Question

In: Biology

PKU is a rare, fully penetrant, autosomal recessive condition in which an affected individual cannot break...

PKU is a rare, fully penetrant, autosomal recessive condition in which an affected individual cannot break down the amino acid phenylalanine. Neither Brian nor Mary has PKU, and they want to have children. They are concerned because Brian's father has PKU, and while neither of Mary's parents has PKU, her brother has PKU. Use symbols A and a.

1) What is the probability that Mary and Brian's first child will have PKU? Show work.
2) Mary and Brian have a child, and it does not have PKU. They decide to have a second child. Does knowing the phenotype of the first child cause you to change the probability you calculated in question 1 to predict the probability the next child will have PKU? If not, why?
3) Consider an alternate scenario where Mary and Brian have a child and it does have PKU. They decide to have another child. Does knowing the phenotype of the first child change the probability you calculated in 1 to predict the probability the next child will have PKU? If not, why?

Expert Solution

Subpart 1:

As PKU is an autosomal recessive trait so the infected individuals must be homozygous for recessive alleles.

Let A is dominant allele and a is recessive allele for the trait'

As Brian's father has PKU it is must that his genotype is aa and Brian has no PKU means his genotype is Aa.

In Mary's family his brother has PKU and her perants dont have means both the parents are heterozygotes. So Mary's genotype could be either AA or Aa.

So the alleles generated from Brian would be A and a, where as Mary's possible alleles would be A and A if she is a homozygous and A & a considering her as a heterozygous for the trait.

the possibilities of genotypes of the children of Brian and Mary are as follows.

The total number of genotypic possibilities Brian and Mary's child may have is 8. Out of which only in 1 case the child will have PKU.

The probability of Brian and Mary's first child having PKU is (1/8)= 0.125

Subpart 2

If the first child does not have PKU then also Mary's genotype can either be homozygous or heterozygous and the probability for a second child to have PKU will remain same.

Subpart 3

If the first child does have PKU, it will be clear Mary's genotyp is heterozygous. So now the genotypic possibilities of Mary and Brian's child will change. Now the possible genotypes will be

So the total possible genotypes will be 4 in which only in one case the child will have PKU.

The probability of the next child having PKU will be 1/4= 0.25.

gladiator answered 2 years ago

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population....

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the allele frequency of the recessive allele? 0.2 0.4 0.64 0.8 Not enough information is provided.

In a population of 16,000 people, 67 are affected with a mild autosomal recessive condition. What...

In a population of 16,000 people, 67 are affected with a mild autosomal recessive condition. What is the estimated number of carriers for this condition in this population?

Phenylketonuria (PKU) is an autosomal recessive disorder in which phenylalanine accumulates in the body, resulting inbrain...

Phenylketonuria (PKU) is an autosomal recessive disorder in which phenylalanine accumulates in the body, resulting inbrain damage, intellectual disabilities, behavioral symptoms, or seizures. PKU affects approximately 1 in 12,000 people living in the United States. What is the frequency of heterozygous carriers of the PKU allele within the U.S. population? (Assume Hardy-Weinberg equilibrium) Group of answer choices 1.80% 0.02% 99.99% 0.90% 0.01%

In humans, hypotrichosis (sparse body hair) is a rare autosomal recessive condition. Let h = hypotrichosis...

In humans, hypotrichosis (sparse body hair) is a rare autosomal recessive condition. Let h = hypotrichosis and H = Normal. Family A consists of two phenotypically normal parents with five children: the first two have hypotrichosis, and the rest are normal. Family B consists of a phenotypically normal man and a hypotrichotic woman with three children: one has hypotrichosis, and the other two are normal. ------------ Family C is formed when a normal child from family A marries a normal...

Albinism is a rare autosomal recessive trait in which there is a complete absence of melanin...

Albinism is a rare autosomal recessive trait in which there is a complete absence of melanin pigment in the skin, eyes, and hair. A normally pigmented woman whose parents are normal has a brother who is albino. She marries a normally pigmented man whose grandfather was albino. What is the probability their first child will be albino? I would like to see the problem worked out. Dont understand why it is 1/12. Thank you in advance

1. A woman is affected with Alpha-1 antitrypsin deficiency, an autosomal recessive condition with a carrier...

1. A woman is affected with Alpha-1 antitrypsin deficiency, an autosomal recessive condition with a carrier frequency of 1 in 13. As they are planning on having children, her partner undergoes carrier testing for the condition. The result was negative, indicating that he was not found to be a carrier. The reported sensitivity of the carrier testing is 95%. What is the chance that their first child will be affected with Alpha-1 antitrypsin deficiency?

Phenylketonuria (PKU) is an autosomal recessive metabolic disorder that can result in mental retardation if left...

Phenylketonuria (PKU) is an autosomal recessive metabolic disorder that can result in mental retardation if left untreated. Two normal parents produce a child with PKU. If the recessive PKU allele is ‘n’ and the dominant normal allele is ‘N’, what are the genotypes of both parents? A. Mother is NN and Father is nn B. Mother is Nn and Father is nn C. Mother is nn and Father is Nn D. Mother and Father are both Nn E. Mother and...

The pedigree from Parts A and B is shown below. Now it shows the inheritance of a rare recessive autosomal condition that affects individuals

The pedigree from Parts A and B is shown below. Now it shows the inheritance of a rare recessive autosomal condition that affects individuals II-3, IV-2, IV-5, and V-3 (denoted by the filled symbols and the genotype aa). Individuals I-1, I-2, I-5, and I-6 have no history of this condition in their families.

(1) Haemochromatosis is a mild autosomal recessive condition. Approximately 1 in 200 Australians have the condition,...

(1) Haemochromatosis is a mild autosomal recessive condition. Approximately 1 in 200 Australians have the condition, and approximately 1 in 7 Australians are carriers. Jane is planning on starting a family, and does not have haemochromatosis, though her mother is affected with the condition. Her father is not affected, and is not a carrier. Jane’s partner Steve does not have the condition either, but his father is a carrier (his mother is not a carrier). (a) What are the chances...

1. you are evaluating autosomal recessive condition A (alleles A and a) and B (alleles B...

1. you are evaluating autosomal recessive condition A (alleles A and a) and B (alleles B and b) which you hypothesize are on different chromosomes. You are given 7 wildtype (homozygous dominant) female flies and 7 male flies that are homozygous recessive. What genotypic ratio would you expect in the F1 generation produced? 2. you mate the F1 females with males that are homozygous recessive for both disease A and B to produce an F2 generation. What phenotypic ratio would...