A woman who is a carrier (heterozygous) for Cystic Fibrosis marries a man who has Cystic Fibrosis, a recessive disease. What are
the genotypes? What is the probability that their children will have the disease? What is the probability that their children will not have the disease? Show a Punnett square
2 answers:
Answer:
The genotype of a woman who is heterozygous for cystic fibrosis Cc.
The genotype of a man who has cystic fibrosis is cc.
The probability that their children will have the disease is 1/2 while the probability that their children will not have the disease is 1/2.
Explanation: If C represents the normal trait and c represents the trait for cystic fibrosis; and if the normal trait is dominant over the trait for cystic fibrosis which is recessive, therefore a woman who is heterozygous for cystic fibrosis will have the genotype Cc while a man who has cystic fibrosis will have the genotype of cc.
In a cross between the man (cc)and the woman (Cc), fifty percent of the children have a genotype (cc), an indication that they have cystic fibrosis while fifty percent of the children have the genotype (Cc) which is heterozygous for cystic fibrosis, an indication that they do not have cystic fibrosis because the trait for cystic fibrosis is recessive.
See the punnett square attached
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as the ratio comes to be 3:1
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