CHAPTER 9 the Molecular Genetics of Gene Expression Assigned: 9.1

CHAPTER 9 The Molecular Genetics of Gene Expression

Assigned: 9.1, 9.2, 9.3, 9.4, 9.9, 9.13,9.14, 9.16

9-1. Only statement (d) is true.

9-2. The 5’ end of the RNA transcript is transcribed first, so (a) through (c) are all wrong. The 5’ end of the RNA transcript is transcribed from the 3’ end of the DNA template, so (e) is wrong. The additional statement in (e) that the amino end of the polypeptide is synthesized first serves as confirmation of the correctness of statement (d).

9-3.The initiation codon is AUG and the stop codons are UAA, UAG, and AGA. In a random sequence, the probability of an initiation codon is (1/4)3 = 1/64, and the probability of a stop codon is 3(1/4)3 = 3/64. The average distance between stop codons is 64/3 = 21.3 codons.

9-4. The mRNA sequence is 5’-AAAAUGCCCUUAAUCUCAGCGUCCUAC- 3’ and the first AUG is the initiation codon; so the resulting gamino acid sequence is Met—Pro—Leu—Ile—Ser—Ala—Ser—Tyr or in the single-letter abbreviations M—P—L—I—S—A—S—Y.

9-5. Because the pairing of codon and anticodon are antiparallel, it is convenient to write the anticodon as 3’-UAI-5’. The first two codon positions are therefore 5’-AU-3’, and the third is A or U or C (because each can pair with I). The possible codons are 5’-AUA-3’, 5’-AUU-3’, and 5’-AUC-3’, all of which code for isoleucine.

9-6. (a) Methionine has 1 codon, histidine 2, and threonine 4, so the total number is 1 × 2 × 4 = 8. The sequence AUGCAYACN encompasses them all. (b) Because arginine has six codons, the possible number of sequences coding for Met–Arg–Thr is 1 × 6 × 4 = 24. The sequences are AUGCGNACN (16 possibilities) and AUGAGRACN (8 possibilities).

n 9-7. The requirement is that 2 ≥ 20, or n = 5. If the codons were of four 4 nucleotides, then only 16 amino acids could be specified because 2 = 16. A 5 codon of five nucleotides does the job because 2 = 32; in fact, it leaves a little room for stop codons and redundancy.

9-8. With a G at the 5’ end, the first codon is GUU, which codes for valine; and with a G at the 3’ end, the last codon is UUG, which codes for leucine. 9-9. The alternating polymer GUGUGU encodes the alternating polypeptide Val—Cys.

9-13.

h j b a, l k m c g mRNA

b DNA f f f a

e e

9-14. (a) The deletion must have fused the amino-coding terminus of the B gene with the carboxyl-coding terminus of the A gene. The nontranscribed strand must therefore be oriented 5’-B–A-3’. (b) The number of bases deleted must be a multiple of three, otherwise the carboxyl terminus would not have the correct reading frame.

9-15. (a) DNA double helix A C T T G A T A C G C A A C C C G T A T C G G T T G A A C T A T G C G T T G G G C A T A G C C A mRNA U G A A C U A U G C G U U G G G C A U A G C C A tRNA anticodon A C U U G A U A C G C A A C C C G U A U C G G U

(b) The top strand is transcribed.

U GAA CUA UGC GUU GGG CAU AGC CA Glu Leu Cys Val Gly His Ser

UG AAC UAU GCG UUG GGC AUA GCC A Asn Tyr Ala Leu Gly Ile Ala

(d) In a Southern blot hybridization, double-stranded DNA is denatured and transferred to a filter, so the probe can be complementary to either strand. Thus either strand of the DNA molecule may be used. (The mRNA could also be used, though this is not usually done in practice because RNA is easily degraded.)

(e) A Northern blot hybridization detects mRNA immobilized on a membrane. Thus the probe should be a strand complementary to the mRNA and should therefore be the top strand of the DNA as written here.

9-16. The wildtype and double-mutant mRNA codons are:

Wildtype: 5’-AAR AAR UAY CAY CAR UGG ACN UGY AAY-3’ Double Mutant: 5’-AAR CAR AUH CCN CCN GUN GAY AUG AAY-3’

Comparing these sequence, it can be deduced that the first frameshift addition is the C at position 4 in the double-mutant gene. Realigning the sequences to take this into account, yields

Wildtype: 5’-AAR AAR UAY CAY CAR UGG ACN UGY AAY-3’ Double Mutant: 5’-AARC ARA UHC CNC CNG UNG AYA UGA AY-3’

Now it is also clear that the single-nucleotide deletion in the wildtype gene must be the Y in the fourth nucleotide position from the 3’ end. Realigning again to take this into account, we have

Wildtype: 5’-AAR AAR UAY CAY CAR UGG ACN UGY AAY-3’ Double Mutant: 5’-AARC ARA UHC CNC CNG UNG AYA UG AAY-3’ Hence the complete sequences, as thoroughly as they can be resolved from the data, are

Wildtype: 5’-AAR AAA UAC CAC CAG UGG ACA UGY AAY-3’ Double Mutant: 5’-AAR CAA AUA CCA CCA GUG GAC AUG AAY-3’

9-17. There are 3 possible reading frames for the alternating polymer GUCGUC, each of which encodes a different repeating polymer. One has repeating Val (GUC), another repeating Ser (UCG), and the third repeating Arg (CGU).

9-18. (a) Lys has the codon formula AAR, hence the next amino acid must have the formula ARN; this includes Asn (AAY), Lys (AAR), Ser (AGY), and Arg (AGR). (b) Met has one codon, AUG, and could be followed by Cys (UGY) or Trp (UGG). (c) Tyr has the formula UAY and could be followed by the (AUU, AUC, AUA), Met (AUG), or Thr (ACN). (d) Tyr has one codon, UGG, and could be followed only by Gly (GGN).

9-19. For an error rate of 10--3, the probability of no errors in 300 amino acids is (1 – 10--3)300 = 0.74. (Alternatively you can use the zero term of the Poisson distribution, noting that the mean number of errors would be 300 x 1 x 10--3 = 0.30, and hence the probability of no error equals e—0.30 = 0.74.) The probability of no errors in a polypeptide chain of 1000 amino acids equals (1 – 10-3)1000 = 0.368 (or, alternatively, e--1 = 0.368), hence the probability of no errors in a complete tetramer is (0.368)4 = 0.018. (Less than 2 percent of tetramers are completely error free).

9-20. The 25 ribonucleotides encode a polypeptide of 22 amino acids because a shift of reading fame occurs each time the circle is traversed until a stop codon is finally encountered. The complete amino acid sequence of the product is Met—Ala—Asp—Ser—Ala—Asp—Leu—Ala—Asp—Gly—Arg—Phe—Gly—Ar g—Phe—Ser-_Arg—Trp—Gln—Ilu—Arg—Glu—Ile, or in the single letter abbreviations. M—A—D—S—A—D—L—A—D—G—R—F—G—R—F—S—R—W—Q—I—R— E—L