Chapter 3

13.  The genotype of the daughter cells will be identical with that of the original cell: (f) A/a ; B/b.

18.  Mitosis produces daughter cells having the same genotype as that of the original cell: A/a ; B/b ; C/c.

21.  His children will have to inherit the satellite-containing 4 (probability = 1/2), the abnormally staining 7 (probability = 1/2), and the Y chromosome (probability = 1/2). To inherit all three, the probability is (1/2)(1/2)(1/2) = 1/8.

26.  With the assumption of independent assortment and simple dominant–recessive relations of all genes, the number of genotypic classes expected from selfing a plant heterozygous for n gene pairs is 3n and the number of phenotypic classes expected is 2n.

29.  a. and b. Cross 2 indicates that purple (G) is dominant over green (g), and cross 1 indicates that cut (P) is dominant over potato (p).

Cross 1: G/g ; P/p × g/g ; P/p

There are 3 cut : 1 potato, and 1 purple : 1 green.

Cross 2: G/g ; P/p × G/g ; p/p

There are 3 purple : 1 green, and 1 cut : 1 potato.

Cross 3: G/G ; P/p × g/g ; P/p

There is no green, and there are 3 cut : 1 potato.

Cross 4: G/g ; P/P × g/g ; p/p

There is no potato, and there is 1 purple : 1 green.

Cross 5: G/g ; p/p × g/g ; P/p

There is 1 cut : 1 potato, and there is 1 purple : 1 green.

34.  The crosses are

Cross 1:

stop-start female × wild-type male → all stop-start progeny

Cross 2:

wild-type female × stop-start male → all wild-type progeny

mtDNA is inherited only from the “female” in Neurospora.

834

40. 

  1. There should be nine classes corresponding to 0, 1, 2, 3, 4, 5, 6, 7, 8 “doses.”

  2. There should be 13 classes corresponding to 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 “doses.”

49.  a. and b. Begin with any two of the three lines and cross them. If, for example, you began with a/a ; B/B ; C/C × A/A ; b/b ; C/C, all the progeny would be A/a ; B/b ; C/C. Crossing two of them would yield

9

A/– ; B/– ; C/C

3

a/a ; B/– ; C/C

3

A/– ; b/b ; C/C

1

a/a ; b/b ; C/C

The a/a ; b/b ; C/C genotype has two of the genes in a homozygous recessive state and is found in 1/16 of the offspring. If that genotype were crossed with A/A ; B/B ; c/c, all the progeny would be A/a ; B/b ; C/c. Crossing two of them (or “selfing”) would lead to a 27:9:9:9:3:3:3:1 ratio, and 1/64 of the progeny would be the desired a/a ; b/b ; c/c.

There are several different routes to obtaining a/a ; b/b ; c/c, but the one just outlined requires only four crosses.

56. 

  1. Let B = brachydactylous, b = normal, T = taster, and t = nontaster. The genotypes of the couple are B/b ; T/t for the male and b/b ; T/t for the female.

  2. For all four children to be brachydactylous, p = (1/2)4 = 1/16.

  3. For none of the four children to be brachydactylous, p = (1/2)4 = 1/16.

  4. For all to be tasters, p = (3/4)4 = 81/256.

  5. For all to be nontasters, p = (1/4)4 = 1/256.

  6. For all to be brachydactylous tasters, p = (1/2 × 3/4)4 = 81/4096.

  7. The probability of not being a brachydactylous taster is 1 − (the probability of being a brachydactylous taster), or 1 − (1/2 × 3/4) = 5/8. The probability that all four children are not brachydactylous tasters is (5/8)4 = 625/4096.

  8. The probability that at least one is a brachydactylous taster is 1 − (the probability of none being a brachydactylous taster), or 1 − (5/8)4.