A man and a woman are each heterozygous carriers of an autosomal recessive mutation of a disorder that is fatal in infancy. They both want to have multiple children, they are concerned about the risk of the disorder appearing in one or more of their children. In separate calculations, determine the probabilities of the couple having five children with 0, 1, 2, 3, 4, and all 5 children being affected by the disorder.

0.2373, 0.0794, 0.0263, 0.0087, 0.0029, 0.0010

Explanation:

Let Aa be the heterozygous genotype for the autosomal recessive character.

Since both parents are heterozygous (Aa) the possibility genotype of one child is AA Aa Aa aa

Since the disease is a recessive disorder, the probability of giving birth to an affected child is 1/4 and the probability of giving birth to unaffected child is 3/4.

If the couple are having 5 children:

1. The probability of 0 Affected

Since 0 is affected it means 5 are unaffected.

(3/4)^5 = 243/1024 = 0.2373

2. The probability of 1 affected children

Since 1 is affected it means 4 are unaffected

1/4 * (3/4)^4 = 1/4 * 81/256 = 81/1024 = 0.0794

3. The probability of 2 affected children

Since 2 are affected it means 3 are unaffected

(1/4)^2 * (3/4)^3 = 1/16 * 27/64 = 27/1024 = 0.0263

4. The probability of 3 affected children

Since 3 are affected it means 2 are unaffected

(1/4)^3 * (3/4)^2 = 1/64 * 9/16 = 9/1024 = 0.0087

5. The probability of 4 affected children

Since 4 are affected it means 1 is unaffected

(1/4)^4 * (3/4)= 1/256 * 3/4 = 3/1024 = 0.0029

6. The probability of all five children affected

(1/4)^5 = 1/1024 = 0.0010

the index of refraction of the second medium is higher. i just took the test trust me !