Question

please provide an explanation for both questions! Thank You!

9. Consider the pedigree to the right, displaying the inheritance of a rare recessive autosomal disease which is fully penetrant. What is the likelihood that the child will be affected? a. 1/2 b. 27/81 c. 1/4 d. 1/36 e. 1/64 10. Consider the following pedigree from a human family containing a male with Klinefelter syndrome (a set of abnormalities seen in XXY individuals; indicated with a shaded box). In each, A and B refer to codominant alleles of the X-linked G6PD gene. The phenotypes of each individual (A, B, or AB) are shown on the pedigree. In which parent did nondisjunction occur, and during which meiotic division? AB a. Mother, Meiosis 1 b. Mother, Meiosis 2 c. Father, Meiosis 1 d. Father, Meiosis 2 e. Either a orb

Answer 1

9) we have to find the probability that IV-1 is affected, for IV-1 to be affected the parents of IV-1 ( III-1 and III-2 ) has to be carriers, for III-1 to a carrier her father has to be a carrier and for III-2 to be a carrier his mother has to be a carrier.

the probability that II-2 is a carrier=?

II-2 and II-4 have an affected sister, but their parents are normal this shows the parents of II-2 and II-4 are carriers.

let the alleles be D- dominant normal, d- recessive

Dd * Dd

	D	d
D	DD	Dd
d	Dd	dd

probability of getting phenotypically normal carrier progenies= number of Dd/total number of progeny with D allele=2/3

probability of getting dd progenies for Dd parents= number of dd/total number=1/4

II-1 and II-5 are married to the family and they don`t have the disease so they are homozygous DD,

DD * Dd

	D	d
D	DD	Dd

probability of getting Dd progeny= number of Dd/total number=1/2

the probability that III-1 is a carrier= probability that II-2 is a carrier $\times$ probability of getting Dd progenies for DD and Dd

= probability of getting phenotypically normal carrier progenies for Dd parents $\times$ probability of getting Dd progenies for DD and Dd

= 2/3$\times$1/2

= 1/3

probability that III-2 is a carrier = probability that II-4 is a carrier $\times$ probability of getting Dd progenies for DD and Dd

= probability of getting phenotypically normal carrier progenies for Dd parents $\times$ probability of getting Dd progenies for DD and Dd

= 2/3$\times$1/2

= 1/3

probability that IV-1 is affected= probability that III-2 is a carrier $\times$ probability that III-1 is a carrier $\times$ probability of getting dd progenies for Dd parents

= 1/3$\times$1/3$\times$1/4

=1/36

so the answer is d) 1/36

10) the affected child has 2 copies of X chromosome and one Y chromosome, G6PD gene is located on the X chromosome. the phenotype of the affected child is A, so the child has inherited the X chromosome from mother only, ( if the child has inherited X chromosome from father it should have produced B phenotype).

mother has 2 X chromosomes one with A allele and the other with B allele, the homologous chromosomes are separated during meiosis I, if non-disjunction occurred in meiosis I both X chromosomes would have moved to one cell so the child would have got both A and B allele.

sister chromatids are separated during meiosis II, here the sister chromatids of the X chromosome with the A allele failed to separate in meiosis II, so one gamete has 2 X chromosomes with A allele which was fertilized with sperm containing Y chromosome so the genotype of the child is XXY.

here non-disjunction occurred in mother in meiosis II.

so the answer is b) mother, meiosis 2.