BIOD 210 Genetics Final Exam Principles of Heredity
Molecular Genetics Actual Exam 2026/2027 – Verified
Answers with Detailed Rationales – Pass Guaranteed
– A+ Graded
======================================================
========== SECTION 1: MENDELIAN GENETICS & INHERITANCE
PATTERNS (12 Questions)
Q1: In pea plants, yellow seeds (Y) are dominant to green seeds (y), and round seeds (R)
are dominant to wrinkled seeds (r). A plant with genotype YyRr is crossed with a plant
with genotype yyrr. What proportion of offspring will have yellow and wrinkled seeds?
A. 1/4 [CORRECT]
B. 1/2
C. 1/8
D. 1/16
Correct Answer: A
Rationale: Key concept: Dihybrid test cross (YyRr × yyrr). Correct answer: Yellow
wrinkled = Y_ rr. From Yy × yy → 1/2 Yy (yellow); from Rr × rr → 1/2 rr (wrinkled).
Product rule: 1/2 × 1/2 = 1/4. B is wrong because that would be probability of yellow
alone OR wrinkled alone. C is wrong because that would be probability of yyrr (green
wrinkled). D is wrong because that is the proportion of yyrr from a YyRr × YyRr cross.
Q2: Snapdragons show incomplete dominance for flower color: red (C^R), white (C^W),
with heterozygotes being pink. If two pink snapdragons are crossed, what is the
expected phenotypic ratio?
A. 3 red : 1 white
B. 1 red : 2 pink : 1 white [CORRECT]
C. 1 red : 1 white
D. All pink
,Correct Answer: B
Rationale: Key concept: Incomplete dominance produces intermediate heterozygote
phenotypes. Correct answer: C^R C^W × C^R C^W produces 1 C^R C^R (red) : 2 C^R C^W
(pink) : 1 C^W C^W (white). A is wrong because it assumes complete dominance. C is
wrong because that would be a test cross ratio of pure lines. D is wrong because it
ignores segregation of homozygotes.
Q3: Labrador coat color is controlled by two genes: the B locus determines pigment type
(B=black, b=brown), and the E locus determines pigment deposition (E=pigmented,
e=yellow, epistatic to B). A dihybrid cross of BbEe × BbEe produces what phenotypic
ratio?
A. 9:3:3:1
B. 9:4:3
C. 9:3:4 [CORRECT]
D. 12:3:1
Correct Answer: C
Rationale: Key concept: Recessive epistasis (ee masks B locus expression). Correct
answer: 9 B_E_ (black) : 3 bbE_ (brown) : 4 (3 B_ee + 1 bbee) (yellow). A is wrong
because it ignores epistasis and shows standard dihybrid ratio. B is wrong because it
reverses the brown and yellow ratios. D is wrong because that represents dominant
epistasis, not recessive.
Q4: Hemophilia is an X-linked recessive disorder. A carrier female (X^H X^h) and a
normal male (X^H Y) have children. What is the probability their son will have
hemophilia?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
Correct Answer: C
Rationale: Key concept: X-linked inheritance and sex-specific probabilities. Correct
answer: Sons inherit X only from mother; 50% chance of receiving X^h. A is wrong
because it assumes sons cannot inherit the trait from carrier mothers. B is wrong
,because it calculates probability across all children rather than specifically for sons. D is
wrong because it assumes the mother is affected rather than a carrier.
Q5: In the ABO blood group system, alleles I^A and I^B are codominant, and both are
dominant to i. A man with type A blood (heterozygous) and a woman with type B blood
(heterozygous) have children. What is the probability of a child with type O blood?
A. 0%
B. 25% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: Key concept: Multiple alleles and codominance inheritance. Correct answer:
I^A i × I^B i produces 25% ii (type O). A is wrong because it ignores the heterozygosity of
both parents. C is wrong because it miscalculates the probability of homozygous
recessive. D is wrong because it confuses with the probability of having type A or B
blood.
Q6: Genes A and B are linked on the same chromosome. A test cross of AB/ab (cis) ×
ab/ab produces the following offspring: AB 460, ab 440, Ab 50, aB 50. What is the
recombination frequency between these genes?
A. 5%
B. 10% [CORRECT]
C. 20%
D. 50%
Correct Answer: B
Rationale: Key concept: Recombination frequency = (recombinants/total) × 100. Correct
answer: (50+50)/(460+440+50+50) = 100/1000 = 10%. A is wrong because it uses only
one recombinant class (50/1000). C is wrong because it uses the wrong denominator or
doubles the recombinant count. D is wrong because it assumes independent
assortment when genes are actually linked.
Q7: Male-pattern baldness is a sex-influenced trait where allele B is dominant in males
but recessive in females. A heterozygous (Bb) man and heterozygous (Bb) woman have
children. What is the probability a son will be bald?
A. 25%
, B. 50%
C. 75% [CORRECT]
D. 100%
Correct Answer: C
Rationale: Key concept: Sex-influenced dominance patterns. Correct answer: In males, B
is dominant, so both BB and Bb genotypes result in baldness; cross yields 3/4 B_ (bald).
A is wrong because it treats B as recessive in males. B is wrong because it counts only
BB homozygotes. D is wrong because it includes bb (non-bald) offspring.
Q8: Sickle cell anemia demonstrates pleiotropy, where a single mutation affects multiple
organ systems. This occurs because:
A. The mutation creates a lethal dominant allele
B. The altered hemoglobin affects blood flow, oxygen delivery, and organ perfusion
throughout the body [CORRECT]
C. Multiple genes are mutated simultaneously
D. The mutation only affects red blood cell shape
Correct Answer: B
Rationale: Key concept: Pleiotropy—one gene affects multiple phenotypic traits. Correct
answer: The hemoglobin defect causes systemic effects including spleen damage, pain
crises, and organ failure. A is wrong because sickle cell is recessive and not necessarily
lethal in heterozygotes. C is wrong because pleiotropy involves one gene, not multiple
gene mutations. D is wrong because it ignores the systemic consequences of the
cellular defect.
Q9: In cattle, coat color alleles for red (C^R) and white (C^W) show codominance,
producing roan (spotted) heterozygotes. Which cross produces 100% roan offspring?
A. Red × White [CORRECT]
B. Roan × Roan
C. Roan × White
D. Roan × Red
Correct Answer: A
Rationale: Key concept: Codominance produces distinct heterozygote phenotype.
Correct answer: C^R C^R × C^W C^W yields all C^R C^W (roan). B is wrong because it
Molecular Genetics Actual Exam 2026/2027 – Verified
Answers with Detailed Rationales – Pass Guaranteed
– A+ Graded
======================================================
========== SECTION 1: MENDELIAN GENETICS & INHERITANCE
PATTERNS (12 Questions)
Q1: In pea plants, yellow seeds (Y) are dominant to green seeds (y), and round seeds (R)
are dominant to wrinkled seeds (r). A plant with genotype YyRr is crossed with a plant
with genotype yyrr. What proportion of offspring will have yellow and wrinkled seeds?
A. 1/4 [CORRECT]
B. 1/2
C. 1/8
D. 1/16
Correct Answer: A
Rationale: Key concept: Dihybrid test cross (YyRr × yyrr). Correct answer: Yellow
wrinkled = Y_ rr. From Yy × yy → 1/2 Yy (yellow); from Rr × rr → 1/2 rr (wrinkled).
Product rule: 1/2 × 1/2 = 1/4. B is wrong because that would be probability of yellow
alone OR wrinkled alone. C is wrong because that would be probability of yyrr (green
wrinkled). D is wrong because that is the proportion of yyrr from a YyRr × YyRr cross.
Q2: Snapdragons show incomplete dominance for flower color: red (C^R), white (C^W),
with heterozygotes being pink. If two pink snapdragons are crossed, what is the
expected phenotypic ratio?
A. 3 red : 1 white
B. 1 red : 2 pink : 1 white [CORRECT]
C. 1 red : 1 white
D. All pink
,Correct Answer: B
Rationale: Key concept: Incomplete dominance produces intermediate heterozygote
phenotypes. Correct answer: C^R C^W × C^R C^W produces 1 C^R C^R (red) : 2 C^R C^W
(pink) : 1 C^W C^W (white). A is wrong because it assumes complete dominance. C is
wrong because that would be a test cross ratio of pure lines. D is wrong because it
ignores segregation of homozygotes.
Q3: Labrador coat color is controlled by two genes: the B locus determines pigment type
(B=black, b=brown), and the E locus determines pigment deposition (E=pigmented,
e=yellow, epistatic to B). A dihybrid cross of BbEe × BbEe produces what phenotypic
ratio?
A. 9:3:3:1
B. 9:4:3
C. 9:3:4 [CORRECT]
D. 12:3:1
Correct Answer: C
Rationale: Key concept: Recessive epistasis (ee masks B locus expression). Correct
answer: 9 B_E_ (black) : 3 bbE_ (brown) : 4 (3 B_ee + 1 bbee) (yellow). A is wrong
because it ignores epistasis and shows standard dihybrid ratio. B is wrong because it
reverses the brown and yellow ratios. D is wrong because that represents dominant
epistasis, not recessive.
Q4: Hemophilia is an X-linked recessive disorder. A carrier female (X^H X^h) and a
normal male (X^H Y) have children. What is the probability their son will have
hemophilia?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
Correct Answer: C
Rationale: Key concept: X-linked inheritance and sex-specific probabilities. Correct
answer: Sons inherit X only from mother; 50% chance of receiving X^h. A is wrong
because it assumes sons cannot inherit the trait from carrier mothers. B is wrong
,because it calculates probability across all children rather than specifically for sons. D is
wrong because it assumes the mother is affected rather than a carrier.
Q5: In the ABO blood group system, alleles I^A and I^B are codominant, and both are
dominant to i. A man with type A blood (heterozygous) and a woman with type B blood
(heterozygous) have children. What is the probability of a child with type O blood?
A. 0%
B. 25% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: Key concept: Multiple alleles and codominance inheritance. Correct answer:
I^A i × I^B i produces 25% ii (type O). A is wrong because it ignores the heterozygosity of
both parents. C is wrong because it miscalculates the probability of homozygous
recessive. D is wrong because it confuses with the probability of having type A or B
blood.
Q6: Genes A and B are linked on the same chromosome. A test cross of AB/ab (cis) ×
ab/ab produces the following offspring: AB 460, ab 440, Ab 50, aB 50. What is the
recombination frequency between these genes?
A. 5%
B. 10% [CORRECT]
C. 20%
D. 50%
Correct Answer: B
Rationale: Key concept: Recombination frequency = (recombinants/total) × 100. Correct
answer: (50+50)/(460+440+50+50) = 100/1000 = 10%. A is wrong because it uses only
one recombinant class (50/1000). C is wrong because it uses the wrong denominator or
doubles the recombinant count. D is wrong because it assumes independent
assortment when genes are actually linked.
Q7: Male-pattern baldness is a sex-influenced trait where allele B is dominant in males
but recessive in females. A heterozygous (Bb) man and heterozygous (Bb) woman have
children. What is the probability a son will be bald?
A. 25%
, B. 50%
C. 75% [CORRECT]
D. 100%
Correct Answer: C
Rationale: Key concept: Sex-influenced dominance patterns. Correct answer: In males, B
is dominant, so both BB and Bb genotypes result in baldness; cross yields 3/4 B_ (bald).
A is wrong because it treats B as recessive in males. B is wrong because it counts only
BB homozygotes. D is wrong because it includes bb (non-bald) offspring.
Q8: Sickle cell anemia demonstrates pleiotropy, where a single mutation affects multiple
organ systems. This occurs because:
A. The mutation creates a lethal dominant allele
B. The altered hemoglobin affects blood flow, oxygen delivery, and organ perfusion
throughout the body [CORRECT]
C. Multiple genes are mutated simultaneously
D. The mutation only affects red blood cell shape
Correct Answer: B
Rationale: Key concept: Pleiotropy—one gene affects multiple phenotypic traits. Correct
answer: The hemoglobin defect causes systemic effects including spleen damage, pain
crises, and organ failure. A is wrong because sickle cell is recessive and not necessarily
lethal in heterozygotes. C is wrong because pleiotropy involves one gene, not multiple
gene mutations. D is wrong because it ignores the systemic consequences of the
cellular defect.
Q9: In cattle, coat color alleles for red (C^R) and white (C^W) show codominance,
producing roan (spotted) heterozygotes. Which cross produces 100% roan offspring?
A. Red × White [CORRECT]
B. Roan × Roan
C. Roan × White
D. Roan × Red
Correct Answer: A
Rationale: Key concept: Codominance produces distinct heterozygote phenotype.
Correct answer: C^R C^R × C^W C^W yields all C^R C^W (roan). B is wrong because it
