Question

1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia (15pts).

a. What are the parental gametes? (2pts)

b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible genotypes. (2 pts)

c. What is the gamete frequency if the two genotypes in (b) were crossed with each other? (4pts)? (4pts)

d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts)

Answer 1

Both the conditions are recessive.

• Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for beta thalessemia. Genotype is LLbb

• Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta-thalessemia. Genotype is llBb

1. The possible gametes produced by these parents:

2. Nikoleta: Lb

3. Husband: lB, lb

Possible genotype of the F1 offspring: LlBb or Llbb

1. Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype should be Llbb.

1. Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb). Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2

2. The following is the phenotype frequency:

LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3

LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3

llBb (Low platelet count and no beta thallasemia) = 1

llbb (Low platelet count and beta thallasemia) = 1

Thus phenotypic ratio stands as:

Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1