Cystic fibrosis is an autosomal recessive disorder (c). Consider a cross between two heterozygote parents. Using...

Cystic fibrosis is an autosomal recessive disorder (c). Consider a cross between two heterozygote parents. Using a Punnett square, demonstrate what percentage of their offspring would have the disease? What are the genotypic ratios of all the possible outcomes from this cross?

Expert Solution

Cystic fibrosis is a genetically inherited disorder caused by a mutation in a gene which is located in chromosomes number 7 (autosomal chromosome). In this disorder, cells producing digestive enzymes, mucus are damaged. As a result, the digestive system and the respiratory system of a person cannot work properly. This disorder may cause death.

Let's suppose allele 'c' is mutated for cystic fibrosis, i.e., causes cystic fibrosis and allele 'C' is normal, i.e., doesn't cause cystic fibrosis. As the disorder is recessive, a person with the homozygous recessive condition, i.e., with 'cc' genotype, will have the cystic fibrosis disorder. On the other hand, a person with heterozygous (Cc) or homozygous dominant (CC) condition will not have the disorder.

In the above question, the parents are heterozygous (Cc). Each of them will produce 50% gametes with 'C' allele and the rest 50% gametes with 'c' allele.

Punnett square -

	C	c
C	CC	Cc
c	Cc	cc

Now we can see that out of four offsprings only one is homozygous recessive (cc) which will have cystic fibrosis disorder.

Hence, the percentage of offsprings with cystic fibrosis disorder = (number of homozygous recessive offspring ÷ total number of offsprings) ×100 = (1÷4)×100 = 25%

The genotypic ratios of the cross are -

CC:Cc:cc = 1:2:1

gladiator answered 2 years ago

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