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TFM02 ACTUAL EXAM QUESTIONS AND CORRECT
DETAILED ANSWERS LATEST UPDATE THIS YEAR
EXAM COVERAGE FOR TFM02
Principles of transmission genetics, molecular inheritance, population genetics, quantitative
traits, linkage and mapping, mutation analysis, epigenetic regulation, genetic drift, selection
models, inbreeding, Hardy-Weinberg equilibrium, gene interactions, penetrance, expressivity,
mitochondrial inheritance, genomic imprinting, recombination frequency, genetic markers, QTL
mapping, heritability estimates, twin studies, assortative mating, founder effects, bottleneck
events, neutral theory, and evolutionary genetics applications.
1. A plant with purple flowers (PP) is crossed with white flowers (pp). All F1 offspring have
purple flowers. What is the expected genotypic ratio in the F2 generation?
A) 1:2:1
B) 3:1
C) 1:1
D) 9:3:3:1
Answer: A — The monohybrid cross of heterozygotes yields 1 PP : 2 Pp : 1 pp.
2. In a population of 500 rabbits, 80 have white fur (recessive). Assuming Hardy-Weinberg
equilibrium, how many are heterozygous?
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A) 160
B) 240
C) 320
D) 400
Answer: B — q² = 80/500 = 0.16, q=0.4, p=0.6, 2pq=0.48, 0.48×500=240.
3. A patient presents with muscle weakness and ragged-red fibers on biopsy. The mother is
affected but father is not. Inheritance pattern?
A) Autosomal dominant
B) Autosomal recessive
C) X-linked
D) Mitochondrial
Answer: D — Mitochondrial disorders are maternally inherited and cause ragged-red fibers.
4. Two genes on the same chromosome show 12% recombination. How far apart are they?
A) 6 cM
B) 12 cM
C) 24 cM
D) 48 cM
Answer: B — Recombination frequency directly equals map distance in centiMorgans.
5. A quantitative trait has broad-sense heritability of 0.8. This means:
A) 80% of trait variance is due to genetic differences
B) 80% of trait is determined by a single gene
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C) Environment explains 80% of variance
D) Heritability is 0.2 in narrow sense
Answer: A — Broad-sense heritability includes all genetic effects (additive, dominant, epistatic).
6. In a genome-wide association study, a SNP has p=5×10⁻⁸. This is considered significant
because:
A) It equals the standard threshold for multiple testing correction
B) It indicates a causal mutation
C) It proves linkage disequilibrium
D) It shows 100% penetrance
Answer: A — The conventional genome-wide significance threshold is 5×10⁻⁸ after Bonferroni
correction.
7. A child has Angelman syndrome. The mutation is a deletion on chromosome 15 inherited
from:
A) Father
B) Mother
C) Either parent equally
D) Both parents
Answer: B — Angelman syndrome results from maternal deletion of 15q11-q13; paternal
deletion causes Prader-Willi.
8. In a population of 1000, the frequency of the AA genotype is 0.49. How many Aa individuals
are expected under HWE?
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A) 210
B) 420
C) 490
D) 700
Answer: B — p²=0.49, p=0.7, q=0.3, 2pq=0.42, 0.42×1000=420.
9. A farmer selects the tallest 10% of corn plants to breed. The next generation’s mean height
increases by 4 cm. The selection differential was 10 cm. What is the realized heritability?
A) 0.2
B) 0.4
C) 0.6
D) 0.8
Answer: B — Realized h² = response/selection differential = 4/10 = 0.4.
10. A pedigree shows affected males in every generation, all daughters of affected males are
affected, but sons of affected males are normal. Pattern?
A) Autosomal dominant
B) X-linked dominant
C) X-linked recessive
D) Y-linked
Answer: B — X-linked dominant: affected males pass to all daughters but no sons.
11. Two true-breeding plants with different flower colors produce F1 all pink. F2 shows 1 red:2
pink:1 white. This is:
TFM02 ACTUAL EXAM QUESTIONS AND CORRECT
DETAILED ANSWERS LATEST UPDATE THIS YEAR
EXAM COVERAGE FOR TFM02
Principles of transmission genetics, molecular inheritance, population genetics, quantitative
traits, linkage and mapping, mutation analysis, epigenetic regulation, genetic drift, selection
models, inbreeding, Hardy-Weinberg equilibrium, gene interactions, penetrance, expressivity,
mitochondrial inheritance, genomic imprinting, recombination frequency, genetic markers, QTL
mapping, heritability estimates, twin studies, assortative mating, founder effects, bottleneck
events, neutral theory, and evolutionary genetics applications.
1. A plant with purple flowers (PP) is crossed with white flowers (pp). All F1 offspring have
purple flowers. What is the expected genotypic ratio in the F2 generation?
A) 1:2:1
B) 3:1
C) 1:1
D) 9:3:3:1
Answer: A — The monohybrid cross of heterozygotes yields 1 PP : 2 Pp : 1 pp.
2. In a population of 500 rabbits, 80 have white fur (recessive). Assuming Hardy-Weinberg
equilibrium, how many are heterozygous?
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A) 160
B) 240
C) 320
D) 400
Answer: B — q² = 80/500 = 0.16, q=0.4, p=0.6, 2pq=0.48, 0.48×500=240.
3. A patient presents with muscle weakness and ragged-red fibers on biopsy. The mother is
affected but father is not. Inheritance pattern?
A) Autosomal dominant
B) Autosomal recessive
C) X-linked
D) Mitochondrial
Answer: D — Mitochondrial disorders are maternally inherited and cause ragged-red fibers.
4. Two genes on the same chromosome show 12% recombination. How far apart are they?
A) 6 cM
B) 12 cM
C) 24 cM
D) 48 cM
Answer: B — Recombination frequency directly equals map distance in centiMorgans.
5. A quantitative trait has broad-sense heritability of 0.8. This means:
A) 80% of trait variance is due to genetic differences
B) 80% of trait is determined by a single gene
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C) Environment explains 80% of variance
D) Heritability is 0.2 in narrow sense
Answer: A — Broad-sense heritability includes all genetic effects (additive, dominant, epistatic).
6. In a genome-wide association study, a SNP has p=5×10⁻⁸. This is considered significant
because:
A) It equals the standard threshold for multiple testing correction
B) It indicates a causal mutation
C) It proves linkage disequilibrium
D) It shows 100% penetrance
Answer: A — The conventional genome-wide significance threshold is 5×10⁻⁸ after Bonferroni
correction.
7. A child has Angelman syndrome. The mutation is a deletion on chromosome 15 inherited
from:
A) Father
B) Mother
C) Either parent equally
D) Both parents
Answer: B — Angelman syndrome results from maternal deletion of 15q11-q13; paternal
deletion causes Prader-Willi.
8. In a population of 1000, the frequency of the AA genotype is 0.49. How many Aa individuals
are expected under HWE?
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A) 210
B) 420
C) 490
D) 700
Answer: B — p²=0.49, p=0.7, q=0.3, 2pq=0.42, 0.42×1000=420.
9. A farmer selects the tallest 10% of corn plants to breed. The next generation’s mean height
increases by 4 cm. The selection differential was 10 cm. What is the realized heritability?
A) 0.2
B) 0.4
C) 0.6
D) 0.8
Answer: B — Realized h² = response/selection differential = 4/10 = 0.4.
10. A pedigree shows affected males in every generation, all daughters of affected males are
affected, but sons of affected males are normal. Pattern?
A) Autosomal dominant
B) X-linked dominant
C) X-linked recessive
D) Y-linked
Answer: B — X-linked dominant: affected males pass to all daughters but no sons.
11. Two true-breeding plants with different flower colors produce F1 all pink. F2 shows 1 red:2
pink:1 white. This is:
