Two people heterozygous for Tay-Sachs (Tt) have two children who are both normal
ID: 99262 • Letter: T
Question
Two people heterozygous for Tay-Sachs (Tt) have two children who are both normal, but carriers (Tt).Tay-Sachs is a recessively inherited disease.What is the percent (%) chance their third child will have Tay-Sachs?Draw a Punnett Square to help explain your answer. (1 pt)
Probability for child with Tay Sachs:
Recessive condition problems
(1 pt) Cystic Fibrosis is a recessively inherited disorder.Individuals must have both recessive genes to have Cystic Fibrosis.A male is a carrier for the Cystic fibrosis gene (Ff).A female does not have the recessive gene (FF).Can they have a child with Cystic Fibrosis?Draw a Punnett Square to help explain your answer. (1 pt)
Cystic Fibrosis is a recessively inherited disorder.Individuals must have both recessive genes to have Cystic Fibrosis.A male and female, both carriers for the Cystic Fibrosis gene, have decided to have a child.What is the probability that the child will have Cystic Fibrosis?(2 pt)
What are the genotypes of the parents?
Male:__________ Female:___________
Draw a Punnett square to determine probability that the child will have Cystic Fibrosis.
Probability:
Explanation / Answer
Answer Part 1.
Punnet Square:
T
t
T
TT
Tt
t
Tt
tt
Normal = 1/4 or 25% (TT)
Carrier = 1/2 or 50% (Tt)
Affected Child = 1/4 or 25% (tt)
Even the third child has the probability of 25%. For every child, it will be 25%.
Probability for child with Tay Sachs: 25%
Answer Part 2.
Punnet Square:
F
f
F
FF
Ff
F
FF
Ff
Normal = 1/2 or 50% (FF)
Carrier = 1/2 or 50% (Ff)
Affected Child = 0, because there is no child with recessive homozygous i.e. “ff”.
Answer Part 3.
Genotype Male: Ff
Genotype Female: Ff
Punnet Square:
F
f
F
FF
Ff
f
Ff
ff
Normal = 1/4 or 25% (FF)
Carrier = 1/2 or 50% (Ff)
Affected Child = 1/4 or 25% (ff)
Probability of affected child is 25%
T
t
T
TT
Tt
t
Tt
tt
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