… and thus polypeptides with the repeat Met-Tyr-Val-Cys will be made. At high Mg2+ concentrations, translation can start anywhere on the RNA. The result is only polypeptides containing Met, Tyr, Val, and Cys in that order, but they will not all begin with Met. (RNAs that are not read starting with AUG are read as UGU AUG UAU GUA… or GUA UGU AUG UAU…)

Answer 4

``` UGA = stop; so UCA, UUA UAG = Stop; so UAU, UAC UAA = stop; so, UCA, UUA, UAC, UAU ```

Answer 5

a. Eight codons exist in RNAs made up of Us and Gs in random order: UUU, UUG, UGU, GUU UGG, GUG GGU, GGG. b. The eight codons in part (a) correspond to six different amino acids: Phe, Leu, Val, Cys, Trp, and Gly. c. The answers to part (a) and (b) are different because the genetic code is degenerate in that the same amino acid is specified by several different codons: UUU = Phe, UUG = Leu, UGU = Cys, GUU or GUG = Val, UGG = Trp, GGU and GGG = Gly d. The frequency of U is 3/4, and the frequency of G is 1/4. Therefore: f(UUU) = (3/4)3 = 27/64; f(UUG) = f(UGU) = f(GUU) = 3/4 × 3/4 × 1/4 = 9/64; f(UGG) = f(GUG) = f(GGU) = 3/4 × 1/4 × 1/4 = 3/64; f(GGG) = (1/4) 3 = 1/64. e. Phe (UUU) = 27/64; Leu (UUG) = 9/64; Cys (UGU) = 9/64; Val (GUU or GUG) = 9/64 + 3/64 = 12/64; Trp (UGG) = 3/64; Gly (GGU or GGG) = 3/64 + 1/64 = 4/64 f. By doing this experiment, you would have been able to determine that UUU specifies Phe; that Val, Leu, and Cys are specified by codons composed of G+2U; and that Trp and Gly are specified by codons composed of 2G+U.

Answer 6

3 reading frames each sequence

Answer 7

a. Yanofsky could have screened for Trp− auxotrophs by growing wild-type bacteria in the presence of a mutagen, and then plating the culture to obtain single colonies on a plate containing minimal medium supplemented with tryptophan. Each colony on the plate would then be replica-plated onto a minimal medium plate; colonies that fail to grow on the replica plate are Trp− auxotrophs. b. trpA− colonies should be able to grow only on minimal medium containing the end product tryptophan or any intermediate downstream of TrpA action (that is, an intermediate whose conversion into tryptophan does not depend on TrpA). The trpA− cells should not grow on minimal medium supplemented with any upstream intermediate.

Answer 8

the initial frameshift mutation changes the reading frame and there is a stop codon in this alternate reading frame. The second frameshift mutation of the opposite sign cannot restore function if it occurs after this stop codon. Because no suppressors can be identified between the original mutation and the premature stop codon, we can predict that the stop codon is likely to be very close to the site of the original frameshift mutation

Answer 9

a. no; a tRNA with the anticodon 5′ IAA cannot exist in an organism obeying the normal wobble rules shown in Figure 8.21b. Such an anticodon would recognize codons 5′ UUU (Phe), 5′ UUC (Phe), and 5′ UUA (Leu). A single tRNA species must carry only a single amino acid, and so the genetic code table tells you that a tRNA with the anticodon 5′ IAA cannot exist because it would not correspond to a single amino acid. b. Yes; a tRNA with the anticodon 5′ xm5s2U would recognize the codons 5′ UUA and 5′ UUG, both of which specify Leu. Therefore, this tRNA could exist and it would carry the amino acid Leu. c. For part (a), the RNA-like strand of the part of the tRNA gene encoding the anticodon is 5′ AAA; remember that A in the 5′ position of the anticodon is enzymatically altered to I after transcription. The template strand sequence is 5′ TTT. For part (b), the RNA-like strand is 5′ TAA, and the template strand is 5′ TTA. Remember that U in the 5′ position of the anticodon is enzymatically altered after transcription, in different ways in different tRNAs.

Answer 10

a. the anticodons that can have I as the wobble base are: 5′ ICG, 5′ ICC, 5′ IGG, 5′ IGA, 5′ IGU, 5′ IGC, 5′ IAG, 5′ IAC, and 5′ IAU. b. the only anticodon seqence that could never be allowed to have a 5' U is 5' UUA c. xm^5U can be the wobble base for all anticodons except for 5′ UUA. anticodons that would be able to have a xo5U as the wobble base are: 5′ UAG, 5′ UAC, 5′ UGA, 5′ UGG, 5′ UGU, 5′ UGC, 5′ UCG, and 5′ UCC. d. the minimal number of tRNAs required in humans is 32.

Answer 11

a. many of the 500 human tRNA genes are functionally redundant b. having many different genes that encode the same tRNA is one way to produce the large quantities of tRNA molecules that cells need. Also, multiple genes for each tRNA molecules protect the organism from the effects of mutation of tRNA genes

Answer 12

a. the minimum length of the coding region is 477 Amino acids x 3 bases/codon = 1431 base pairs b. XXX template strand 5′ GTAAGTTAACTTTCGACTAGTCCAGGGT 3′ RNA-like strand 3′ CATTCAATTGAAAGCTGATCAGGTCCCA 5′ XOX mRNA 5′ ACCCUGGACUAGUCGAAAGUUAACUUAC 3′ c. N…Pro Trp Thr Ser Arg Lys Leu Thr Tyr...C

Answer 13

a. The bottom strand is the RNA-like strand, so the top strand is the template. The RNA polymerase moves 3′-to-5′ along the template; that is, in the right-to-left direction with respect to the written sequence. b.5′ GCC AUG UAC AG|G GGG GCA UAG GGG 3′ c. Thr at the C terminus of the truncated protein could occur if the G base on the bottom strand that just precedes the junction between the intron and the first exon The sequence of the mutant mRNA is: 5′ GCC AUG UAC ACG UAA GUG UUU UGA UCC UCC CCC AGG GGG GCA UAG GGG 3′

Answer 14

a. very severe effect as there will be no functional protein b. probably mild effect if none of the last few amino acids are important for function c. very severe effect as most of the amino acids in the protein will be incorrect d. probably mild effect as only the last few amino acids will be affected e. no effect because by definition, a silent mutation maintains the same amino acid f. mild to no effect as the replacement is with an amino acid with similar chemical properties g. severe effect as the mutation is likely to destroy the protein′s activity h. could be severe if it affects the protein structure enough to hinder the action of the protein, or could be mild if the protein′s function can tolerate the substitution

Answer 15

a. The anticodon in this nonsense suppressor tRNA is either 5′ CUA 3′ or 5′ xm5UUA 3 b. the sequence of the template strand of DNA for the wild-type tRNAGln gene is either 5′ CAG 3′ or 5′ CAA 3′. c. at least two tRNAGln genes must exist in a wild-type B. adonis cell. One would code for the tRNAGln (anticodon 5′ CUG 3′ or 5′ xm5UUG 3′) that was changed into a nonsense suppressor, and the other gene would code for the tRNAGln (anticodon of 5′ xm5s2UUG 3′) that recognizes both normal Gln codons.

Answer 16

a. the original wild-type proline codon must have been 5′ CCG, so the sequence of the DNA in this region must have been: 5′ CCG 3′ 3′ GGC 5′ b. Trp at position 5 is compatible with the function of the enzyme encoded by the gene Ser at position 5 is not compatible with enzyme function - it is a nonconservative substitution c. Strain C does not have any detectable protein, so it is likely to be a nonsense mutation that stops translation after only a few amino acids have been added. d. Because the reversion mutation does not map to the enzyme locus, it must be a mutation in a tRNA gene that would encode a nonsense suppressing tRNA.

Answer 17

a. . A tRNA with the anticodon 5′ UUA 3′ (U modified to xm5s2U) could recognize both nonsense codons b. This nonsense suppressing tRNA would suppress 5′ UAA 3′ and 5′ UAG 3′. c. no d. Gln, Glu, Lys, Leu, Ser, Tyr, and possibly Trp.

Answer 18

a. The mutant gene was a tRNA whose anticodon could recognize the stop codon 5′ UAG 3′. b. Each M protein had a single amino acid change in its polypeptide sequence; the amino acid with which the nonsense suppressor tRNA was charged was inserted in place of the amino acid specified by the normal codon that was changed to UAG by the nonsense mutation