Question

Using Punnett squares, calculate for each of the scenarios below, the risk each couple has to have the affected, carrier, and normal offspring. Assume that the disease is cystic fibrosis (6 points):

Carrier mother, Normal father;

1. ? Risk of: __% normal daughter; __ % normal son; __ % carrier daughter; __ % carrier son; __ % normal daughter; __ % normal son

Normal mother, Affected father;

2. ? Risk of: __% normal daughter; __ % normal son; __ % carrier daughter; __ % carrier son; __ % normal daughter; __ % normal son

Carrier mother, Affected father;

3. ? Risk of: __% normal daughter; __ % normal son; __ % carrier daughter; __ % carrier son; __ % normal daughter; __ % normal son

Answer 1

Answer:

Normal = C; cystic fibrosis = c

Normla allele (C) is dominant over mutant allele (c). Cystic fibrosis is an autosomal recessive trait. Hence, the probability of the trait is same in both male and female.

1). Carrier mother (Cc) x (CC) Normal father –Parents

	C
C	CC (normal)-50%
c	Cc (carrier)-50%

Rist of = 0% (no cystic fibrosis)

normal daughter= 50 %

normal son=50%

carrier daughter= 50%

carrier son=50 %

Affected = 0%

2).

Normal mother (CC) x (cc) affected father –Parents

	C
C	Cc (carrier)-50%

Rist of = 0% (no cystic fibrosis)

normal daughter= 0 %

normal son=0%

carrier daughter= 100%

carrier son=100 %

3).

Carrier mother (Cc) x (cc) affected father –Parents

	c
C	Cc (carrier)-50%
c	cc (affected)-50%

Rist of = 50% (cystic fibrosis)

normal daughter= 0 %

normal son=0%

carrier daughter= 50%

carrier son=50 %

Affected daughter = 50%

Affected son = 50%