Albinism in humans is caused by homozygosity for mutant recessive alleles of the tyr gene, which is located on chromosome 11. alkaptonuria in humans is caused by homozygosity for mutant recessive alleles of the hgd gene, which is located on chromosome 3. suppose that a man and a woman who are each heterozygous carriers of mutant recessive alleles of both the tyr and hgd genes have five children. what is the probability that at least one of their children will have albinsim and/or alkaptonuria?

Exactly 989527/1048576, or approximately 94.37%  
Since each trait is carried on a different chromosome, the two traits are independent of each other. Since both parents are heterozygous for the trait, each parent can contribute 1 of a possible 4 combinations of the alleles. So there are 16 possible offspring. I'll use "a", "A", "b", "B" to represent each allele and the possible children are aabb, aabB, aaBb, aaBB, aAbb, aAbB, aABb, aABB, Aabb, AabB, AaBb, AaBB, AAbb, AAbB, AABb, and AABB 
Of the above 16 possibilities, there are 7 that are homozygous in an undesired traint and 9 that don't exhibit the undesired trait. So let's first calculate the probability of "what are the chances that all 5 children not exhibiting an undesired trait?" and then subtract that result from 1. So 
1-(9/16)^5 = 1 - 59049/1048576 = 989527/1048576 which is approximately 0.943686485 = 94.3686485%  
So the answer is exactly 989527/1048576, or approximately 94.37%

The answer is c hope this

Idont understand i'm sorry