pierceessentials3e_answers1

Chapter 5

1. ¼ wild-type eyes, wild-type wings; ¼ red eyes, wild-type wings; ¼ wild-type eyes, white-banded wings; ¼ red eyes, white-banded wings.
2. The recombinant progeny are the 19 with red eyes, wild-type wings and the 16 with wild-type eyes, white-banded wings.
  
  recombination frequency = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0%
  
  The distance between the genes is 4 map units.
The genes are linked and have not assorted independently.
Because the genes for leaf shape and fruit spines are 32.6 m.u. apart, we expect that 32.6% of the progeny of the cross will be recombinants. There will be two types of recombinants (those with heart-shaped leaves and few spines and those with normal-shaped leaves and numerous spines), so each recombinant phenotype will constitute 16.3% of the progeny. The nonrecombinant progeny (those with heart-shaped leaves and numerous spines and those with normal-shaped leaves and few spines) will make up the remainder of the progeny (100% – 32.6% = 67.4%), which will be equally divided between the two nonrecombinant phenotypes (67.4%/2% = 33.7% each).

Heart-shaped, numerous spines 33.7%

Normal-shaped, few spines 33.7%

Heart-shaped, few spines 16.3%

Normal-shaped, numerous spines 16.3%

Genotype	Body color	Eyes	Bristles	Proportion
(a) e⁺ ro⁺ f⁺	normal	normal	normal	20%
e⁺ ro⁺ f	normal	normal	forked	20%
e ro f⁺	ebony	rough	normal	20%
e ro f	ebony	rough	forked	20%
e⁺ ro f⁺	normal	rough	normal	5%
e⁺ ro f	normal	rough	forked	5%
e ro⁺ f⁺	ebony	normal	normal	5%
e ro⁺ f	ebony	normal	forked	5%
(b) e⁺ ro⁺ f⁺	normal	normal	normal	5%
e⁺ ro⁺ f	normal	normal	forked	5%
e ro f⁺	ebony	rough	normal	5%
e ro f	ebony	rough	forked	5%
e⁺ ro f⁺	normal	rough	normal	20%
e⁺ ro f	normal	rough	forked	20%
e ro⁺ f⁺	ebony	normal	normal	20%
e ro⁺ f	ebony	normal	forked	20%

Gene f is unlinked to either of these groups; it is on a third linkage group.

1. V is the middle gene.
2. The Wx–V distance = 7 m.u. and the Sh–V distance = 30 m.u. The Wx–Sh distance is the sum of these two distances, or 37 m.u.
3. Coefficient of coincidence = 0.80; interference = 0.20.

1. All the progeny receive p sh-1 Hb² from the male parent, shown as the lower chromosome in each progeny type. The upper chromosomes are the products of meiosis in the heterozygous parent, and are identified in the table. The nonrecombinants have p sh-1 Hb² or P Sh-1 Hb¹; the double crossovers have p Sh-1 Hb² or P sh-1 Hb¹. The two classes differ in the Sh-1 locus; therefore, Sh-1 is the middle locus.
2. P and Sh-1: Recombinants have P sh-1 or p Sh-1. Recombination frequency = (57 + 45 + 1)/708 = 0.145 = 14.5 m.u.
  
  Sh-1 and Hb: Recombinants have Sh-1 Hb² or sh-1 Hb¹. Recombination frequency = (6 + 5 + 1)/708 = 0.017 = 1.7 m.u.
3. Expected double crossovers = RF1 × RF2 × total progeny = 0.145(0.017)(708) = 1.7
  
  Coefficient of coincidence = number of observed double crossovers/number of expected double crossovers = 1/1.7 = 0.59
  
  Interference = 1 − coefficient of coincidence = 0.41

Genotype	Body	Eyes	Wings	Proportion
B⁺ pr⁺ vg⁺	normal	normal	normal	40.7%
b pr vg	black	purple	vestigial	40.7%
b⁺ pr⁺ vg	normal	normal	vestigial	6.3%
b pr vg⁺	black	purple	normal	6.3%
b⁺ pr vg	normal	purple	vestigial	2.8%
b pr⁺ vg⁺	black	normal	normal	2.8%
b⁺ pr vg⁺	normal	purple	normal	0.2%
b pr⁺ vg	black	normal	vestigial	0.2%