2. (2 pts) Renal rickets is a rare kidney condition that is inherited as an auto
ID: 144899 • Letter: 2
Question
2. (2 pts) Renal rickets is a rare kidney condition that is inherited as an autosomal recessive trait. Examine the pedigree of a family with a history of the disease. List the genotypes of all individuals to the extent possible. Assume people marrying into the family are homozygous nomal, unless there is data to show otherwise 123 456 A. If individuals -1 and 111-2 marry, what is the probability that their first child has renal rickets? Suppose their first child was indeed affected. What is the probability that their second child is also affected? 2 B. 3. (3 pts) Look at the following pedigree A What is the mode of inheritance for this pedigree? Assign as many genotypes as possible B. What is the genotype of individuals 3 and 4? C lf individual 9 marries a normal man and has a child with him, what is the chance for the child to have the disease?Explanation / Answer
2. Autosomal recessive genes are those which express themselves only if present in pairs. A single gene cannot present the symptoms of the disease.
Let the normal gene for renal rickets be K and recessive gene be k
The genotype of parents I-1 and I-2 must be Kk because one of their offsprings is suffering from the disease and his genotype must be kk.
The genotype of individuals I-1, I-2, I-3, and I-4 is Kk i.e. they are carriers.
The genotypes of individuals II-3 and II-4 is KK [ it is mentioned that they are homozygous normal]
The genotype of individuals II-1 and II-6 is kk because they suffer from the disease.
The genotypes of individual II-2 and II-5 might be KK [normal] or Kk [ carrier ].
Likewise, genotypes of individuals III-1 and III-2 can be KK [normal] or Kk [ carrier ] because one of their parents can be a carrier.
If both of them are Kk then --
A] Thus, the probability that their first child will have renal rickets is 25%
That it will be a carrier is 50% and that it will be normal is 25%.
If one of the parents is normal KK and one is carrier Kk, then the following combination can occur-
50% are normal and 50% are carriers.
B] That their second child will be affected by the disease is also the same because genetic recombination is random.
3] A] The gene for the disease is heterozygous dominant.
We can assume that the affected indviduals are heterozygous for the gene and the unaffected individuals are not carrying the defective gene.
B] Individuals 3 and 4 are carrying the dominant gene and are affected by the disease.
C] Individual 9 is heterozygous and marries a normal man.
The chances that their offspring will be affected is 50% because even one gene is capable of expressing the disease.
K k K KK Kk k Kk kkRelated Questions
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