Genetic Interactions Practice Problems (page 3)

based on 5 ratings
By — McGraw-Hill Professional
Updated on Sep 20, 2011

Practice 8

At least three loci are known to govern coat colors in mice. The genotype C- will allow pigment to be produced at the other two loci. The recessive genotype cc does not allow pigment production, resulting in "albino." The "agouti" pattern depends upon the genotype A-, and nonagouti upon the recessive aa. The color of the pigment may be black (B-) or chocolate (bb). Five coat colors may be produced by the action of alleles at these three loci:

Genetic Interactions Solved Problems

  1. What phenotypic frequencies are expected in the F2 from crosses of pure black with albinos of type AAbbcc?
  2. A cinnamon male is mated to a group of albino females of identical genotype and among their progeny were observed 43 wild type, 40 cinnamon, 39 black, 41 chocolate, and 168 albino. What are the most probable genotypes of the parents?

Solution 8

  1. Genetic Interactions Solved Problems
  2. Genetic Interactions Solved Problems
  3. In part (b), the cinnamon progeny, A-bbC-, indicate b in the female parents. The black progeny, aaB-C-, indicate a in both parents, and B in the female parents. The chocolate progeny, aabbC-, indicate a in both parents, and b in the females. The albinos indicate c in themale. The genotype of themale is now known to be AabbCc. But the genotype of the albino females is known only to be a-Bbcc. They could be either AaBbcc or aaBbcc

    Genetic Interactions Solved Problems

Practice 9

Lewis-a blood group substance appears on the human red blood cell when the dominant gene Le is present, but is absent if the dominant gene of the "secretor" locus Se is present. Suppose that from a number of families where both parents are Lewis-a negative of genotype LeleSese, we find that most of them have 3 Lewis-a positive : 13 Lewis-a negative children. In a few other families, suppose we find 2 Lewis-a negative : 1 Lewis-a positive. Furthermore, in families where both parents are secretors of genotype Sese, we find most of them exhibit a ratio of 3 secretor : 1 nonsecretor, but a few of them show 9 secretor : 7 nonsecretor. Propose a hypothesis to account for these results.

Solution 9

If only two loci are interacting, the dominant Se gene can suppress the expression of Le, resulting in Lewis-a negative blood type. When both parents are dihybrid, we expect a 13 : 3 ratio in the progeny, characteristic of dominant and recessive interaction.

Genetic Interactions Solved Problems

The 9 : 7 ratio found in some families for the secretor trait indicates that two factors are again interacting. This is the ratio produced by duplicate recessive interaction; i.e., whenever the recessive alleles at either of two loci are present, a nonsecretor phenotype results. Let us symbolize the alleles of the second locus by X and x.

Genetic Interactions Solved Problems

If we assume the x gene to be relatively rare, then most families will have only the dominant gene X, but in a few families both parents will be heterozygous Xx. Let us assume that this is the case in those families that produce 2 Lewis-a negative : 1 Lewis-a positive.>

Genetic Interactions Solved Problems

If Se suppresses Le, but xx suppresses Se, then only the genotypes marked with an asterisk (*) will be Lewis-a positive, giving a ratio of 21 Lewis-a positive : 43 Lewis-a negative. This is very close to a 1 : 2 ratio and indeed would appear to be such with limited data.

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
150 Characters allowed