Chapter 2, Problem 43P

People with nail-patella syndrome have poorly developed or absent kneecaps and nails. Individuals with alkaptonuria have arthritis as well as urine that darkens when exposed to air. Both nail-patella syndrome and alkaptonuria are rare phenotypes. In the following pedigree, vertical red lines indicate individuals with nail-patella syndrome, while horizontal green lines denote individuals with alkaptonuria.

a.	What are the most likely modes of inheritance of nail-patella syndrome and alkaptonuria? What genotypes can you ascribe to each of the individuals in the pedigree for both of these phenotypes?
b.	In a mating between IV-2 and IV-5, what is the chance that the child produced would have both nail-patella syndrome and alkaptonuria? Nail-patella syndrome alone? Alkaptonuria alone? Neither defect?

 

Summary Introduction

a.

To determine:

The mode of inheritance of nail-patella syndrome and alkaptonuria, and genotype of each individual that is given in the pedigree.

Introduction:

Nail patella syndrome affects the nail, kidney and skeletal muscles. This kind of syndrome is autosomal dominant because only one copy of the defective gene may cause disease in the person.

Explanation of Solution

Two diseases are discussed in the problem, one is nail-patella (NN), and another is alkaptonuria (aa). It can be observed from the given pedigree that children who are affected with nail-patella have at least one affected parent. Thus, it can be concluded that nail-patella is dominant. When the individuals of a population are crossed in a consanguineous manner, then it results in the appearance of character for alkaptonuria. The characters for alkaptonuria appear in IV- 5 and IV-6 so, it shows the occurrence of these characters in parents also.

The genotype of each individual is as follows:

Individuals	Genotypes
I-1	nn Aa
I-2	Nn AA
II-1	nn AA
II-2	nn Aa
II-3	Nn AA or Nn Aa
II-4	nn AA or nn Aa
II-5	Nn Aa
II-6	nn AA
III-1	nn AA
III-2	nn AA or nn Aa
III-3	nn Aa
III-4	Nn Aa
III-5	nn AA or nn Aa
III-6	nn AA or nn Aa
IV-1	nn AA or nn Aa
IV-2	nn AA or nn Aa
IV-3	Nn AA or Nn Aa
IV-4	nn AA or nn Aa
IV-5	Nn aa
IV-6	nn aa
IV-7	nn Aa or nn AA

Summary Introduction

b.

To determine:

The chance that the produced child by mating IV-2 and IV-5 would have both nail-patella syndrome and alkaptonuria, Nail-patella syndrome alone, alkaptonuria alone and neither defect.

Introduction:

Alkaptonuria is a disease which can be inherited from parents to offsprings, and in this condition, several amino acids are not processed like phenylalanine and tyrosine. In this condition, the colour of the urine turns into black.

Explanation of Solution

The chance that the produced child by mating IV-2 and IV-5 would have both nail-patella syndrome and alkaptonuria.

In this problem, the cross is made between IV-2 and IV-5. The genotype, which is possible for IV-2 is nn AA or nn Aa because the genotype of his father is uncertain. Thus, both the possible genotypes will be considered for solving the problem. If both the syndromes, nail-patella syndrome and alkaptonuria, are inherited to the child than it can be said that the contributed gametes for IV-2 will be ‘n a’ the probability has the individual IV-2 has the genotype nn Aa is 1/4. Genotype nn Aa in IV-2 is possible when the gametes ‘n’ and ‘a’ are inherited from parent III-2. The genotype of III-2 is ‘nn AA’ or ‘nn Aa’ so, the probability of donating each type of gamete is 1/4. The probability of delivering gametes ‘n a’ by IV-2 to its offsprings is 1/2 and the probability of IV-5 providing ‘N a’ gametes is 1/2.

Thus, the probability that the child will have both the diseases can be calculated from above probabilities. Therefore the probability of a child having disease is as follows:

$\text{Probability}=\left(\frac{\text{1}}{\text{4}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)=\left(\frac{\text{1}}{\text{16}}\right)$

The genotype ‘nn AA’ will not be considered for IV-2 because an individual with this genotype will not donate gametes for alkaptonuria.

The chance that the produced child by mating IV-2 and IV-5 would have the only nail-patella syndrome.

The genotype ‘N- A-’ determines that the child will have the only nail-patella syndrome. This genotype is only possible if the IV-2 has the genotype ‘nn Aa’ and its probability is 1/4. The probability that the gamete ‘A’ will be donated by IV-2 is 1/2 and for IV-5 to deliver gametes ‘N a’ is 1/2.

Thus, if the genotype of IV-2 is ‘nn Aa’ then the probability that the child will have only nail-patella syndrome is as follows:

$\begin{array}{rll}\text{Probability}& =& \left(\frac{\text{1}}{\text{4}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{1}}{\text{16}}\right)\end{array}$

Similarly, the probability that IV-2 will have genotype nn AA is 3/4. The likelihood that the gametes ‘nA’ will be donated by IV-2 is 1, and the probability that gametes ‘N a’ will be donated by IV-5 is 1/2.

Thus, if the genotype of IV-2 is ‘nn AA’ then the probability that the child will have only nail-patella syndrome is as follows:

$\begin{array}{rll}\text{Probability}& =& \left(\frac{3}{\text{4}}\right)\times 1\times \left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{3}{8}\right)\end{array}$

The addition of the above probability can determine the actual probability of a child having the only nail-patella syndrome. Thus, the probability is as follows:

$\begin{array}{rll}\text{Probability}& =& \left(\frac{3}{16}\right)+\left(\frac{\text{1}}{8}\right)\\ & =& \left(\frac{7}{\text{16}}\right)\end{array}$

The chance that the produced child by mating IV-2 and IV-5 would have only alkaptonuria.

If the children have the genotype ‘nn aa’ then these children will have only alkaptonuria. Children will have genotype ‘nn aa’ if their parents (IV-2) also has genotype ‘nn aa’. The probability that IV-2 has genotype ‘nn aa’ is 1/4. The probability for donating gamete ‘n a’ by IV-2 and IV-5 is 1/2.

Thus, the probability that the child will have the only alkaptonuria is as follows:

$\begin{array}{rll}\text{Probability}& =& \left(\frac{\text{1}}{\text{4}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)\times \left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{1}}{\text{16}}\right)\end{array}$

Parents with genotype ‘nn AA’ cannot deliver gametes for alkaptonuria. So the parents with genotype ‘nn AA’ will not be considered to obtain the desired probability.

The chance that the produced child by mating IV-2 and IV-5 would be normal.

The probability of a child having neither disease can be calculated by subtracting the summation of the above three probabilities from 1. Thus, the probability of a child having neither syndrome is as follows:

Probability = 1- (probability of having both the syndrome + probability of having only nail-paella + probability of having only alkaptonuria).

$\begin{array}{rll}\text{Probability}& =& 1-\left[\left(\frac{1}{16}\right)+\left(\frac{7}{16}\right)+\left(\frac{1}{16}\right)\right]\\ & =& \left(\frac{7}{16}\right)\end{array}$