Assume gene pairs located on different pairs of homologous chromosomes unless ot

Question

Assume gene pairs located on different pairs of homologous chromosomes unless otherwise stated.

4. Mendel observed a 9:3:3:1 ratio of offspring when studying different traits. What would he have concluded if he had observed a 3:1 ratio for two traits?

5. How many different types of gametes are produced by genotype YyRr if Y and R are on different chromosomes? What are they?

6. How many different types of gametes are produced by genotype YyRr if Y and R are on the same chromosome (assume the dominant YR and recessive yr traits are together)? What are they?

8. Using the previous table, if A and B show incomplete dominance, how many offspring would exhibit the homozygous dominant, heterozygous, and homozygous recessive phenotypes for both traits together?

9. In a cross between AaBbCcDD and AabbCCDd, what is the probability of an offspring with the phenotypes or genotypes below? Show math.

Phenotype Dominant A, Dominant B, Dominant C, and Dominant D

Phenotype Recessive A, Recessive B, Dominant C, Dominant D

Genotype aaBbCcDD

Genotype AAbbCCDd

Genotype AabbCCdd

Explanation / Answer

4. The presence of one dominant gene influcences the other gene expression. That is, if one dominant gene is present, the other gene also expresses the dominant trait.

5. The number of gametes produced is proportional to the number of genes and the number of loci that are heterozygous. It can be given by the simple formula 2n where n represents the number of loci that are heterozygous. There are two heterozygous loci in the given case, and hence, there are 4 possible gametes of the genotypes YR, Yr, yR, and yr.

6.

6. If the two genes are separated far apart, the chromosomes can undergo recombination and produce the same four gametes YR, Yr, yR, and yr. If they are close and linked, they produce the gametes YR and yr only.

9.

The probability that the offspring is dominant A, dominant B, dominant C, and dominant D is as follows:

Parent genotypes: AaBbCcDD × AabbCCDd

The probability of dominant A (AA, Aa) = ¾

The probability of dominant B (BB, Bb) = ½

The probability of dominant C (CC, Cc) = 1

The probability of dominant D (DD, Dd) = 1

Therefore, the probability is ¼ × ¾ × 1× 1 = 3/8.

The probability of the given phenotype is 3/8 or 0.375

The probability of recessive a, recessive B, recessive C, and recessive D:

The probability of recessive A (aa) = 1/4

The probability of recessive B (bb) = 1/2

The probability of recessive C (cc) = 0

The probability of recessive D (dd) = 0

The probability of said recessive phenotypes is 0.

Genotype aaBbCcDD = ¼ × ½ × ¼ × ½ = 1/64

Genotype AAbbCCDd = ¼ × ½ × ½ × ½ = 1/32

Genotype AabbCCdd = 0 since there is no chance of recessive dd.

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