AP Biology: Unit 5, Non-Mendelian Inheritance (Drill 19)

Question 1. Two pink-flowered plants (RW genotype, incomplete dominance) are crossed. What phenotypic ratio is expected in the offspring?

• A) All pink, because both parents are pink
• B) 3 pink : 1 white
• C) 1 red : 2 pink : 1 white ✓
• D) 1 red : 1 pink : 1 white : 1 colorless

Explanation: Correct: (C) In incomplete dominance, the heterozygote (RW) shows an intermediate phenotype (pink). Crossing RW x RW produces 1 RR : 2 RW : 1 WW. Because neither allele is fully dominant, each genotype produces a distinct phenotype: RR = red, RW = pink, WW = white. The phenotypic ratio mirrors the genotypic ratio: 1 red : 2 pink : 1 white. This differs from standard dominant-recessive inheritance (which gives 3:1) because genotype and phenotype are in 1:1 correspondence. Option B (3 pink : 1 white) incorrectly treats one allele as dominant.

Question 2. A woman with blood type A (genotype I^Ai) and a man with blood type B (genotype I^Bi) have children. Which of the following correctly lists all possible blood types their children could have?

• A) AB and O only, because neither parent has type A or B in homozygous form
• B) A and B only, because the codominant alleles I^A and I^B will always be expressed in offspring
• C) AB only, because crossing type A with type B always produces type AB offspring
• D) A, B, AB, and O, because the cross I^Ai x I^Bi produces all four possible genotypic combinations in equal proportions. ✓

Explanation: Correct: (D) Using a Punnett square for I^Ai x I^Bi: offspring genotypes are I^AI^B (blood type AB), I^Ai (blood type A), I^Bi (blood type B), and ii (blood type O), each with 1/4 probability. This cross produces all four ABO blood types in equal proportions. It demonstrates both codominance (I^AI^B → type AB) and simple dominance (I^A and I^B each dominant over i). The ability of type A and type B parents to produce type O children illustrates how recessive alleles can be hidden in parents but expressed in offspring.

Question 3. Red-green color blindness is X-linked recessive. A woman with normal vision whose father was color blind has children with a man who has normal vision. What is the probability that their son will be color blind?

• A) 50%, because the mother is a carrier (X^AX^a) and passes the X^a allele to half her sons, who will be hemizygous and express the trait. ✓
• B) 0%, because the father has normal vision and cannot pass color blindness to sons.
• C) 25%, because 1/4 of all children are expected to be color blind.
• D) 100%, because the mother's father was color blind, ensuring all sons inherit color blindness.

Explanation: Correct: (A) The woman has normal vision but her father was color blind (X^aY). She received X^a from her father and X^A from her normal-vision mother, so her genotype is X^AX^a (carrier). The father has normal vision (X^AY). The cross X^AX^a x X^AY produces: X^AX^A (normal daughter), X^AX^a (carrier daughter), X^AY (normal son), and X^aY (color-blind son), each at 1/4. Among sons only, 1/2 are color blind. Sons inherit their single X exclusively from their mother; since half the mother's X chromosomes carry X^a, 50% of sons will be color blind. Option C (25%) is the probability among all children, not among sons specifically.

Question 4. In Labrador retrievers, a cross between two black dogs (BbEe x BbEe) produces a litter of 16 puppies. Based on the epistasis model described in the passage, how many puppies are expected to be yellow?

• A) 1
• B) 3
• C) 4 ✓
• D) 9

Explanation: Correct: (C) The E gene controls whether pigment is deposited in the coat: ee dogs are yellow regardless of genotype at the B locus. P(ee) from Ee x Ee = 1/4. Therefore 1/4 of offspring are expected to be yellow: 1/4 x 16 = 4 yellow puppies. This is recessive epistasis. The remaining 12 puppies are either black (9/16, genotype B_E_) or chocolate/brown (3/16, genotype bbE_). The overall phenotypic ratio is 9 black : 3 chocolate : 4 yellow.

Question 5. A cross between two black mice produces offspring in the ratio 9 black : 3 brown : 4 white. Which of the following best explains the 4 white offspring?

• A) A second gene controls pigment production; when homozygous recessive at this locus, animals are albino (no pigment), masking the effects of the color gene entirely. ✓
• B) The white offspring result from incomplete dominance at the color gene, producing an intermediate (white) phenotype in heterozygotes.
• C) The white offspring represent a new mutation that arose spontaneously during the cross.
• D) The white offspring carry two different recessive alleles at the same locus, and this compound heterozygous state prevents pigment expression.

Explanation: Correct: (A) The 9:3:4 ratio is the classic signature of recessive epistasis in a two-gene system. This pattern arises when a second gene controls whether any pigment is produced at all: cc animals cannot produce pigment and are albino, regardless of alleles at the color gene (B/b). When both parents are heterozygous at this epistatic locus (Cc), 1/4 of offspring are cc and therefore white. These cc individuals would have been black or brown if they carried one or two C alleles. The 9:3:4 ratio results from merging the 3 colored-recessive-class (green round equivalent) and 1 double-recessive class from the 9:3:3:1 framework into a single "4" white class.