Chapter 2

15.  PFGE separates DNA molecules by size. When DNA is carefully isolated from Neurospora (which has seven different chromosomes), seven bands should be produced with the use of this technique. Similarly, the pea has seven different chromosomes and will produce seven bands (homologous chromosomes will comigrate as a single band).

18.  The key function of mitosis is to generate two daughter cells genetically identical with the original parent cell.

22.  As cells divide mitotically, each chromosome consists of identical sister chromatids that are separated to form genetically identical daughter cells. Although the second division of meiosis appears to be a similar process, the “sister” chromatids are likely to be different from each other. Recombination in earlier meiotic stages will have swapped regions of DNA between sister and nonsister chromosomes such that the two daughter cells of this division are typically not genetically identical.

26.  Yes. Half of our genetic makeup is derived from each parent, half of each parent’s genetic makeup is derived from half of each of their parents’, etc.

30.  (5) Synapsis (chromosome pairing)

35.  The progeny ratio is approximately 3 : 1, indicating classic heterozygous-by-heterozygous mating. Because Black (B) is dominant over white (b),

Parents:

B/b × B/b

Progeny:

3 black : 1 white (1 B/B : 2 B/b : 1 b/b)

39.  The fact that about half of the F1 progeny are mutant suggests that the mutation that results in three cotyledons is dominant and the original mutant was heterozygous. If C = the mutant allele and c = the wild-type allele, the cross is as follows:

P

C/c × c/c

F1

C/c three cotyledons

c/c two cotyledons

44.  p (child has galactosemia) = p (John is G/g) × p (Martha is G/g) × p (both parents passed g to the child) = (2/3)(1/4)(1/4) = 2/48 = 1/24

50. 

  1. The disorder appears to be dominant because all affected individuals have an affected parent. If the trait were recessive, then I-1, II-2, III-1, and III-8 would all have to be carriers (heterozygous for the rare allele).

  2. With the assumption of dominance, the genotypes are

    I:   

    d/d, D/d

    II:   

    D/d, d/d, D/d, d/d

    III:   

    d/d, D/d, d/d, D/d, d/d, d/d, D/d, d/d

    IV:   

    D/d, d/d, D/d, d/d, d/d, d/d, d/d, D/d, d/d

  3. The probability of an affected child (D/d) equals 1/2, and the probability of an unaffected child (d/d) equals 1/2. Therefore, the chance of having four unaffected children (since each is an independent event) is (1/2) × (1/2) × (1/2) × (1/2) = 1/16.

56. 

  1. Sons inherit the X chromosome from their mothers. The mother has free earlobes; the son has attached earlobes. If the allele for free earlobes is dominant and the allele for attached earlobes is recessive, then the mother could be heterozygous for this trait and the gene could be X linked.

  2. It is not possible from the data given to decide which allele is dominant. If attached earlobes is dominant, then the father would be heterozygous and the son would have a 50% chance of inheriting the dominant attached earlobes allele. If attached earlobes is recessive, then the trait could be autosomal or X linked, but, in either case, the mother would be heterozygous.

60.  Let H = hypophosphatemia and h = normal. The cross is H/Y × h/h, yielding H/h (females) and h/Y (males). The answer is 0%.

65. 

  1. XC/Xc, Xc/Xc

  2. p (color-blind) × p (male) = (1/2)(1/2) = 1/4

  3. The girls will be 1 normal (XC/Xc) : 1 color-blind (Xc/Xc).

  4. The cross is XC/Xc × Xc/Y, yielding 1 normal : 1 color-blind for both sexes.

73. 

  1. The pedigree suggests that the allele causing red hair is recessive because most red-haired individuals are from parents without this trait.

  2. Observation of those around us makes the allele appear to be somewhat rare.

77.  Note that only males are affected and that, in all but one case, the trait can be traced through the female side. However, there is one example of an affected male having affected sons. If the trait is X linked, this male’s wife must be a carrier. Depending on how rare this trait is in the general population, that could be unlikely, suggesting that the disorder is caused by an autosomal dominant allele with expression limited to males.