Chapter 6

1. Red

2. Purple

3. 9	M1/– ; M2/–	purple
   3	m1/m1 ; M2/–	blue
   3	M1/– ; m2/m2	red
   1	m1/m1 ; m2/m2	white

4. The mutant alleles do not produce functional enzyme. However, enough functional enzyme must be produced by the single wild-type allele of each gene to synthesize normal levels of pigment.

1. The original cross was a dihybrid cross. Both oval and purple must represent an incomplete dominant phenotype.

2. A long, purple × oval, purple cross is as follows:

   

1. The sex ratio is expected to be 1 : 1.

2. The female parent was heterozygous for an X-linked recessive lethal allele, which would result in 50% fewer males than females.

3. Half of the female progeny should be heterozygous for the lethal allele, and half should be homozygous for the nonlethal allele. Individually mate the F₁ females and determine the sex ratio of their progeny.

1. The mutations are in two different genes because the heterokaryon is prototrophic (the two mutations complemented each other).

2. leu1⁺ ; leu2⁻ and leu1⁻ ; leu2⁺

3. With independent assortment, expect

   

1. P  A/a (frizzle) × A/a (frizzle)

   F₁  1 A/A (normal) : 2 A/a (frizzle) : 1 a/a (woolly)

2. If A/A (normal) is crossed with a/a (woolly), all offspring will be A/a (frizzle).

1. The genotypes are

   P	B/B ; i/i × b/b ; I/I
   F1	B/b ; I/i	hairless
   F2	9 B/– ; I/–	hairless
   	3 B/– ; i/i	straight
   	3 b/b ; I/–	hairless
   	1 b/b ; i/i	bent

   837

2. The genotypes are B/b ; I/i × B/b ; i/i.

1. Cross-feeding is taking place, whereby a product made by one strain diffuses to another strain and allows growth of the second strain.

2. For cross-feeding to take place, the growing strain must have a block that is present earlier in the metabolic pathway than the block in the strain from which the growing strain is obtaining the product for growth.

3. The data suggest that the metabolic pathway is

   trpE → trpD → trpB

4. Without some tryptophan, there would be no growth at all, and the cells would not have lived long enough to produce a product that could diffuse

1. The best explanation is that Marfan’s syndrome is inherited as a dominant autosomal trait.

2. The pedigree shows both pleiotropy (multiple affected traits) and variable expressivity (variable degree of expressed phenotype).

3. Pleiotropy indicates that the gene product is required in a number of different tissues, organs, or processes. When the gene is mutant, all tissues needing the gene product will be affected. Variable expressivity of a phenotype for a given genotype indicates modification by one or more other genes, random noise, or environmental effects.

1. This type of gene interaction is called epistasis. The phenotype of e/e is epistatic to the phenotypes of B/– or b/b.

2. The inferred genotypes are as follows:

   I	1 (B/b E/e)	2 (B/b E/e)
   II	1 (b/b E/e)	2 (B/b E/e)	3 (–/– e/e)	4 (b/b E/–)
   	5 (B/b E/e)	6 (b/b E/e)
   III	1 (B/b E/–)	2 (–/b e/e)	3 (b/b, E/–)	4 (B/b E/–)
   	5 (b/b E/–)	6 (B/b E/–)	7 (–/b e/e)

1. A multiple-allelic series has been detected: superdouble > single > double.

2. Although the explanation for part a does rationalize all the crosses, it does not take into account either the female sterility or the origin of the superdouble plant from a double-flowered variety.

1. A trihybrid cross would give a 63 : 1 ratio. Therefore, there are three R loci segregating in this cross.

2. P	R1/R1 ; R2/R2 ; R3/R3 × r1/r1 ; r2/r2 ; r3/r3
   F1	R1/r1 ; R2/r2 ; R3/r3
   F2	27	R1/–; R2/–; R3/–	red
   	9	R1/–; R2/–; r3/r3	red
   	9	R1/–; r2/r2 ; R3/–	red
   	9	r1/r1 ; R2/–; R3/–	red
   	3	R1/–; r2/r2 ; r3/r3	red
   	3	r1/r1 ; R2/–; r3/r3	red
   	3	r1/r1 ; r2/r2 ; R3/–	red
   	1	r1/r1 ; r2/r2 ; r3/r3	white

3. (1) To obtain a 1 : 1 ratio, only one of the genes can be heterozygous. A representative cross is R₁/r₁ ; r₂/r₂ ; r₃/r₃ × r₁/r₁ ; r₂/r₂ ; r₃/r₃.

   (2) To obtain a 3 red : 1 white ratio, two alleles must be segregating and they cannot be within the same gene. A representative cross is R₁/r₁ ; R₂/r₂ ; r₃/r₃ × r₁/r₁ ; r₂/r₂ ; r₃/r₃.

   (3) To obtain a 7 red : 1 white ratio, three alleles must be segregating, and they cannot be within the same gene. The cross is R₁/r₁ ; R₂/r₂ ; R₃/r₃ × r₁/r₁ ; r₂/r₂ ; r₃/r₃.

4. The formula is 1 − ( )n, where n = the number of loci that are segregating in the representative crosses in part c.

1. Intercrossing mutant strains that all have a common recessive phenotype is the basis of the complementation test. This test is designed to identify the number of different genes that can mutate to a particular phenotype. In this problem, if the progeny of a given cross still express the wiggle phenotype, the mutations fail to complement and are considered alleles of the same gene; if the progeny are wild type, the mutations complement and the two strains carry mutant alleles of separate genes.

2. These data identify five complementation groups (genes).

3. 

   Conclusion: 1 and 5 are both mutant for gene A.

   (The relevant cross a⁺/a⁺ · b²/b² × a²/a² · b⁵/b⁵ gives the following hybrid.)

   

   Conclusion: 2 and 5 are mutant for different genes.