Biology Quiz: Use Probability For Inheritance Predictions

20 questions · exam conditions

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Use Probability For Inheritance PredictionsQuestion 1 of 20

In a plant, red fruit (A) is dominant over yellow fruit (a). Two heterozygous plants are crossed: $Aa \times Aa$ . What is the probability an offspring will be heterozygous ( $Aa$ )?

25\%

33\%

50\%

75\%

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Biology Quiz

Biology Quiz: Use Probability For Inheritance Predictions

Practice Use Probability For Inheritance Predictions in Biology 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 Use Probability For Inheritance Predictions, giving you a quick way to practice the rules, question types, and explanations that matter most for Biology.

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 plant, red fruit (A) is dominant over yellow fruit (a). Two heterozygous plants are crossed: $Aa \times Aa$ . What is the probability an offspring will be heterozygous ( $Aa$ )?

$25\%$
$33\%$
$50\%$ (correct answer)
$75\%$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Aa × Aa cross, the Punnett square shows gametes A (50%) and a (50%) from each parent, resulting in offspring genotypes: AA (1/4), Aa (2/4), aa (1/4); the probability of heterozygous Aa is 2/4 or 50%. Choice C correctly calculates the inheritance probability by properly setting up the Punnett square and counting the 2 boxes out of 4 for Aa. Choice D is incorrect because 75% is the dominant phenotype probability, not specifically heterozygous—heterozygotes are only the Aa, not including AA! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Aa, Parent 2 is Aa. (2) DETERMINE possible gametes: Parent 1 can make A or a gametes (50% each). Parent 2 can make A or a gametes (50% each). (3) SET UP Punnett square: Put parent 1 gametes on top (A, a). Put parent 2 gametes on left (A, a). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left = A + A = AA. Top-right = A + a = Aa. Bottom-left = a + A = Aa. Bottom-right = a + a = aa. Result: 1 AA, 2 Aa, 1 aa. (5) COUNT for probability: Want probability of Aa? Count Aa boxes = 2. Probability = 2/4 = 50%. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 2

In pea plants, purple flowers (P) are dominant over white flowers (p). Two heterozygous plants are crossed: $Pp \times Pp$ . What is the probability that an offspring will have white flowers?

$75\%$
$50\%$
$25\%$ (correct answer)
$0\%$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Pp × Pp cross, the Punnett square shows gametes P (50%) and p (50%) from each parent, resulting in offspring genotypes: PP (1/4), Pp (2/4), pp (1/4); since white flowers require the recessive pp genotype, the probability is 1/4 or 25%. Choice C correctly calculates the inheritance probability by properly setting up the Punnett square and counting the 1 box out of 4 for the pp (white) outcome. Choice A is incorrect because it represents the probability of the dominant purple phenotype (3/4 or 75%), perhaps from mistakenly counting the boxes for purple instead of white—remember to focus on the specific trait asked! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Pp, Parent 2 is Pp. (2) DETERMINE possible gametes: Parent 1 can make P or p gametes (50% each). Parent 2 can make P or p gametes (50% each). (3) SET UP Punnett square: Put parent 1 gametes on top (P, p). Put parent 2 gametes on left (P, p). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left box = P + P = PP. Top-right = P + p = Pp. Bottom-left = p + P = Pp. Bottom-right = p + p = pp. Result: 1 PP, 2 Pp, 1 pp. (5) COUNT for probability: Want probability of white (pp)? Count pp boxes = 1. Total boxes = 4. Probability = 1/4 = 25%. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 3

In rabbits, black fur (B) is dominant over white fur (b). A heterozygous black rabbit is crossed with a white rabbit: $Bb \times bb$ . What is the probability an offspring will have genotype $Bb$ ?

$\frac{1}{4}$
$\frac{1}{2}$ (correct answer)
$\frac{3}{4}$
$1$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Bb × bb cross, the Punnett square shows gametes B (50%) and b (50%) from the first parent and b (100%) from the second, resulting in offspring genotypes: Bb (2/4), bb (2/4); the probability of Bb is 2/4 or 1/2. Choice B correctly calculates the inheritance probability by properly setting up the Punnett square and counting the 2 boxes out of 4 for the Bb genotype. Choice A is incorrect because 1/4 might come from confusing this with a dihybrid cross or miscounting gametes—remember, the homozygous bb parent only contributes b, so half the outcomes are Bb! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Bb, Parent 2 is bb. (2) DETERMINE possible gametes: Parent 1 can make B or b gametes (50% each). Parent 2 can make b gametes (100%). (3) SET UP Punnett square: Put parent 1 gametes on top (B, b). Put parent 2 gametes on left (b, b). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left box = B + b = Bb. Top-right = b + b = bb. Bottom-left = B + b = Bb. Bottom-right = b + b = bb. Result: 2 Bb, 2 bb. (5) COUNT for probability: Want probability of Bb? Count Bb boxes = 2. Total boxes = 4. Probability = 2/4 = 1/2. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 4

In pea plants, purple flowers (P) are dominant over white flowers (p). Two heterozygous plants are crossed: $Pp \times Pp$ . What is the probability that an offspring will have white flowers?

$\frac{3}{4}$ (75%)
$\frac{1}{2}$ (50%)
$\frac{1}{4}$ (25%) (correct answer)
$1$ (100%)

Question 5

In pea plants, purple flowers (P) are dominant over white flowers (p). Two heterozygous plants are crossed: $Pp \times Pp$ . What is the probability that an offspring will have white flowers?

$75\%$
$25\%$ (correct answer)
$50\%$
$0\%$

Question 6

In a bird species, allele G is dominant over allele g. Two heterozygous parents are crossed: $Gg \times Gg$ . What is the expected phenotype ratio (dominant : recessive) among the offspring?

$1:1$
$3:1$ (correct answer)
$1:2:1$
$2:1$

Question 7

In a plant species, tall (T) is dominant over short (t). A homozygous tall plant is crossed with a heterozygous tall plant: $TT \times Tt$ . What is the probability that an offspring will be short?

$50\%$
$25\%$
$0\%$ (correct answer)
$100\%$

Question 8

In a fish species, striped (S) is dominant over solid (s). Two solid fish are crossed: $ss \times ss$ . What is the probability that an offspring will be striped?

$1$ (100%)
$\frac{1}{2}$ (50%)
$\frac{1}{4}$ (25%)
$0$ (0%) (correct answer)

Question 9

In cats, short hair (H) is dominant over long hair (h). Two heterozygous cats are crossed: $Hh \times Hh$ . What is the probability that an offspring will have genotype Hh?

$\frac{1}{4}$ (25%)
$\frac{1}{2}$ (50%) (correct answer)
$\frac{3}{4}$ (75%)
$\frac{2}{3}$ (66.7%)

Question 10

In a certain plant, red fruit (R) is dominant over yellow fruit (r). Two plants are crossed: $Rr \times rr$ . What percentage of offspring are expected to have red fruit?

$25\%$
$0\%$
$50\%$ (correct answer)
$75\%$

Question 11

In a plant species, tall (T) is dominant over short (t). Two short plants are crossed: $tt \times tt$ . What is the probability that an offspring will be tall?

$100\%$
$75\%$
$25\%$
$0\%$ (correct answer)

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this tt × tt cross, the Punnett square shows only t gametes (100%) from both parents, resulting in all offspring tt (short); the probability of tall (requiring at least one T) is 0/4 or 0%. Choice D correctly calculates the inheritance probability by recognizing that homozygous recessive parents produce only recessive offspring. Choice A is incorrect because 100% would apply if both parents were TT, not tt—double-check the parent genotypes to avoid this mix-up! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is tt, Parent 2 is tt. (2) DETERMINE possible gametes: Parent 1 can make t gametes (100%). Parent 2 can make t gametes (100%). (3) SET UP Punnett square: Put parent 1 gametes on top (t). Put parent 2 gametes on left (t). Creates 1×1 = 1 box (but often expanded to 4 for consistency). (4) FILL boxes: Combine gametes. Box = t + t = tt. Result: all tt. (5) COUNT for probability: Want probability of tall (Tt or TT)? Count such boxes = 0. Total boxes = 4 (if expanded). Probability = 0/4 = 0%. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 12

In mice, normal tail (A) is dominant over tailless (a). Two heterozygous mice are crossed: $Aa \times Aa$ . What is the probability an offspring will have the dominant phenotype (normal tail)?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$ (correct answer)
$\frac{1}{3}$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Aa × Aa cross, the Punnett square shows gametes A (50%) and a (50%) from each parent, resulting in offspring genotypes: AA (1/4), Aa (2/4), aa (1/4); the dominant phenotype (normal tail, AA or Aa) occurs in 3/4 of outcomes. Choice C correctly calculates the inheritance probability by properly setting up the Punnett square and counting the 3 boxes out of 4 for AA or Aa. Choice A is incorrect because 1/4 is the probability of the recessive aa only, not the dominant—always add up all genotypes that show the dominant trait! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Aa, Parent 2 is Aa. (2) DETERMINE possible gametes: Parent 1 can make A or a gametes (50% each). Parent 2 can make A or a gametes (50% each). (3) SET UP Punnett square: Put parent 1 gametes on top (A, a). Put parent 2 gametes on left (A, a). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left box = A + A = AA. Top-right = A + a = Aa. Bottom-left = a + A = Aa. Bottom-right = a + a = aa. Result: 1 AA, 2 Aa, 1 aa. (5) COUNT for probability: Want probability of dominant phenotype (AA or Aa)? Count AA + Aa boxes = 1 + 2 = 3. Probability = 3/4. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 13

In a plant species, $T$ (tall) is dominant over $t$ (short). A homozygous dominant plant is crossed with a heterozygous plant: $TT \times Tt$ . What is the probability that an offspring will be short?

$0\%$ (correct answer)
$25\%$
$50\%$
$100\%$

Question 14

In a certain animal, the allele $A$ is dominant to $a$ . A heterozygous parent is crossed with a homozygous recessive parent: $Aa \times aa$ . What is the probability that an offspring will have genotype $Aa$ ?

$\frac{1}{4}$
$\frac{1}{2}$ (correct answer)
$\frac{3}{4}$
$1$

Question 15

In mice, normal ears (E) are dominant over folded ears (e). Two heterozygous mice are crossed: $Ee \times Ee$ . What is the probability that an offspring will show the dominant phenotype (normal ears)?

$\frac{1}{4}$ (25%)
$\frac{1}{2}$ (50%)
$\frac{3}{4}$ (75%) (correct answer)
$\frac{1}{3}$ (33.3%)

Question 16

In a certain bird species, green feathers (G) are dominant over yellow feathers (g). Two heterozygous birds are crossed: $Gg \times Gg$ . What is the expected genotype ratio of offspring ( $GG : Gg : gg$ )?

$3:1:0$
$1:2:1$ (correct answer)
$1:1:2$
$1:1:1$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Gg × Gg cross, the Punnett square shows gametes G (50%) and g (50%) from each parent, resulting in offspring genotypes: GG (1/4), Gg (2/4), gg (1/4), so the ratio GG : Gg : gg is 1:2:1. Choice B correctly calculates the inheritance probability by properly setting up the Punnett square and counting the boxes to get the 1:2:1 genotype ratio. Choice A is incorrect because 3:1:0 might confuse the phenotype ratio (3:1 dominant to recessive) with genotypes or omit gg—ratios must include all possible genotypes! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Gg, Parent 2 is Gg. (2) DETERMINE possible gametes: Parent 1 can make G or g gametes (50% each). Parent 2 can make G or g gametes (50% each). (3) SET UP Punnett square: Put parent 1 gametes on top (G, g). Put parent 2 gametes on left (G, g). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left = G + G = GG. Top-right = G + g = Gg. Bottom-left = g + G = Gg. Bottom-right = g + g = gg. Result: 1 GG, 2 Gg, 1 gg. (5) COUNT for probability: Ratio = 1:2:1. Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 17

In a fish species, striped pattern (R) is dominant over plain pattern (r). Two heterozygous fish are crossed: $Rr \times Rr$ . If the first offspring is plain ( $rr$ ), what is the probability the second offspring will also be plain ( $rr$ )?

$\frac{1}{4}$ (correct answer)
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{3}{4}$

Explanation: This question tests your ability to use Punnett squares and probability to predict the likelihood of specific genotypes or phenotypes in offspring from parents with known genotypes. Calculating inheritance probabilities uses Punnett squares as a tool to visualize all possible offspring outcomes: set up the square by putting one parent's possible gametes across the top (if parent is Aa, can contribute A or a—two possibilities, each 50% chance) and the other parent's possible gametes down the left side, then fill in boxes by combining gametes (top gamete + left gamete = offspring genotype in that box). Each box represents one equally likely outcome, so PROBABILITY = (number of boxes with desired outcome) / (total number of boxes). Example: Aa × Aa cross creates 4-box Punnett: box 1 = AA (A from parent 1, A from parent 2), box 2 = Aa (A from 1, a from 2), box 3 = Aa (a from 1, A from 2), box 4 = aa (a from 1, a from 2). For probability of aa: 1 box out of 4 = 1/4 = 25% chance. For probability of Aa: 2 boxes out of 4 = 2/4 = 1/2 = 50% chance. For probability of dominant phenotype (AA or Aa if A is dominant): 3 boxes out of 4 = 3/4 = 75% chance. Simple counting from Punnett square gives all probabilities! For this Rr × Rr cross, the Punnett square shows overall probabilities of RR (1/4), Rr (2/4), rr (1/4); since each offspring is independent, even if the first is rr, the second still has 1/4 probability of rr. Choice A correctly calculates the inheritance probability by recognizing that offspring events are independent and using the Punnett square count of 1/4 for rr. Choice B is incorrect because 1/3 might come from wrongly conditioning on the first outcome and thinking of remaining possibilities—remember, probabilities reset for each independent offspring! The Punnett square probability recipe: (1) WRITE parent genotypes: Parent 1 is Rr, Parent 2 is Rr. (2) DETERMINE possible gametes: Parent 1 can make R or r gametes (50% each). Parent 2 can make R or r gametes (50% each). (3) SET UP Punnett square: Put parent 1 gametes on top (R, r). Put parent 2 gametes on left (R, r). Creates 2×2 = 4 boxes. (4) FILL boxes: Combine gametes. Top-left = R + R = RR. Top-right = R + r = Rr. Bottom-left = r + R = Rr. Bottom-right = r + r = rr. Result: 1 RR, 2 Rr, 1 rr. (5) COUNT for probability: Want probability of rr? Count rr boxes = 1. Total boxes = 4. Probability = 1/4 (independent for each child). Quick probability shortcuts for common crosses: Aa × Aa: offspring probabilities = 1/4 AA (25%), 1/2 Aa (50%), 1/4 aa (25%). Phenotype: 3/4 dominant (75%), 1/4 recessive (25%). The "3:1 ratio" parents! Aa × aa: offspring probabilities = 1/2 Aa (50%), 1/2 aa (50%). Phenotype: 1/2 dominant, 1/2 recessive (1:1 ratio). The "test cross"! AA × aa: offspring probabilities = 100% Aa (all heterozygous). Phenotype: 100% dominant if A dominant. AA × AA or aa × aa: 100% same as parents (homozygous × homozygous = all homozygous). Memorizing these common crosses saves time, but you can always draw Punnett square to derive them! Remember: each CHILD is independent event—if two Aa parents have one child with aa (1/4 probability), their NEXT child STILL has 1/4 probability of aa (doesn't change based on first child). Probabilities are per offspring, not per family!

Question 18

In a certain animal, allele A produces a dominant trait and allele a produces a recessive trait. A heterozygous parent is crossed with a homozygous recessive parent: $Aa \times aa$ . What fraction of the offspring are expected to have genotype $aa$ ?

$\frac{1}{4}$
$\frac{3}{4}$
$\frac{1}{2}$ (correct answer)
$1$

Question 19

In a certain species, allele R is dominant to allele r. Two heterozygous parents ( $Rr \times Rr$ ) have children. If their first child has genotype $rr$ , what is the probability that their second child will also have genotype $rr$ ?

$\frac{1}{4}$ (correct answer)
$\frac{1}{2}$
$\frac{3}{4}$
$0$

Question 20

In a certain species, allele A is dominant over allele a. Two parents are crossed: $Aa \times Aa$ . What is the probability that an offspring will show the dominant phenotype (have at least one A allele)?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$ (correct answer)
$1$