Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Modified. For the following crosses, determine as accurately as possible the gen

ID: 215316 • Letter: M

Question

Modified.

For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-clotting disorder, are X-linked recessive traits. In each case, assume the parents have normal karyotypes (see the table given below) Table Human Aneuploidiesand Frequenciesat Birth Frequency at Birth Aneuploidy Autosomal Aneuploidy Trisomy 13 Trisomy 18 Trisomy 21 Syndrome Syndrome Characteristics Mental retardation and developmental delay, possible deafness, major organ Patau syndrome 1 in 15,000 Edward syndrome 1 in 8000 Down syndrome 1 in 1500 abnormalities, early death Mental retardation and developmental delay, skull and facial abnormalities, early death Mental retardation and developmental delay, characteristic facial abnormalities, short stature, variable life span Sex-Chromosome Aneuploidy 1 in 1000 (males) 1 in 1000 (males) 1 in 1000 Variable secondary sexual characteristics, infertility, frequent breast swelling, no impact on mental capacity Klinefelter 47, XXY syndrome Jacob syndrome Triple X syndrome (females) Tall stature common; possible reduction but not loss of fertility; no impact on mental capacity 47, XYY Tall stature common; possible reduction of fertility; menstrual irregularity; no impact on mental capacity 1 in 5000 (females) No secondary sexual characteristics; infertility, short stature; webbed neck common; no impact on mental capacity 45, XO Turner syndrome

Explanation / Answer

A. The man is XHY, the woman is XhXH, the child is XhXhY and nondisjunction occurred in the woman during meiosis I.

B. The man is XcY, the woman is XhXH, the child is XhYY and nondisjunction must have occurred in the man during meosis II

C. The man is XcY, the woman has at least one X chromosome with the wild type C allele (XcXC), and the child is XCO non disjunction must have occurred in the woman at meiosis I or II

D. The man is XchY, and the woman is XhCXH, the child could be XchXhCXhC

Where the nondisjunction has occurred in the woman during meiosis I or the child could be XchXchXCh where nondisjunction has occurred in the man during meiosis II

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote