I know the answer is meiosis II in the male, but I cannot understand the relatio

Question

I know the answer is  meiosis II in the male, but I cannot understand the relationship of Cp/Ct/Cz. Could you explain it with picture and label the Cp/Ct/Cz? (There is no picture for this question. I wonder if you could draw one for me to show how it happen.)

Three alleles exist for coat color in rabbits: Cp, Ct, and Cz. A male Cp/Cz is crossed with a female Ct/Ct. Occasionally some rare genotypes occur. For each of the following rare genotypes, explain how they occurred: state which meiotic stage nondisjunction occurred and in which parent (male vs. female) the event occurred (example: female meiosis I). What caused the following rare genotype: Cp/Cz/Ct

Explanation / Answer

Ans. Note the following points-

I. Since the female is homozygous (Ct/Ct), non-disjunction in female at any stage would produce gametes with one or more copies of Ct allele only. So, the female can’t cause Cp/Ct/Cz phenotype because there is only one allele Ct contributed by the female.

So, let’s focus on non-disjunction in male.

II. During S-phase DNA synthesis occurs. The duplicated chromosomes are held together by cohesin rings- the paired chromosomes at this stage are called sister chromatids. So, after S-phase there are two sister chromatids- (Cp-Cp), and (Cz-Cz) – note that each pair two replicated chromosomes.

III. During prophase I, the homologous chromosomes pair.

So, the overall homologous structure is- [(Cp-Cp)---(Cz-Cz)].

IV. Meiosis I non-disjunction: During anaphase I of meiosis I, the homologous chromosomes (consisting of 2 pair of sister chromatids – one pair of maternal and other pair of fraternal chromosomes) separate from each other.

It forms two daughter cells - one with (Cp-Cp), and the other with (Cz-Cz).

# In case of non-disjunction, the resultant cells would be (Cp) and (Cp—Cz-Cz).

# If the (Cp—Cz-Cz) undergoes normal disjunction in meiosis II, the resultant male gametes would be (Cp-Cz) and (Cz).

It is the (Cp-Cz) male gamete that fuses with normal female gamete (Ct) and would give (Cp-Cr-Cz) phenotype.

Therefore, the abnormal phenotype (Cp-Cr-Cz) is due to meiosis I non-disjunction.

Or, correct answer is: Male, Meiosis I

V. Meiosis II non-disjunction: The sister chromatids (two replicated chromosomes held together) separate during anaphase II of meiosis II.

# If the cell undergoes normal disjunction in meiosis I, the resultant daughter cells would be- one with (Cp-Cp), and the other with (Cz-Cz).

# Now, if there is non-disjunction in meiosis II, the resultant gametes would be either (Cp-Cp) or (Cz-Cz) BUT NEVER (Cp-Cz). So, fusion of either of the two male gametes with normal female gamete can never lead to abnormal pehnotype because none of the male gametes simultaneously carry Cpa and Cz alleles.

Therefore, meiosis II non-disjunction can’t lead to abnormal (Cp-Cr-Cz) phenotype.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.