ella syndrome have poorly-developed or absent kneecaps and nails. Individuals ar
ID: 135581 • Letter: E
Question
ella syndrome have poorly-developed or absent kneecaps and nails. Individuals arthritis as well as urine that darkens when exposed to air. Both nail-patella syndrome 7. People with nail-pat patella syndrome andtes. In the following pedigree, vertical lines indicate individuals with nail- and alkaptonuria are rare phenotypes. In the followinls ns with alkaptonuria. 4. (4 points) What is the most likely mode of inheritance for nail-patella syndrome? and horizontal lines indicate individuals alkaptonuria B. (4 points) What is the most likely mode of inheritance for alkaptonuria? C. (4 points) In a mating between IV-2 and IV-5, what is the probability that the first child would have neither condition?Explanation / Answer
I = Vertical line = Nail Patella Syndrome
__ = Horizontal line = Alkaptonuria Syndrome
A. Nail Patella is a rare disorder. So we can assume that the people who are marrying each other in this pedigree are Homozygous Normal. Since all the children affected with this disease have one affected parent, Nail-Patella syndrome is Dominant.
B. For Alkaptonuria, the affected children are the result of mating between two individuals who are related as second cousins or closer (Consanguineous Marriage). III-3 & III-4 both are unaffected individuals. Therefore, Alkaptonuria is Recessive.
C. Let us represent Nail-Patella syndrome with N and Alkaptonuria with A.
The genotype of IV-2 is nn A- and that of IV-5 is Nn aa. The genotype of IV-A is unknown because of the uncertainty of her father's genotype (III-2).
To calculate the probability of their first child having neither of the diseases, we should first calculate the probability of the child:-
1. Having only Nail-Patella syndrome
2. Having only Alkaptonuria syndrome
3. Having both the syndromes
We shall then add all the three above probabilities and subtract from 1, to obtain our final probability of the child having neither of the diseases.
1. Having only Nail-Patella syndrome: (N-A-)
IV-2 would have to provide an n-A- gamete and IV-5 would have to provide an N-a gamete. This could occur if IV-2 were nnAa, the probablity of this happening is 1/4 (Probability IV-2 is Aa) x 1/2 (Probability of A if IV-2 is Aa) x 1/2 (Probability of N a gamete from IV-5). So, 1/4 x 1/2 x 1/2 = 1/16. This could also occur if IV-2 were nnAA, the probability of this happening is 3/4 (Probability IV-2 is nnAA) x 1 ( probability of n A gamete if IV-2 is nnAA) x 1/2 (Probability of N a gamete from IV-5). So, 3/4 x 1 x 1/2 = 3/8.
1/16 + 3/8 =7/16
2. Having Alkaptonuria syndrome (nn aa)
IV-2 should contribute n a gamete. This could only occur if IV-2 were nn Aa. The probability of this happening is 1/4 (Probability IV-2 is Aa) x 1/2 (Probability of A if IV-2 is Aa) x 1/2 (Probability of N a gamete from IV-5). So, 1/4 x 1/2 x 1/2 = 1/16.
3. Having both the syndromes
This could only occur if IV-2 were nn Aa. The probability of this happening is 1/4 (Probability IV-2 is Aa) x 1/2 (Probability of A if IV-2 is Aa) x 1/2 (Probability of N a gamete from IV-5). So, 1/4 x 1/2 x 1/2 = 1/16
The probability of their first child having neither of the diseases is 1 - (Sum of all the three possibilities).
1 - ( 7/16 + 1/16 +1/16) = 7/16
This is possible since there are only four possible outcomes.
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