Phenylketonuria is a metabolic disease of humans that results from an autosomal recessive gene. If the frequency of phenylketonurics in the population is 9/10,000, what is the probability that two normal individuals will produce a diseased child?

(9/10,000)? Is that the frequency of a phenyketonuria child, or the frequency of the recessive gene. The problem statement is uncertain.

If the 9 tenthousandts is the frequency of phenylketonuria people, then that is the frequency two normal individuals will produce a diseased child.

If the 9 ten thousandths is the frequence of the recessive gene, then the frequency of children is the square of that.

I think its the frequency of the gene this is a Hardy-Weignberg problem

Thanks!

To calculate the probability that two normal individuals will produce a diseased child, we need to understand the inheritance pattern of phenylketonuria (PKU) and apply basic principles of probability.

Given that PKU is an autosomal recessive disorder, both parents must carry at least one copy of the recessive gene for their child to inherit the disease. Let's break down the steps to calculate the probability:

Step 1: Find the probability that an individual is a carrier of the disease:
Since PKU occurs with a frequency of 9/10,000 in the population, we can assume that the gene frequency is the square root of this value (as it is a recessive gene). Therefore, the gene frequency is √(9/10,000) = 3/100.

Step 2: Calculate the probability that two normal individuals will both be carriers:
The probability of an individual being a carrier is 3/100. Since the inheritance of these alleles from each parent is independent, we can multiply the probabilities together.

So, the probability that both parents are carriers would be (3/100) * (3/100) = 9/10,000.

Step 3: Ultimately, the probability of two normal individuals producing a diseased child is the probability that both parents are carriers multiplied by the probability that the child inherits the disease.

Since offspring of two carriers have a 1/4 chance of inheriting the disease, we can multiply the probability from Step 2 (9/10,000) by 1/4 to get the final probability.

Therefore, the probability that two normal individuals will produce a diseased child is (9/10,000) * (1/4) = 9/40,000.

Please note that these calculations are based on certain assumptions and simplifications. In reality, genetic inheritance can be more complex, and the frequencies mentioned may vary in different populations.