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# The Biochemical Basis of Heredity Practice Problems

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By — McGraw-Hill Professional
Updated on Aug 19, 2011

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## The Biochemical Basis of Heredity Practice Problems

#### Practice 1

The percentage of nucleotide A in DNA isolated from human liver is observed to be 30.7%. What is the expected percentage of (a) T (b) G (c) C?

#### Solution 1

Because the amount of A always equals T, the percentage of T is expected to be very close to 30.7%. G and C together would make up the remainder, or 100 – (30.7 + 30.7) = 38.6; thus, the percentage of C and G separately would be expected to equal half of 38.6, or 19.3.

#### Practice 2

How many triplet codons can be made from the four ribonucleotides A, U, G, and C containing (a) no uracils, (b) one or more uracils?

#### Solution 2

1. Since uracil represents 1 among 4 nucleotides, the probability that uracil will be the first letter of the codon is 1/4; and the probability that U will not be the first letter is 3/4. The same reasoning holds true for the second and third letters of the codon. The probability that none of the three letters of the codon are uracils is (3/4)3 = 27/64.
2. The number of codons containing at least one uracil is 1 – (27/64) = 37/64.

#### Practice 3

A synthetic polyribonucleotide is produced from a mixture containing U and C in the relative frequencies of 5 : 1, respectively. Assuming that the ribonucleotides form in a random linear array, predict the relative frequencies in which the various triplets are expected to be formed.

#### Solution 3

The frequencies with which the different triplets are expected to be formed by chance associations can be predicted by combining independent probabilities through multiplication as follows:

UUU should occur with a frequency of (5/6) (5/6) (5/6) = 125/216.
Codons with 2U and IC = (5/6)2 (1/6) = 25/216 each (UUC, UCU, CUU)
Codons with 1U and 2C = (5/6)(1/6)2 = 5/216 each (UCC, CUC, CCU)
CCC = (1/6)3 = 1/216

#### Practice 4

Four single mutant strains of Neurospora are unable to grow on minimal medium unless supplemented by one or more of the substances A–F. In the following table, growth is indicated by + and no growth by 0. In addition, both strains 2 and 4 grow if E and F, or C and F, are added to minimal medium. Diagram a biochemical pathway consistent with the data involving all six metabolites, indicating where the mutant block occurs in each of the four strains.

#### Solution 4

Strain 1 will grow only if given substance A. Therefore, the defective enzyme produced by the mutant gene in this strain must act sometime prior to the formation of substance A and after the formation of substances B, C, D, E, and F. In other words, this mutation is probably causing a metabolic block in the last step of the biochemical sequence in the synthesis of substance A.

Strain 2 grows if supplemented by eitherAorD, but not by B. Therefore, the metabolic block in strain 2 must occur after B but before A. Furthermore, since the dual addition of substances E and F, or C and F, allows strain 2 to grow, we can infer that the intervenient pathway is split, with E andCin one line and F in the other. SubstanceDcould be at one of two positions as shown below.

Strain 3 grows if supplemented by A or C, but not by D. Therefore, D cannot immediately precede A (as shown above), and the metabolic block in strain 3 must precede the formation of C but not E.

Strain 4 can grow if given dual supplementation of E and F, or C and F, but not if given D alone. The mutation in strain 4 apparently cannot split D into E and F.

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