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Modern Genetic Analysis Answers to Selected Problems
Chapter 3*

1. | There are three codons for isoleucine: 5-AUU-3, 5-AUC-3, and 5-AUA-3. Possible anticodons are 3-UAA-5 (complementary), 3-UAG-5 (complementary), and 3-UAI-5 (wobble). 5-UAU-3, although complementary, would also base pair with 5-AUG-3 (methionine) because of wobble and therefore would not be an acceptable alternative. See question

4. | a and b. 5-UUG GGA AGC-3 c and d. Assume that the reading frame starts at the first base: See question

7. | a. b. 5 4 2 1 3 indicates where in the pathway each mutant is blocked See question

9. | (a). White; (b). blue. See question

11. | a. Yes; transcript yes, polypeptide maybedepends on mutation type. b. Yes; transcript yes, polypeptide no. c. Yes; transcript no, polypeptide no. d. Yes; transcript yes; polypeptide no. See question

12. | Protein function can be destroyed by a mutation that causes the substitution of a single amino acid even though the protein has the same immunological properties. For example, an enzyme requires very specific amino acids in exact positions within its active site. A substitution of one of these key amino acids might have no effect on the overall size and shape of the protein while completely destroying the enzymatic activity. See question

14.

a. The allele sn will show dominance over sf because there will be only 40 units of square factor in the heterozygote.
b. No. Here the functional allele is recessive.
c. The allele sf may become dominant over time by one of several mechanisms: it could mutate in such a way that it produces 50 units; other genes might mutate to lower the cells to 40 units; or other genes might mutate and increase the production of sf.
See question

17.

a. Recessive. The normal allele provides enough enzyme to be sufficient for normal function (the definition of haplosufficient).
b. There are many ways in which to mutate a gene to destroy enzyme function. One possible mutation might be a frameshift mutation within an exon of the gene. If a single base pair were assumed to have been deleted, the mutation would completely alter the translational product 3 to the mutation.
c. Hormone replacement could be given to the patient.
See question

20. | Mutant 1: null mutation; no product is made. Mutant 2: leaky mutation; not enough product is made for normal function. Mutant 3: nonsense mutation/frameshift; a truncated protein product is made. Mutant 4: Substitution mutation; protein is the same size but a single altered amino acid destroys its function. See question

22. | Transcription (mRNA/DNA); translation (tRNA/mRNA); translation initiation (rRNA/mRNA); splicing. See question

© 1999 by W. H. Freeman and Company. All rights reserved.