Problem 12 - Biochemistry

Question

Phenylketonuria is caused by a recessive genetic mutation, whereas acute intermittent porphyria is caused by a dominant genetic mutation. a. What is the biochemical difference in terms of protein function that distinguishes most recessive mutations from dominant mutations? b. What is the probability that a father with phenylketonuria and a homozygous normal mother will have a child that is a phenylketonuria carrier? c. What is the probability that a heterozygous mother with acute intermittent porphyria will have a child with the disease if the father is disease free?

Step-by-Step Solution

Answer

Additionally, what is the probability of a child being a carrier of Phenylketonuria if their father has the condition and their mother is homozygous normal? And lastly, what is the probability of a child having acute intermittent porphyria if their mother is heterozygous and their father is disease-free? Answer: The biochemical difference between recessive and dominant mutations is that recessive mutations often result in non-functional or loss of function proteins, while dominant mutations typically result in functional proteins with altered activity or gain of function. The probability of a child being a carrier of Phenylketonuria if their father has the condition and mother is homozygous normal is 100%. The probability of a child having acute intermittent porphyria if their mother is heterozygous and their father is disease-free is 50%.

Step 1: Question A: Biochemical Difference in Terms of Protein Function

The main difference that distinguishes recessive mutations from dominant mutations in terms of protein function is that recessive mutations often result in non-functional or loss of function proteins, while dominant mutations typically result in functional proteins with altered activity or gain of function. In a recessive mutation, the affected gene must be inherited from both parents in order for the condition to be manifest, whereas in a dominant mutation, inheriting the condition requires only one affected gene from one parent.

Step 2: Question B: Phenylketonuria Carrier Probability

To find the probability of a child being a Phenylketonuria (PKU) carrier i.e., heterozygous for the recessive mutation (Pp), we will use the Punnett square method: Father with phenylketonuria: pp Homozygous normal mother: PP The Punnett Square: P | P ------|------ p | Pp | Pp ------|------ p | Pp | Pp Based on the square above, the probability for the child to be a carrier (Pp) is 100%. Therefore, there is a 100% chance that the child will be a carrier of Phenylketonuria if the father has the condition and the mother is homozygous normal (PP).

Step 3: Question C: Acute Intermittent Porphyria Probability

Given that acute intermittent porphyria is a dominant mutation, a heterozygous mother has one affected allele (D for disease) and one allele that is disease-free (d). The father is disease-free (dd). We will again use the Punnett square method to determine the probability of their child having the disease. Heterozygous mother: Dd Disease-free father: dd The Punnett Square: D | d ------|------ d | Dd | dd ------|------ d | Dd | dd Looking at the square, there is a 50% chance that the child inherits the disease (Dd) from the heterozygous mother and the disease-free father.