Consider two gene loci in the mouse, locus 1 and locus 2. Each locus is on a dif

Question

Consider two gene loci in the mouse, locus 1 and locus 2. Each locus is on a different, nonhomologous autosomal chromosome. Alleles at locus 1 influence fur color. Allele A is black and allele a is for white. Allele A is completely dominant to allele a. Aleles at locus 2 influence stature; allele B is for tall and allele b is for short. Allele B is completely dominant to allele b.

Mouse 1, genotype Aa Bb, mates with female mice, genotype Aa Bb; 100 progeny are produced.

1. What fraction of the gametes from mouse 1 is expected to be genotype ab?

2. What fraction of the progeny is expected to be true-breeding for both tall stature and fur color?

3. How many progeny are expected to be short with white fur? (show calculations)

4. How many progeny are expected to be homozygous at both loci? (show calculations)

5. If you randomly choose two progeny mice, one at a time, what is the probability that both mice are tall and black? (show calculations)

Explanation / Answer

1. Mouse 1 gametes would be AB, Ab, aB and ab. 1/4 would be fraction for gamete ab.

2. Out of 16 progeny, 9 will be true breeding for tall stature and fur color. 9/16.

3. AaBb x AaBb

Progeny: AABB, AABb, AaBB, AaBb, AABb, AAbb, AaBb, Aabb, AaBB, AaBb, aaBB, aaBb, AaBb, Aabb, aaBb, aabb. Only 1 out of 16 would be with short with wihite fur. The fraction woule be 1/16.

4. AaBb x AaBb

Progeny: AABB, AABb, AaBB, AaBb, AABb, AAbb, AaBb, Aabb, AaBB, AaBb, aaBB, aaBb, AaBb, Aabb, aaBb, aabb. 4 out of 16 would be homozygoues at both the loci.

5. The probablility of first mice to be black and tall mouse would 9/16 and second mice would be 8/16.

So the total probability would be 9/16 x 8/16= 9/32

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