Genetics Quiz: Dihybrid Crosses
20 questions · exam conditions
0:00
Dihybrid CrossesQuestion 1 of 20

In a certain plant, two unlinked genes affect flower viability. The presence of at least one dominant A allele and at least one dominant B allele is required for a flower to be fertile. All other genotypes (A_bb, aaB_, and aabb) result in sterile flowers. If two AaBb plants are crossed, what is the expected ratio of fertile to sterile plants in the progeny?

9:3:3:1
1:1
15:1
9:7
← Back to quizzes

Genetics Quiz

Genetics Quiz: Dihybrid Crosses

Practice Dihybrid Crosses in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Dihybrid Crosses, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In a certain plant, two unlinked genes affect flower viability. The presence of at least one dominant A allele and at least one dominant B allele is required for a flower to be fertile. All other genotypes (A_bb, aaB_, and aabb) result in sterile flowers. If two AaBb plants are crossed, what is the expected ratio of fertile to sterile plants in the progeny?

  1. 9:3:3:1
  2. 1:1
  3. 15:1
  4. 9:7 (correct answer)

Explanation: This problem describes a case of complementary gene action, a type of epistasis, where two dominant alleles are required to produce a specific phenotype (fertility).

  1. Start with the standard phenotypic outcome of a dihybrid cross between two heterozygotes (AaBb x AaBb), which is 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.
  2. Apply the rule for fertility given in the stem:
    • Plants are fertile only if they have the genotype A_B_.
    • Plants with genotypes A_bb, aaB_, or aabb are sterile.
  3. Group the phenotypic classes based on this rule:
    • The proportion of fertile plants is the proportion of A_B_ genotypes, which is 9/16.
    • The proportion of sterile plants is the sum of the proportions of the other genotypes: P(A_bb) + P(aaB_) + P(aabb) = 3/16 + 3/16 + 1/16 = 7/16.
  4. The resulting ratio of fertile to sterile plants is 9/16 : 7/16, which simplifies to 9:7.
Distractor Rationale:
  • A) 9:3:3:1 is the standard Mendelian ratio for phenotypes, which is modified by the epistatic interaction described.
  • B) 1:1 is a common ratio from a test cross, but not applicable here.
  • C) 15:1 is the ratio for a different epistatic interaction (duplicate dominant epistasis), where a dominant allele at either of two loci is sufficient to produce the dominant phenotype.

Question 2

Two genes in a plant control height (T=tall, t=dwarf) and flower color (P=purple, p=white) and are located on different chromosomes. A plant with genotype TtPp is self-fertilized. An F2 offspring is chosen at random that has the dominant phenotype for both traits. What is the probability that this plant is homozygous for the height gene (TT)?

  1. 1/9
  2. 1/4
  3. 1/3 (correct answer)
  4. 2/3

Explanation: This is a conditional probability problem. We are given that the offspring has the dominant phenotype for both traits (T_P_), and we want to find the probability that it has the TT genotype.

  1. Start with the standard TtPp x TtPp cross.
    • The probability of the dominant phenotype for height (T_) is 3/4.
    • The probability of the dominant phenotype for color (P_) is 3/4.
    • The probability of having both dominant phenotypes (T_P_) is 3/4 * 3/4 = 9/16. This is our condition, the denominator in the conditional probability formula.
  2. Now, find the probability of the event we're interested in, which is being homozygous for height (TT) AND having the double dominant phenotype (T_P_). This simplifies to the genotype TTP_.
    • P(TT) = 1/4.
    • P(P_) = 3/4.
    • P(TTP_) = P(TT) * P(P_) = 1/4 * 3/4 = 3/16. This is the numerator.
  3. Calculate the conditional probability: P(TT | T_P_) = P(TT and T_P_) / P(T_P_).
    • P(TT | T_P_) = (3/16) / (9/16) = 3/9 = 1/3.
Alternatively, since the genes are independent, knowing the phenotype for flower color gives no information about the genotype for height. The question reduces to: Given that a plant from a Tt x Tt cross has the tall phenotype (T_), what is the probability its genotype is TT? The possible genotypes for a tall plant are TT and Tt, with probabilities 1/4 and 1/2, respectively. The total probability of being tall is 1/4 + 1/2 = 3/4. The conditional probability is P(TT | T_) = P(TT) / P(T_) = (1/4) / (3/4) = 1/3. Distractor Rationale:
  • A) 1/9 is the probability that the plant is homozygous dominant for both genes, given it has the double dominant phenotype (P(TTPP | T_P_)).
  • B) 1/4 is the unconditional probability of being TT, P(TT), ignoring the condition that the plant is phenotypically dominant.
  • D) 2/3 is the conditional probability of being heterozygous (Tt) given a dominant phenotype (T_), P(Tt | T_).

Question 3

A plant of genotype AABB is crossed with a plant of genotype aabb. An F1 individual is then crossed with a plant of genotype Aabb. What is the probability that an offspring of this second cross is heterozygous for both genes?

  1. 1/4 (correct answer)
  2. 1/16
  3. 1/2
  4. 3/4

Explanation: This question tests your understanding of dihybrid crosses and probability calculations in genetics. When you see crosses involving two genes, break down the problem step by step and track each gene separately. First, determine the F1 genotype. AABB × aabb produces F1 offspring that are all AaBb (heterozygous for both genes). Next, you need to analyze the cross AaBb × Aabb. To find the probability of offspring heterozygous for both genes (AaBb), multiply the individual probabilities for each gene. For gene A: AaBb × Aabb gives you Aa offspring with probability 12\frac{1}{2} (from the Punnett square: AA, Aa, Aa, ab). For gene B: Bb × bb gives you Bb offspring with probability 12\frac{1}{2} (from BB, Bb, bb, bb). Therefore, the probability of AaBb is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Looking at the wrong answers: B) 116\frac{1}{16} would be correct if you were looking for a specific genotype in an F2 cross between two dihybrids, but that's not this scenario. C) 12\frac{1}{2} might tempt you if you only considered one gene instead of both. D) 34\frac{3}{4} could result from incorrectly adding probabilities instead of multiplying them. Study tip: In genetics problems involving multiple genes, always work with each gene independently first, then multiply the probabilities together. Remember that "and" means multiply, while "or" means add in probability calculations.

Question 4

In snapdragons, flower color exhibits incomplete dominance (RR=red, Rr=pink, rr=white), while leaf shape has complete dominance (B=broad, b=narrow). A pink-flowered, broad-leafed plant is crossed with a white-flowered, narrow-leafed plant. Of the offspring that have pink flowers, what proportion is expected to have broad leaves?

  1. 1/4
  2. 1/2 (correct answer)
  3. 3/4
  4. 1

Explanation: This is a conditional probability problem that is simplified by the independent assortment of genes. The incomplete dominance is extra information designed to test if the student can focus on the relevant gene.

  1. Determine parental genotypes. A pink-flowered, broad-leafed plant has genotype RrB_. A white-flowered, narrow-leafed plant is rrbb. Since the cross produces offspring with narrow leaves (bb), the broad-leafed parent must carry the b allele. Therefore, its genotype is RrBb. The cross is RrBb x rrbb.
  2. The question asks for the proportion of broad-leafed offspring among the pink-flowered offspring. This can be stated as P(Broad | Pink).
  3. Because the genes assort independently, the flower color of an offspring does not influence its leaf shape. Therefore, we only need to consider the cross for the leaf shape gene: Bb x bb.
  4. The cross Bb x bb produces offspring with genotypes 1/2 Bb and 1/2 bb. The corresponding phenotypes are 1/2 broad leaves (Bb) and 1/2 narrow leaves (bb).
  5. Thus, regardless of flower color, the proportion of offspring with broad leaves is 1/2. The same is true for the subset of offspring with pink flowers.
Distractor Rationale:
  • A) 1/4 is the overall probability of an offspring being both pink and broad (P(RrBb) = P(Rr) * P(Bb) = 1/2 * 1/2 = 1/4). This answer fails to account for the conditional nature of the question.
  • C) 3/4 is the proportion of dominant phenotype from a standard heterozygous cross (Bb x Bb), which is not the cross performed here.
  • D) 1 (or 100%) would be chosen if the student incorrectly assumes that because the parent was broad-leafed, all its dominant-phenotype offspring must also be broad-leafed.

Question 5

In pea plants, purple flowers (P) are dominant to white (p), and yellow seeds (Y) are dominant to green (y). A cross is performed between a plant of genotype PpYy and a plant of genotype Ppyy. What is the probability that an offspring will have the same phenotype as at least one of its parents?

  1. 1/2
  2. 3/8
  3. 3/4 (correct answer)
  4. 1

Explanation: The first parent (PpYy) has the phenotype purple flowers, yellow seeds. The second parent (Ppyy) has the phenotype purple flowers, green seeds. An offspring's phenotype will match at least one parent if it is either (purple, yellow) or (purple, green). First, determine the probabilities of offspring phenotypes:

  • Cross Pp x Pp gives 3/4 purple (P_) and 1/4 white (pp).
  • Cross Yy x yy gives 1/2 yellow (Y_) and 1/2 green (yy).
Next, use the product rule to find the probabilities of the combined phenotypes:
  • P(purple, yellow) = P(P_) * P(Y_) = 3/4 * 1/2 = 3/8. This phenotype matches the first parent.
  • P(purple, green) = P(P_) * P(yy) = 3/4 * 1/2 = 3/8. This phenotype matches the second parent.
  • P(white, yellow) = P(pp) * P(Y_) = 1/4 * 1/2 = 1/8. This phenotype matches neither parent.
  • P(white, green) = P(pp) * P(yy) = 1/4 * 1/2 = 1/8. This phenotype matches neither parent.
Finally, use the sum rule for mutually exclusive events. The probability of matching at least one parent is P(purple, yellow) + P(purple, green) = 3/8 + 3/8 = 6/8 = 3/4. Distractor Rationale:
  • A) 1/2 is the probability of an offspring having the same genotype as at least one parent (P(PpYy) + P(Ppyy) = (1/21/2) + (1/21/2) = 1/4 + 1/4 = 1/2). This confuses genotype with phenotype.
  • B) 3/8 is the probability of matching only one of the parental phenotypes, failing to sum the probabilities for matching either parent.
  • D) 1 represents certainty and is an unlikely outcome in a dihybrid cross of this nature; it reflects a conceptual error in applying probability rules.

Question 6

In fruit flies, red eyes (R) are dominant to sepia (r) and normal wings (W) are dominant to vestigial (w). A geneticist crosses a red-eyed, normal-winged fly with a sepia-eyed, vestigial-winged fly. The cross produces 200 offspring with the following approximate phenotypic distribution:

  • 52 red, normal
  • 49 red, vestigial
  • 51 sepia, normal
  • 48 sepia, vestigial

Based on the passage, what is the probability of obtaining an offspring from this cross that is homozygous for both genes?

  1. 0
  2. 1/4 (correct answer)
  3. 1/2
  4. 9/16

Explanation: This is a two-step problem. First, deduce the genotypes of the parents from the offspring data. Second, calculate the requested probability from the deduced cross. Step 1: Deduce parental genotypes.

  • The second parent is described as sepia-eyed, vestigial-winged, which is the double recessive phenotype, so its genotype must be rrww.
  • Analyze the ratios for each trait separately to determine the first parent's genotype (P_W_).
  • Eye color ratio: (red) : (sepia) = (52+49) : (51+48) = 101 : 99 ≈ 1:1. A 1:1 phenotypic ratio results from a cross between a heterozygote and a homozygous recessive (Rr x rr).
  • Wing shape ratio: (normal) : (vestigial) = (52+51) : (49+48) = 103 : 97 ≈ 1:1. A 1:1 phenotypic ratio results from a cross between a heterozygote and a homozygous recessive (Ww x ww).
  • Therefore, the first parent's genotype must be RrWw.
Step 2: Calculate the probability.
  • The cross is RrWw x rrww.
  • The question asks for the probability of an offspring that is homozygous for both genes. The possible double-homozygous genotypes are RRWW, RRww, rrWW, and rrww.
  • From the Rr x rr cross, P(RR) = 0 and P(rr) = 1/2.
  • From the Ww x ww cross, P(WW) = 0 and P(ww) = 1/2.
  • Therefore, the only possible double-homozygous offspring genotype is rrww.
  • The probability is P(rrww) = P(rr) * P(ww) = 1/2 * 1/2 = 1/4.
Distractor Rationale:
  • A) 0 is incorrect. While homozygous dominant (RRWW) offspring are impossible, homozygous recessive (rrww) offspring are possible.
  • C) 1/2 could result from incorrectly applying the sum rule (1/2 + 1/2) or by only considering a single trait.
  • D) 9/16 is the probability of the double-dominant phenotype in a standard RrWw x RrWw cross, a memorized ratio that is incorrectly applied to this test cross scenario.

Question 7

In mice, black coat (B) is dominant to brown (b), and short tail (S) is dominant to long (s). A cross is made between a BbSs mouse and a bbss mouse. What is the probability that in a litter of three pups, at least one pup is brown with a long tail (bbss)?

  1. 1/64
  2. 27/64
  3. 3/4
  4. 37/64 (correct answer)

Explanation: This is an "at least one" probability problem. The easiest approach is to calculate the probability of the complementary event (no pups are brown with a long tail) and subtract this from 1.

  1. First, determine the probability of a single pup from this cross being brown with a long tail (bbss). The cross is a dihybrid test cross: BbSs x bbss.
    • For the coat color gene (Bb x bb), P(bb) = 1/2.
    • For the tail length gene (Ss x ss), P(ss) = 1/2.
    • Using the product rule, P(bbss) = P(bb) * P(ss) = 1/2 * 1/2 = 1/4.
  2. Next, determine the probability that a single pup is NOT bbss.
    • P(not bbss) = 1 - P(bbss) = 1 - 1/4 = 3/4.
  3. Now, calculate the probability that all three pups in the litter are not bbss. Since each birth is an independent event, we use the product rule.
    • P(all 3 are not bbss) = (3/4) * (3/4) * (3/4) = (3/4)³ = 27/64.
  4. Finally, the probability of at least one pup being bbss is 1 minus the probability that none are bbss.
    • P(at least one bbss) = 1 - 27/64 = 37/64.
Distractor Rationale:
  • A) 1/64 is the probability that all three pups are bbss ((1/4)³), which is a different scenario.
  • B) 27/64 is the probability of the complementary event (that none of the pups are bbss). This is a common error in "at least one" problems.
  • C) 3/4 is the probability that a single pup is not bbss. This answer ignores the fact that there are three pups in the litter.

Question 8

In dragons, fire-breathing (F) is dominant to non-fire-breathing (f), and green scales (G) are dominant to gold scales (g). A cross between two FfGg dragons is performed. If an offspring is a fire-breather, what is the probability that it also has gold scales?

  1. 3/16
  2. 1/4 (correct answer)
  3. 1/3
  4. 3/4

Explanation: This is a conditional probability problem. We want to find the probability of an offspring having gold scales (gg) given that it is a fire-breather (F_). The formula is P(gg | F_) = P(F_ and gg) / P(F_). From the FfGg x FfGg cross:

  1. The probability of being a fire-breather (F_) is P(FF) + P(Ff) = 1/4 + 1/2 = 3/4.
  2. The probability of having gold scales (gg) is 1/4.
  3. Since the genes assort independently, the probability of being a fire-breather AND having gold scales is P(F_ and gg) = P(F_) * P(gg) = 3/4 * 1/4 = 3/16.
  4. Now, apply the conditional probability formula: P(gg | F_) = (3/16) / (3/4) = (3/16) * (4/3) = 12/48 = 1/4.
Alternatively, because the two genes assort independently, knowing the phenotype for the fire-breathing trait provides no information about the scale color trait. Therefore, the probability of the offspring having gold scales (gg) is simply its independent probability from the Gg x Gg cross, which is 1/4. Distractor Rationale:
  • A) 3/16 is the joint probability P(F_ and gg), which a student might choose if they do not understand how to calculate conditional probability.
  • C) 1/3 is the probability of being homozygous dominant for a trait, given a dominant phenotype (e.g., P(GG | G_)). This is a misapplication of a different conditional probability scenario.
  • D) 3/4 is the probability of being a fire-breather, P(F_). This answer ignores the second part of the question concerning scale color.

Question 9

In a dihybrid cross of two true-breeding parents (AABB x aabb), the F1 generation (AaBb) is self-fertilized. What is the probability that a randomly selected F2 offspring has the same genotype as one of its F1 parents?

  1. 1/4 (correct answer)
  2. 1/16
  3. 1/2
  4. 9/16

Explanation: When tackling dihybrid cross problems, focus on what specific genotype you're looking for and systematically work through the Punnett square or use probability rules. The F1 generation has genotype AaBb. To find the probability that an F2 offspring matches this exact genotype, you need to determine how often AaBb appears when AaBb × AaBb. For a dihybrid cross, treat each gene independently. For the A gene: Aa × Aa produces AA, Aa, Aa, aa (probability of Aa = 2/4 = 1/2). For the B gene: Bb × Bb similarly gives a 1/2 probability of Bb. Since genes assort independently, multiply these probabilities: 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Looking at the wrong answers: B) 1/16 represents the probability of getting any single specific genotype in a complete 4×4 dihybrid Punnett square, but this ignores that AaBb appears multiple times. C) 1/2 would be the probability if you were only considering one gene (like Aa from Aa × Aa), but you need both genes to match. D) 9/16 is the classic ratio for dominant phenotypes in a dihybrid cross (A_B_), but this question asks about genotype, not phenotype. The correct answer is A) 1/4. Study tip: For dihybrid crosses, remember that genotype probabilities multiply across independent genes. Don't confuse genotype ratios with phenotype ratios—phenotype questions often involve those familiar 9:3:3:1 ratios, while genotype questions require more careful counting of specific allele combinations.

Question 10

In Labrador retrievers, the B/b locus determines pigment (B=black, b=brown), but the E/e locus is epistatic, where genotype ee masks pigment expression, resulting in a yellow coat. A black lab known to be heterozygous for both genes (BbEe) is crossed with a yellow lab that is heterozygous for the pigment gene (Bbee). What is the probability of producing a puppy with a brown coat?

  1. 1/8 (correct answer)
  2. 0
  3. 1/4
  4. 3/8

Explanation: This question tests epistasis, where one gene masks the expression of another. When you see coat color genetics involving multiple loci, always identify which gene is epistatic (controlling) before analyzing phenotypes. Let's work through this cross: BbEe × Bbee. The E locus is epistatic - only dogs with at least one E allele can express pigment from the B locus. Dogs with ee genotype are always yellow regardless of their B genotype. For a brown coat, you need: bbE_ (bb for brown pigment AND at least one E to express it). Setting up the cross:

  • From BbEe parent: possible gametes are BE, Be, bE, be (each 25%)
  • From Bbee parent: possible gametes are Be, be (each 50%)
The combinations that produce brown coats (bbE_):
  • bE (from first parent) × Be (from second parent) = bbEe (brown)
  • Probability: 14×12=18\frac{1}{4} × \frac{1}{2} = \frac{1}{8}
Answer A (1/8) is correct - this represents the single combination producing brown offspring. Answer B (0) incorrectly assumes no brown offspring are possible, missing that the yellow parent still carries the E allele needed for pigment expression. Answer C (1/4) likely calculated the probability of bb genotype alone, forgetting about the epistatic requirement for the E allele. Answer D (3/8) may have incorrectly included some yellow (ee) genotypes in the brown category. Study tip: In epistasis problems, always identify the controlling gene first, then determine what genotype combinations actually produce each visible phenotype. Don't just focus on one locus at a time.

Question 11

A cross is performed between two pea plants of genotypes GgWw and Ggww. The genes are unlinked. What is the probability that an offspring will have a genotype that is homozygous for one gene and heterozygous for the other?

  1. 1/4
  2. 3/8
  3. 1/2 (correct answer)
  4. 5/8

Explanation: This question requires calculating the probabilities of several mutually exclusive genotypic outcomes and then using the sum rule. The required condition is (homozygous for G and heterozygous for W) OR (heterozygous for G and homozygous for W).

  1. Analyze the cross for each gene:
    • Gg x Gg -> P(GG) = 1/4 (homo), P(gg) = 1/4 (homo), P(Gg) = 1/2 (hetero).
    • Ww x ww -> P(Ww) = 1/2 (hetero), P(ww) = 1/2 (homo).
  2. Calculate the probability of the first part of the condition: homozygous for G (GG or gg) AND heterozygous for W (Ww).
    • P(homo G) = P(GG) + P(gg) = 1/4 + 1/4 = 1/2.
    • P(hetero W) = P(Ww) = 1/2.
    • P(homo G and hetero W) = P(homo G) * P(hetero W) = 1/2 * 1/2 = 1/4.
  3. Calculate the probability of the second part of the condition: heterozygous for G (Gg) AND homozygous for W (ww).
    • P(hetero G) = P(Gg) = 1/2.
    • P(homo W) = P(ww) = 1/2.
    • P(hetero G and homo W) = P(hetero G) * P(homo W) = 1/2 * 1/2 = 1/4.
  4. Use the sum rule to find the total probability, as these two outcomes are mutually exclusive.
    • Total P = P(homo G, hetero W) + P(hetero G, homo W) = 1/4 + 1/4 = 1/2.
Distractor Rationale:
  • A) 1/4 is the probability of only one of the two conditions (e.g., homozygous for G and heterozygous for W), representing an incomplete analysis.
  • B) 3/8 is the probability of being heterozygous for G and showing the dominant phenotype for W (P(Gg) * P(W_) - but there is no W_ phenotype from Ww x ww). It can also be P(G_ww) from GgWw x Ggww cross.
  • D) 5/8 could arise from various calculation errors, such as incorrectly summing probabilities without applying the product rule first.

Question 12

In humans, brown eyes (B) are dominant to blue (b). The ability to taste phenylthiocarbamide (PTC) (T) is dominant to non-tasting (t). These genes assort independently. A blue-eyed woman who is a taster marries a brown-eyed man who is a non-taster. They have a blue-eyed, non-taster child. What is the probability that their next child will be a brown-eyed taster?

  1. 1/8
  2. 1/4 (correct answer)
  3. 3/8
  4. 9/16

Explanation: This problem requires first deducing the parental genotypes based on their phenotypes and the phenotype of their child, and then calculating the probability for their next child.

  1. Deduce parental genotypes:
    • Woman: blue-eyed (bb), taster (T_). Her genotype is bbT_.
    • Man: brown-eyed (B_), non-taster (tt). His genotype is B_tt.
    • Child: blue-eyed (bb), non-taster (tt). Genotype is bbtt.
  2. Refine parental genotypes using the child's information:
    • The child is bb, so it must have inherited a 'b' allele from each parent. Since the man is brown-eyed (B_), he must be heterozygous (Bb).
    • The child is tt, so it must have inherited a 't' allele from each parent. Since the woman is a taster (T_), she must be heterozygous (Tt).
    • Therefore, the finalized parental genotypes are: Woman (bbTt) and Man (Bbtt).
  3. Calculate the probability for the next child:
    • The question asks for the probability of a brown-eyed taster child (B_T_).
    • Perform the cross: bbTt x Bbtt.
    • For eye color (bb x Bb), the probability of a brown-eyed child (Bb) is 1/2.
    • For PTC tasting (Tt x tt), the probability of a taster child (Tt) is 1/2.
    • Using the product rule, the probability of a brown-eyed, taster child (B_T_) is P(Bb) * P(Tt) = 1/2 * 1/2 = 1/4.
Distractor Rationale:
  • A) 1/8 could be obtained by miscalculating one of the monohybrid probabilities as 1/4 instead of 1/2 (e.g., 1/4 * 1/2).
  • C) 3/8 could be obtained if one parent was incorrectly assumed to be heterozygous for both traits, leading to a cross like BbTt x bbtt, giving P(B_T_) = 1/2 * 1/2 = 1/4. Where would 3/8 come from? Maybe P(B_) from BbxBb (3/4) * P(T_) from TtxTt (1/2) = 3/8. No, Tt x tt is 1/2. P(B_) from Bbxbb (1/2) * P(T_) from TtxTt (3/4) = 3/8. This assumes the wrong cross for one trait.
  • D) 9/16 is the classic probability for the double-dominant phenotype from a dihybrid cross between two double heterozygotes (e.g., BbTt x BbTt), which is not the case here.

Question 13

In an organism, Gene A controls pigment production (A=pigment, a=albino), and Gene B controls pigment color (B=black, b=brown). Gene A is epistatic to Gene B, as an organism must have at least one 'A' allele to produce any pigment. A cross between two AaBb parents produces 320 offspring. How many of these offspring are expected to be brown?

  1. 20
  2. 60 (correct answer)
  3. 80
  4. 180

Explanation: This is a two-step problem involving recessive epistasis. First, find the probability of the brown phenotype, then calculate the expected number of offspring.

  1. Determine the genotype for the brown phenotype.
    • To be brown, an organism must be able to produce pigment, so it needs at least one 'A' allele (A_).
    • It must also have the genetic instruction for brown pigment, which is the 'bb' genotype.
    • Therefore, the genotype for a brown organism is A_bb.
  2. Calculate the probability of this genotype from a cross of two AaBb parents.
    • The cross is AaBb x AaBb.
    • The probability of an A_ offspring from Aa x Aa is 3/4.
    • The probability of a bb offspring from Bb x Bb is 1/4.
    • Using the product rule, P(A_bb) = P(A_) * P(bb) = 3/4 * 1/4 = 3/16.
  3. Calculate the expected number of brown offspring out of a total of 320.
    • Expected number = Total offspring * Probability of being brown
    • Expected number = 320 * (3/16)
    • Expected number = (320 / 16) * 3 = 20 * 3 = 60.
Distractor Rationale:
  • A) 20 is the expected number of albino, brown-genotype offspring (aabb). This is calculated as 320 * (1/16).
  • C) 80 is the expected number of all albino offspring (aa__). This is calculated as 320 * P(aa__) = 320 * (1/4) = 80.
  • D) 180 is the expected number of black offspring (A_B_). This is calculated as 320 * P(A_B_) = 320 * (9/16) = 20 * 9 = 180.

Question 14

In a species of beetle, black body (B) is dominant to brown (b), and long antennae (L) are dominant to short (l). A cross is made between a beetle of genotype BbLl and one of genotype bbLl. What is the probability that an offspring will exhibit at least one dominant phenotype?

  1. 1/8
  2. 3/8
  3. 1/2
  4. 7/8 (correct answer)

Explanation: The most straightforward way to solve for "at least one" is to calculate the probability of the complementary event (exhibiting no dominant phenotypes, i.e., being double recessive) and subtract it from 1.

  1. Determine the probability of the double recessive phenotype (brown body, short antennae), which corresponds to the genotype bbll.
  2. For the body color gene, the cross is Bb x bb. The probability of a bb offspring is 1/2.
  3. For the antennae length gene, the cross is Ll x Ll. The probability of an ll offspring is 1/4.
  4. Using the product rule, the probability of a bbll offspring is P(bb) * P(ll) = 1/2 * 1/4 = 1/8.
  5. The probability of an offspring having at least one dominant phenotype is 1 - P(bbll) = 1 - 1/8 = 7/8.
Alternatively, using the sum rule: P(B_ or L_) = P(B_) + P(L_) - P(B_ and L_). P(B_) = 1/2. P(L_) = 3/4. P(B_ and L_) = 1/2 * 3/4 = 3/8. So, P(at least one dominant) = 1/2 + 3/4 - 3/8 = 4/8 + 6/8 - 3/8 = 7/8. Distractor Rationale:
  • A) 1/8 is the probability of the complementary event (double recessive phenotype). This is a common error where the student calculates the opposite of what is asked.
  • B) 3/8 is the probability of exhibiting both dominant phenotypes (P(B_ and L_)), resulting from misinterpreting "at least one" as "both".
  • C) 1/2 is the probability of having only one dominant phenotype (P(B_ll) + P(bbL_) = (1/21/4) + (1/23/4) = 1/8 + 3/8 = 4/8 = 1/2), an incomplete application of the sum rule.

Question 15

In pea plants, purple flowers (P) are dominant to white (p), and yellow seeds (Y) are dominant to green (y). A cross is performed between a plant of genotype PpYy and a plant of genotype Ppyy. What is the probability that an offspring will have the same phenotype as at least one of its parents?

  1. 1/2
  2. 3/8
  3. 3/4 (correct answer)
  4. 1

Explanation: The first parent (PpYy) has the phenotype purple flowers, yellow seeds. The second parent (Ppyy) has the phenotype purple flowers, green seeds. An offspring's phenotype will match at least one parent if it is either (purple, yellow) or (purple, green). First, determine the probabilities of offspring phenotypes:

  • Cross Pp x Pp gives 3/4 purple (P_) and 1/4 white (pp).
  • Cross Yy x yy gives 1/2 yellow (Y_) and 1/2 green (yy).
Next, use the product rule to find the probabilities of the combined phenotypes:
  • P(purple, yellow) = P(P_) * P(Y_) = 3/4 * 1/2 = 3/8. This phenotype matches the first parent.
  • P(purple, green) = P(P_) * P(yy) = 3/4 * 1/2 = 3/8. This phenotype matches the second parent.
  • P(white, yellow) = P(pp) * P(Y_) = 1/4 * 1/2 = 1/8. This phenotype matches neither parent.
  • P(white, green) = P(pp) * P(yy) = 1/4 * 1/2 = 1/8. This phenotype matches neither parent.
Finally, use the sum rule for mutually exclusive events. The probability of matching at least one parent is P(purple, yellow) + P(purple, green) = 3/8 + 3/8 = 6/8 = 3/4. Distractor Rationale:
  • A) 1/2 is the probability of an offspring having the same genotype as at least one parent (P(PpYy) + P(Ppyy) = (1/21/2) + (1/21/2) = 1/4 + 1/4 = 1/2). This confuses genotype with phenotype.
  • B) 3/8 is the probability of matching only one of the parental phenotypes, failing to sum the probabilities for matching either parent.
  • D) 1 represents certainty and is an unlikely outcome in a dihybrid cross of this nature; it reflects a conceptual error in applying probability rules.

Question 16

In mice, black coat (B) is dominant to brown (b), and short tail (S) is dominant to long (s). A cross is made between a BbSs mouse and a bbss mouse. What is the probability that in a litter of three pups, at least one pup is brown with a long tail (bbss)?

  1. 1/64
  2. 27/64
  3. 3/4
  4. 37/64 (correct answer)

Explanation: This is an "at least one" probability problem. The easiest approach is to calculate the probability of the complementary event (no pups are brown with a long tail) and subtract this from 1.

  1. First, determine the probability of a single pup from this cross being brown with a long tail (bbss). The cross is a dihybrid test cross: BbSs x bbss.
    • For the coat color gene (Bb x bb), P(bb) = 1/2.
    • For the tail length gene (Ss x ss), P(ss) = 1/2.
    • Using the product rule, P(bbss) = P(bb) * P(ss) = 1/2 * 1/2 = 1/4.
  2. Next, determine the probability that a single pup is NOT bbss.
    • P(not bbss) = 1 - P(bbss) = 1 - 1/4 = 3/4.
  3. Now, calculate the probability that all three pups in the litter are not bbss. Since each birth is an independent event, we use the product rule.
    • P(all 3 are not bbss) = (3/4) * (3/4) * (3/4) = (3/4)³ = 27/64.
  4. Finally, the probability of at least one pup being bbss is 1 minus the probability that none are bbss.
    • P(at least one bbss) = 1 - 27/64 = 37/64.
Distractor Rationale:
  • A) 1/64 is the probability that all three pups are bbss ((1/4)³), which is a different scenario.
  • B) 27/64 is the probability of the complementary event (that none of the pups are bbss). This is a common error in "at least one" problems.
  • C) 3/4 is the probability that a single pup is not bbss. This answer ignores the fact that there are three pups in the litter.

Question 17

In fruit flies, red eyes (R) are dominant to sepia (r) and normal wings (W) are dominant to vestigial (w). A geneticist crosses a red-eyed, normal-winged fly with a sepia-eyed, vestigial-winged fly. The cross produces 200 offspring with the following approximate phenotypic distribution:

  • 52 red, normal
  • 49 red, vestigial
  • 51 sepia, normal
  • 48 sepia, vestigial

Based on the passage, what is the probability of obtaining an offspring from this cross that is homozygous for both genes?

  1. 0
  2. 1/4 (correct answer)
  3. 1/2
  4. 9/16

Explanation: This is a two-step problem. First, deduce the genotypes of the parents from the offspring data. Second, calculate the requested probability from the deduced cross. Step 1: Deduce parental genotypes.

  • The second parent is described as sepia-eyed, vestigial-winged, which is the double recessive phenotype, so its genotype must be rrww.
  • Analyze the ratios for each trait separately to determine the first parent's genotype (P_W_).
  • Eye color ratio: (red) : (sepia) = (52+49) : (51+48) = 101 : 99 ≈ 1:1. A 1:1 phenotypic ratio results from a cross between a heterozygote and a homozygous recessive (Rr x rr).
  • Wing shape ratio: (normal) : (vestigial) = (52+51) : (49+48) = 103 : 97 ≈ 1:1. A 1:1 phenotypic ratio results from a cross between a heterozygote and a homozygous recessive (Ww x ww).
  • Therefore, the first parent's genotype must be RrWw.
Step 2: Calculate the probability.
  • The cross is RrWw x rrww.
  • The question asks for the probability of an offspring that is homozygous for both genes. The possible double-homozygous genotypes are RRWW, RRww, rrWW, and rrww.
  • From the Rr x rr cross, P(RR) = 0 and P(rr) = 1/2.
  • From the Ww x ww cross, P(WW) = 0 and P(ww) = 1/2.
  • Therefore, the only possible double-homozygous offspring genotype is rrww.
  • The probability is P(rrww) = P(rr) * P(ww) = 1/2 * 1/2 = 1/4.
Distractor Rationale:
  • A) 0 is incorrect. While homozygous dominant (RRWW) offspring are impossible, homozygous recessive (rrww) offspring are possible.
  • C) 1/2 could result from incorrectly applying the sum rule (1/2 + 1/2) or by only considering a single trait.
  • D) 9/16 is the probability of the double-dominant phenotype in a standard RrWw x RrWw cross, a memorized ratio that is incorrectly applied to this test cross scenario.

Question 18

A plant of genotype AABB is crossed with a plant of genotype aabb. An F1 individual is then crossed with a plant of genotype Aabb. What is the probability that an offspring of this second cross is heterozygous for both genes?

  1. 1/4 (correct answer)
  2. 1/16
  3. 1/2
  4. 3/4

Explanation: This question tests your understanding of dihybrid crosses and probability calculations in genetics. When you see crosses involving two genes, break down the problem step by step and track each gene separately. First, determine the F1 genotype. AABB × aabb produces F1 offspring that are all AaBb (heterozygous for both genes). Next, you need to analyze the cross AaBb × Aabb. To find the probability of offspring heterozygous for both genes (AaBb), multiply the individual probabilities for each gene. For gene A: AaBb × Aabb gives you Aa offspring with probability 12\frac{1}{2} (from the Punnett square: AA, Aa, Aa, ab). For gene B: Bb × bb gives you Bb offspring with probability 12\frac{1}{2} (from BB, Bb, bb, bb). Therefore, the probability of AaBb is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Looking at the wrong answers: B) 116\frac{1}{16} would be correct if you were looking for a specific genotype in an F2 cross between two dihybrids, but that's not this scenario. C) 12\frac{1}{2} might tempt you if you only considered one gene instead of both. D) 34\frac{3}{4} could result from incorrectly adding probabilities instead of multiplying them. Study tip: In genetics problems involving multiple genes, always work with each gene independently first, then multiply the probabilities together. Remember that "and" means multiply, while "or" means add in probability calculations.

Question 19

In a certain plant, two unlinked genes affect flower viability. The presence of at least one dominant A allele and at least one dominant B allele is required for a flower to be fertile. All other genotypes (A_bb, aaB_, and aabb) result in sterile flowers. If two AaBb plants are crossed, what is the expected ratio of fertile to sterile plants in the progeny?

  1. 9:3:3:1
  2. 1:1
  3. 15:1
  4. 9:7 (correct answer)

Explanation: This problem describes a case of complementary gene action, a type of epistasis, where two dominant alleles are required to produce a specific phenotype (fertility).

  1. Start with the standard phenotypic outcome of a dihybrid cross between two heterozygotes (AaBb x AaBb), which is 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.
  2. Apply the rule for fertility given in the stem:
    • Plants are fertile only if they have the genotype A_B_.
    • Plants with genotypes A_bb, aaB_, or aabb are sterile.
  3. Group the phenotypic classes based on this rule:
    • The proportion of fertile plants is the proportion of A_B_ genotypes, which is 9/16.
    • The proportion of sterile plants is the sum of the proportions of the other genotypes: P(A_bb) + P(aaB_) + P(aabb) = 3/16 + 3/16 + 1/16 = 7/16.
  4. The resulting ratio of fertile to sterile plants is 9/16 : 7/16, which simplifies to 9:7.
Distractor Rationale:
  • A) 9:3:3:1 is the standard Mendelian ratio for phenotypes, which is modified by the epistatic interaction described.
  • B) 1:1 is a common ratio from a test cross, but not applicable here.
  • C) 15:1 is the ratio for a different epistatic interaction (duplicate dominant epistasis), where a dominant allele at either of two loci is sufficient to produce the dominant phenotype.

Question 20

In a dihybrid cross of two true-breeding parents (AABB x aabb), the F1 generation (AaBb) is self-fertilized. What is the probability that a randomly selected F2 offspring has the same genotype as one of its F1 parents?

  1. 1/4 (correct answer)
  2. 1/16
  3. 1/2
  4. 9/16

Explanation: When tackling dihybrid cross problems, focus on what specific genotype you're looking for and systematically work through the Punnett square or use probability rules. The F1 generation has genotype AaBb. To find the probability that an F2 offspring matches this exact genotype, you need to determine how often AaBb appears when AaBb × AaBb. For a dihybrid cross, treat each gene independently. For the A gene: Aa × Aa produces AA, Aa, Aa, aa (probability of Aa = 2/4 = 1/2). For the B gene: Bb × Bb similarly gives a 1/2 probability of Bb. Since genes assort independently, multiply these probabilities: 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Looking at the wrong answers: B) 1/16 represents the probability of getting any single specific genotype in a complete 4×4 dihybrid Punnett square, but this ignores that AaBb appears multiple times. C) 1/2 would be the probability if you were only considering one gene (like Aa from Aa × Aa), but you need both genes to match. D) 9/16 is the classic ratio for dominant phenotypes in a dihybrid cross (A_B_), but this question asks about genotype, not phenotype. The correct answer is A) 1/4. Study tip: For dihybrid crosses, remember that genotype probabilities multiply across independent genes. Don't confuse genotype ratios with phenotype ratios—phenotype questions often involve those familiar 9:3:3:1 ratios, while genotype questions require more careful counting of specific allele combinations.