Part 6 — Meiosis Disorders, Genetic Consequences, Advanced Checkpoints & Clinical Correlations (Q126–150)

Q126.

Non-disjunction during meiosis I leads to gametes with:
A. All normal chromosome numbers
B. Two with n+1, two with n–1 ✅
C. All with n chromosomes
D. All with 2n chromosomes

Explanation:

**A. Normal only if proper separation occurs.
**B. Correct — failure of homologues to separate produces half gametes with extra, half with missing chromosome.
**C. n only = normal meiosis.
**D. 2n gametes arise from failure of meiosis entirely.

Q127.

Non-disjunction during meiosis II results in:
A. 4 abnormal gametes
B. 2 normal, 1 n+1, 1 n–1 ✅
C. 3 normal, 1 abnormal
D. All normal

Explanation:

**A. Not all four are abnormal.
**B. Correct — sister chromatids fail to separate → 2 abnormal gametes.
C/D. Incorrect counts.

Q128.

Down syndrome is an example of:
A. Monosomy
B. Trisomy ✅
C. Polyploidy
D. Euploidy

Explanation:

**A. Monosomy = single chromosome loss.
**B. Correct — Trisomy 21 = Down syndrome.
**C. Polyploidy = multiple sets.
**D. Euploidy = normal set.

Q129.

Turner’s syndrome individuals are:
A. XXY
B. XO ✅
C. XXX
D. XY

Explanation:

**A. XXY = Klinefelter’s.
**B. XO (Correct): Monosomy X → Turner’s.
**C. XXX = Triple X.
**D. XY = normal male.

Q130.

Klinefelter’s syndrome is caused by:
A. XO
B. XXY ✅
C. XXX
D. XYY

Explanation:

**A. XO = Turner’s.
**B. XXY (Correct): male with extra X → sterility, gynecomastia.
**C. XXX = Triple X.
**D. XYY = supermale.

Q131.

Edwards syndrome results from trisomy of:
A. Chromosome 13
B. Chromosome 18 ✅
C. Chromosome 21
D. Chromosome X

Explanation:

**A. Trisomy 13 = Patau syndrome.
**B. Correct — trisomy 18 = Edwards syndrome.
**C. Trisomy 21 = Down syndrome.
**D. Sex chromosome disorders are different.

Q132.

Patau syndrome results from trisomy of:
A. Chromosome 13 ✅
B. Chromosome 18
C. Chromosome 21
D. Chromosome 22

Explanation:

**A. Correct — Trisomy 13 = Patau.
**B. Trisomy 18 = Edwards.
**C. Trisomy 21 = Down.
**D. Trisomy 22 not classical syndrome.

Q133.

The spindle assembly checkpoint ensures:
A. Chromosomes replicated fully
B. Kinetochores properly attached to spindle ✅
C. DNA damage repaired before S phase
D. Cytokinesis is completed

Explanation:

**A. Replication = G2/M.
**B. Correct — prevents premature anaphase.
**C. G1/S monitors DNA.
**D. Cytokinesis occurs after mitosis.

Q134.

Meiotic recombination occurs during:
A. Pachytene ✅
B. Diplotene
C. Leptotene
D. Diakinesis

Explanation:

**A. Correct — crossing over occurs at pachytene.
B–D: Different substages.

Q135.

Crossing over failure results in:
A. Genetic diversity increase
B. Reduced variability in gametes ✅
C. Increased ploidy
D. Enhanced replication

Explanation:

**A. Failure reduces, not increases variation.
**B. Correct — no new allele combinations formed.
**C. Ploidy unaffected.
**D. Replication independent.

Q136.

The protein cohesin ensures:
A. Spindle fiber attachment
B. Sister chromatid cohesion ✅
C. Crossing over
D. Telomere protection

Explanation:

**A. Attachment is kinetochore role.
**B. Correct — holds sister chromatids until anaphase.
**C. Mediated by recombination proteins, not cohesin.
**D. Telomeres protected by shelterin proteins.

Q137.

The enzyme separase functions by:
A. Joining Okazaki fragments
B. Cleaving cohesin ✅
C. Synthesizing DNA
D. Attaching kinetochores

Explanation:

**A. Ligase joins Okazaki fragments.
**B. Correct — separase cuts cohesin, triggering chromatid separation.
C/D. Not its roles.

Q138.

Failure of spindle assembly checkpoint may lead to:
A. Polyploidy
B. Aneuploidy ✅
C. DNA mutations
D. Crossing over defects

Explanation:

**A. Polyploidy = whole sets.
**B. Correct — mis-segregation → abnormal numbers (aneuploidy).
**C. Mutation: base-level, not checkpoint.
**D. Crossing over occurs earlier.

Q139.

The experimental use of colchicine blocks:
A. DNA replication
B. Microtubule polymerization ✅
C. Cyclin degradation
D. Cohesin cleavage

Explanation:

**A. No effect on DNA replication.
**B. Correct — spindle microtubule assembly inhibited.
**C. Cyclins degraded by ubiquitination.
**D. Cohesin cleavage via separase.

Q140.

If cytokinesis fails after mitosis, the result is:
A. Aneuploidy
B. Polyploidy ✅
C. Trisomy
D. Monosomy

Explanation:

**A. Aneuploidy = abnormal chromosome numbers, not whole genome duplication.
**B. Correct — nucleus divides but cytoplasm doesn’t → polyploid cell.
C/D. Specific chromosome errors.

Q141.

Gametes formed from nondisjunction fertilization may result in:
A. Normal diploid zygote
B. Aneuploid zygote ✅
C. Polyploid zygote
D. Haploid embryo

Explanation:

**A. Diploid only if both gametes normal.
**B. Correct — abnormal gamete fuses with normal → zygote with +/– chromosomes.
**C. Polyploid from entire set duplication.
**D. Haploid embryo not viable in humans.

Q142.

Checkpoint failure leading to persistent DNA damage accumulation may cause:
A. Cell senescence
B. Apoptosis
C. Cancer ✅
D. Differentiation

Explanation:

A/B: Possible, but if checkpoint fails, DNA-damaged cells survive → cancer.
**C. Correct — unchecked proliferation of damaged cells.
**D. Differentiation unrelated.

Q143.

Telomerase reactivation in somatic cells often leads to:
A. Limited cell lifespan
B. Cancer cell immortality ✅
C. Apoptosis
D. Senescence

Explanation:

**A. Limited lifespan occurs when telomerase is off.
**B. Correct — reactivation stabilizes telomeres, allowing indefinite divisions (hallmark of cancer).
C/D. Opposite effects.

Q144.

Which checkpoint prevents entry into mitosis with unreplicated DNA?
A. G1/S
B. G2/M ✅
C. Spindle
D. Cytokinesis

Explanation:

**A. G1/S: before replication.
**B. G2/M (Correct): ensures replication complete before mitosis.
**C. Spindle: metaphase-anaphase.
**D. Not standard.

Q145.

In meiosis, genetic recombination ensures:
A. DNA repair
B. New allele combinations ✅
C. Telomere elongation
D. Cytokinesis completion

Explanation:

**A. Some repair happens, but main function = variation.
**B. Correct — recombination increases genetic diversity.
**C. Telomere elongation = telomerase.
**D. Cytokinesis unrelated.

Q146.

ATM and ATR kinases are activated by:
A. Telomere shortening
B. DNA damage ✅
C. Chromosome alignment
D. Cyclin degradation

Explanation:

**A. Not telomeres specifically.
**B. Correct — ATM responds to double-strand breaks, ATR to replication stress.
C/D. Not their role.

Q147.

A triploid organism results from:
A. Monosomy
B. Fertilization of diploid and haploid gametes ✅
C. Non-disjunction only
D. Gene mutation

Explanation:

**A. Monosomy: one chromosome missing.
**B. Correct — 2n gamete + n gamete → 3n.
**C. Non-disjunction → aneuploid, not triploid.
**D. Mutation = sequence change.

Q148.

The BRCA1 and BRCA2 genes normally function in:
A. Cell cycle checkpoints
B. DNA double-strand break repair ✅
C. Cohesin cleavage
D. Telomere addition

Explanation:

**A. Not checkpoints directly.
**B. Correct — homologous recombination repair.
**C. Separase acts on cohesin.
**D. Telomerase does telomeres.

Q149.

Which meiotic error could cause gametes with 24 chromosomes in humans?
A. Normal meiosis
B. Non-disjunction of one chromosome ✅
C. Crossing over failure
D. Cytokinesis failure

Explanation:

**A. Normal meiosis = 23.
**B. Correct — nondisjunction → n+1 gamete = 24.
**C. Crossing over absence doesn’t change count.
**D. Cytokinesis failure → polyploid, not single extra.

Q150.

If a human sperm with 24 chromosomes fertilizes a normal egg, the result is:
A. 45 chromosomes
B. 46 chromosomes
C. 47 chromosomes ✅
D. 23 chromosomes

Explanation:

**A. 45 = monosomy.
**B. 46 = normal.
**C. 47 (Correct): extra chromosome → trisomy.
**D. 23 = haploid, not zygote.

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Part 6 — Meiosis Disorders, Genetic Consequences, Advanced Checkpoints & Clinical Correlations (Q126–150)