Question

Please provide answers with explanations for these questions.

1. Tay Sachs disease is an inherited disorder. Children who are born with Tay Sachs have progressive neurological problems and typically die before age 4. It is inherited in an autosomal recessive manner Below are pedigrees from two families where some members of a family have Tay Sachs disease. Neither Hillary nor Justin have Tay Sachs disease. If Hillary and Justin have a child, what is the chance this child will have Tay Sachs disease? A a Lè Jus ae A Hillary A. 0 A AA Aa C1/9 D. 9/64 /2. Two different genes are located on the same chromosome, but their exact position is not known. Individual 1 and 2 have the genotypes shown below. The parallel lines represent homologous chromosomes. Individual #1: Individual #2 How likely is it that these individuals will have an offspring with the following genotype: A. More likely if the two genes are close together on the chromosome B. More likely if the two genes are far apart on the chromosome Equally likely no matter where the genes are located on the chromosome Need more information before this can be determined

Answer 1

Please find the answers below:

Answer 1: According to the information, the parents are not suffering from the disease but one of the child is suffering. Sicne the disease is inherited in autosomal recessive manner, both the unaffected parents must be heterozygous in nature in order to transmit this disease to the offspring. Thus, the genotype of both parents will be given by AtA and AtA.

Thus, the Punnett square can be formulated as below:

	At	A
At	AtAt	AtA
A	AtA	AA

Thus, genotype: AtAt : AtA : AtA : AA :: 1:2:1

Phenotype: Diseased : Carrier : Normal non -carrier :: 1:2:1

Thus, total affected individuals : 1/4 or 25% chances.

Since there are two pedigrees involved and probabiltiy for each child to be suffering is 1/4 each, the total probability will be 1/4 *1/4 or 1/16.

Thus, choice B is correct.

Answer 2: Choice B is correct.

Reason: According to the information, a test-cross is conducted between these two organisms where one parent is heterozygous and another is homozygous recessive. This type of crosses give the phenotypic and genotypic ratio of 1:1 like parental forms and thus, in order to obtain such results, the genes must be located far enough so that recombination might take place. If the genes are located casually anywhere on chromosome, they can be found close to each other and linkage might appear.