Genetic Interactions Practice Problems (page 2)

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By — McGraw-Hill Professional
Updated on Sep 20, 2011

Solution 4

  1. Ninety-four percent of 16 is ~ 15 (0:94 × 16 = 15:04; or 15/16 × 100 = 93:75) and 6% of 16 is ~ 1, thus, the ratio represented here is 15 triangular : 1 ovoid
  2. The genotype is the double recessive, aabb, which occurs 1/16 of the time. The dominant condition A- or B- is epistatic to either recessive condition, so A-B- (9/16), A-bb (3/16), aaB- (3/16) all exhibit the dominant phenotype (9 + 3 + 3 = 15).

Practice 5

Two white-flowered strains of the sweet pea (Lathyrus odoratus) were crossed, producing an F1 with only purple flowers. Random crossing among the F1 produced 96 progeny plants, 53 exhibiting purple flowers, and 43 with white flowers. (a) What phenotypic ratio is approximated by the F2? (b) What type of interaction is involved? (c) What were the probable genotypes of the parental strains?

Solution 5

  1. To determine the phenotypic ratio in terms of familiar sixteenths, the following proportion for white flowers may be made: 43/96 = x/16, from which x = 7.2. That is, 7.2 white : 8.8 purple, or approximately a 7 : 9 ratio. We might just as well have arrived at the same conclusion by establishing the proportion for purple flowers: 53/96 = x/16, from which x = 8.8 purple.
  2. A 7 : 9 ratio is characteristic of duplicate recessive genes where the recessive genotype at either or both of the loci produces the same phenotype.
  3. If aa or bb or both could produce white flowers, then only the genotype A-B- could produce purple. For two white parental strains (pure lines) to be able to produce an all-purple F1, they must be homozygous for different dominant-recessive combinations. Thus,
  4. Genetic Interactions Solved Problems

Practice 6

A dominant gene S in Drosophila produces a peculiar eye condition called "star." Its recessive allele S+ produces the normal eye of wild type. The expression of S can be suppressed by the dominant allele of another locus, Su-S. The recessive allele of this locus, Su-S+,has no effect on S+. (a) What type of interaction is operative? (b) When a normal-eyed male of genotype Su-S / Su-S, S/S is crossed to a homozygous wild-type female of genotype Su-S+ / Su-S+, S+ / S+, what phenotypic ratio is expected in the F1 and F2?

Solution 6

  1. Dominant and recessive interaction; both dominant and recessive alleles interact to result in an eye phenotype
  2. Genetic Interactions Solved Problems
  3. All the F1 offspring are wild type, due to the presence of the dominant Su-S allele. F2: 3/16 S/-, Su-S+ / Su-S+ star eye: 13/16 wild type.

Practice 7

The following pedigree involving two-factor epistasis shows the transmission of swine coat colors through three generations:

Genetic Interactions Solved Problems

Assume the offspring of II5 × II6 shown in this pedigree occur in the ratio expected from the genotypes represented by their parents. How are these colors most likely inherited?

Solution 7

Notice first that three phenotypes are expressed in this pedigree. This rules out epistatic combinations producing only two phenotypes such as those expressed in dominant and recessive interaction (13 : 3), duplicate dominant genes (15 : 1), and duplicate recessive genes (9 : 7). Epistatic gene interactions producing three phenotypes are those expressed in dominant epistasis (12 : 3 : 1), recessive epistasis (9 : 3 : 4), and dominant genes with cumulative action (9 : 6 : 1). Let us proceed to solve this problem by making an assumption and then applying it to the pedigree to see if the phenotypes shown there can be explained by our hypothesis.

Case 1.   Assume dominant epistasis is operative. The genotypes responsible for the three phenotypes may be represented as follows: A-B- and A-bb = first phenotype, aaB- = second phenotype, and aabb = third phenotype. We must now determine which of the phenotypes represented in this pedigree corresponds to each of the genotypic classes. Obviously the only pure-line phenotype is the third one. Offspring of the mating aabb × aabb would all be phenotypically identical to the parents. The mating I4 × I5 appears to qualify in this respect and we shall tentatively assume that white coat color is represented by the genotype aabb. Certain matings between individuals with the dominant epistatic gene A could produce three phenotypically different types of offspring (e.g., AaBb × AaBb). Such a mating is observed between II2 and II3. Therefore, we might assume red color to be represented by the genotype A-. Sandy color must then be represented by genotype aaB-. Matings between sandy individuals could produce only sandy (aaB-) or white (aabb) progeny. However, sandy parents I1 × I2 produce white and red progeny (II1, II2). Therefore the assumption of dominant interaction must be wrong.

Case 2.   Assume recessive epistasis to be operative. The genotypes responsible for the three phenotypes in this case may be represented as follows: A-B- as first phenotype, A-bb as second phenotype, and aaB- and aabb as third phenotype. As pointed out in case 1, matings between individuals of genotype AaBb are the only kind among identical phenotypes that are capable of producing all three phenotypes in the progeny. Thus A-B- should represent red (e.g., II2 × II3). The aa genotypes breed true, producing only white individuals (I4 × I5). Sandy is produced by genotype A-bb. Sandy × sandy (I1 × I2) could not produce the red offspring (II2). Therefore the assumption of recessive interaction must be wrong.

Case 3.   Assume that duplicate genes with cumulative action are interacting. The genotypes responsible for the three phenotypes in this case may be represented as follows: A-B- as first phenotype, A-bb and aaB- as second phenotype, and aabb as third phenotype. As explained in the previous two cases, A-B- must be red and aabb must be white. If we assume that any dominant genotype at either the A locus or B locus contributes one unit of pigment to the phenotype, then either the genotype aaB- or A-bb could be sandy; we further assume that the presence of both dominant genes (A-B-) would contribute two units of pigment to produce a red phenotype. Thus, the mating II5 (AaBb) red × II6 (aaBb) sandy would be expected to produce offspring phenotypes in the following proportions:

    Genetic Interactions Solved Problems

The same phenotypic ratio would be expected if II6 were Aabb. These expectations correspond to the ones given in the pedigree (III4–III11) and therefore the hypothesis of duplicate genes with cumulative action is consistent with the data.

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