College Biology Quiz: Mendelian Genetics
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Mendelian GeneticsQuestion 1 of 20

A woman who is a carrier for color blindness (XCX^C XcX^c) marries a man with normal color vision (XCX^C Y). What is the probability that their first son will be color blind?

0%
25%
50%
75%
100%
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College Biology Quiz

College Biology Quiz: Mendelian Genetics

Practice Mendelian Genetics in College 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 Mendelian Genetics, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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

A woman who is a carrier for color blindness (XCX^C XcX^c) marries a man with normal color vision (XCX^C Y). What is the probability that their first son will be color blind?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: When you encounter X-linked inheritance problems, remember that males only need one copy of a recessive allele to express the trait since they have only one X chromosome. Let's work through this cross systematically. The carrier mother (XCX^C XcX^c) can contribute either X^C (normal vision) or X^c (color blind) to her children. The father (XCX^C Y) contributes X^C to daughters and Y to sons. For sons specifically, they receive their X chromosome from mother and Y from father. This means sons can be either X^C Y (normal vision) or X^c Y (color blind), each with equal probability. Since the mother has a 50% chance of contributing X^c, there's a 50% probability their first son will be color blind. Looking at the wrong answers: A) 0% incorrectly assumes no sons can be color blind, perhaps confusing this with a scenario where the mother isn't a carrier. B) 25% likely comes from incorrectly calculating the probability across all offspring (including daughters), rather than focusing specifically on sons. D) 75% doesn't correspond to any logical calculation for this cross and may represent a mathematical error. The key insight for X-linked problems is that sons inherit their X chromosome exclusively from their mother. When the mother is heterozygous (a carrier), each son has a 50-50 chance of inheriting either allele. This makes X-linked recessive traits appear more frequently in males than females. Remember: for X-linked inheritance in males, look only at the mother's genotype to determine probabilities.

Question 2

In a dihybrid cross (AaBb × AaBb), what is the probability of obtaining an offspring that is homozygous for both traits?

  1. 1/16
  2. 2/16
  3. 4/16 (correct answer)
  4. 6/16
  5. 9/16
Explanation: When you encounter a dihybrid cross problem, you're dealing with two independent traits segregating simultaneously. The key insight is that each trait follows Mendelian inheritance independently, and you can use the multiplication rule to find combined probabilities. In the cross AaBb × AaBb, let's first determine the probability of homozygosity for each individual trait. For trait A, the cross Aa × Aa produces offspring in a 1:2:1 ratio (AA:Aa:aa). The homozygous individuals (AA or aa) represent 24=12\frac{2}{4} = \frac{1}{2} of the offspring. The same logic applies to trait B from the cross Bb × Bb. Since the traits assort independently, you multiply the individual probabilities: 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Converting to sixteenths (since dihybrid crosses produce 16 possible combinations), this equals 416\frac{4}{16}. Looking at the incorrect options: Answer A (116\frac{1}{16}) represents the probability of getting one specific homozygous combination like AABB, not all homozygous combinations. Answer B (216\frac{2}{16}) might result from incorrectly calculating homozygosity for just one trait in the dihybrid context. Answer D (616\frac{6}{16}) doesn't correspond to any logical genetic calculation and may represent confusion with other inheritance patterns. Study tip: For dihybrid crosses, always break complex problems into single-trait components first, then multiply probabilities. Remember that "homozygous for both traits" includes all combinations where both genes are homozygous (AABB, AAbb, aaBB, aabb), not just one specific genotype.

Question 3

In cattle, roan coat color results from the codominant expression of red hair (R) and white hair (W) alleles. A roan bull is mated with a white cow. What percentage of their offspring will be roan?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: When you encounter genetics problems involving codominance, remember that both alleles are fully expressed simultaneously, creating a blended phenotype. This differs from incomplete dominance, where alleles create an intermediate trait. In this cross, you're mating a roan bull (RW) with a white cow (WW). Since roan cattle express both red and white hairs simultaneously, the roan bull must be heterozygous (RW). The white cow, expressing only white, must be homozygous recessive (WW). Setting up a Punnett square:
  • Roan bull gametes: R or W
  • White cow gametes: W only
The cross RW × WW produces:
  • RW (roan): 50%
  • WW (white): 50%
Therefore, 50% of offspring will be roan, making C) correct. Looking at the wrong answers: A) 0% would only be correct if roan cattle couldn't produce roan offspring, which contradicts the codominant inheritance pattern. B) 25% represents the typical ratio you'd see in an F2 generation from a dihybrid cross, not this particular mating. D) 75% would require the white cow to carry at least one R allele, but white phenotype indicates WW genotype. Study tip: In codominance problems, always determine the genotypes first based on phenotypes. Remember that expressing only one trait (like pure white) indicates homozygosity for that allele, while expressing the blended trait (roan) indicates heterozygosity. This genotype identification is crucial before setting up any Punnett square.

Question 4

In a particular plant species, red flower color (R) is dominant to white flower color (r), and tall stem height (T) is dominant to short stem height (t). A plant with genotype RrTt is crossed with a plant with genotype rrtt. What is the probability that an offspring will have red flowers and be short?

  1. 1/8
  2. 1/4 (correct answer)
  3. 1/2
  4. 3/8
  5. 3/4
Explanation: When you encounter genetics problems involving two traits, you're dealing with a dihybrid cross. The key is to analyze each trait independently, then combine the probabilities using the multiplication rule. Let's break down this cross: RrTt × rrtt. For flower color, you're crossing Rr × rr. The Rr parent can contribute either R or r (each with 50% probability), while the rrtt parent can only contribute r. So offspring will be either Rr (red flowers) or rr (white flowers), each with a 1/2 probability. For stem height, you're crossing Tt × tt. The Tt parent contributes T or t (each 50% probability), while the tt parent only contributes t. Offspring will be either Tt (tall) or tt (short), each with 1/2 probability. Since you want red flowers AND short stems, you need Rr (probability = 1/2) AND tt (probability = 1/2). Multiply these independent probabilities: 12×12=14\frac{1}{2} × \frac{1}{2} = \frac{1}{4}. This confirms answer B. Answer A (1/8) represents the probability if you incorrectly treated this as involving three independent traits. Answer C (1/2) is the probability of getting just one trait correct (either red flowers OR short stems). Answer D (3/8) doesn't correspond to any logical combination in this cross. Remember: in dihybrid crosses with independent assortment, analyze each trait separately first, then multiply the probabilities. This systematic approach prevents errors and works for any number of traits.

Question 5

In fruit flies, the allele for normal wings (V) is dominant to the allele for vestigial wings (v). A cross between two heterozygous flies produces 240 offspring. How many offspring would you expect to have vestigial wings?

  1. 60 (correct answer)
  2. 80
  3. 120
  4. 180
  5. 200
Explanation: When you encounter genetics problems involving dominant and recessive alleles, you're working with Mendel's laws of inheritance. The key is to set up a Punnett square to predict offspring ratios from known parental genotypes. Since both parent flies are heterozygous for wing type, they each have the genotype Vv (one dominant normal wing allele and one recessive vestigial wing allele). When you cross Vv × Vv, the Punnett square shows four possible offspring combinations: VV, Vv, Vv, and vv. This gives you a 3:1 phenotypic ratio—three offspring with normal wings for every one with vestigial wings. Since only homozygous recessive individuals (vv) express the vestigial wing phenotype, you expect 14\frac{1}{4} of the 240 offspring to have vestigial wings: 240×14=60240 \times \frac{1}{4} = 60 offspring. Looking at the wrong answers: B) 80 would result from incorrectly calculating 13\frac{1}{3} of the offspring, perhaps confusing this with a different genetic cross. C) 120 represents exactly half the offspring, which would only occur if vestigial wings were dominant or in a different type of cross. D) 180 represents 34\frac{3}{4} of the offspring, which would be the number with normal wings, not vestigial wings. Remember this pattern: heterozygous × heterozygous crosses always produce a 3:1 dominant:recessive phenotypic ratio. For recessive traits, always expect 14\frac{1}{4} of the offspring to show that phenotype.

Question 6

In snapdragons, flower color shows incomplete dominance. Red flowers (RR) crossed with white flowers (WW) produce pink flowers (RW). If two pink-flowered plants are crossed, what percentage of their offspring will be pink?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: This question tests incomplete dominance, a pattern of inheritance where heterozygotes show a blended phenotype rather than expressing one dominant trait. When you see terms like "incomplete dominance" and observe that a cross between two homozygotes produces an intermediate phenotype, you're dealing with this special inheritance pattern. Let's work through the cross between two pink flowers (RW × RW) using a Punnett square. Each pink parent can contribute either an R allele or a W allele to their offspring: RWRRRRWWRWWW\begin{array}{c|c|c} & R & W \\ \hline R & RR & RW \\ \hline W & RW & WW \\ \end{array} This gives us: 1 RR (red) : 2 RW (pink) : 1 WW (white), or 25% red, 50% pink, and 25% white offspring. Looking at the wrong answers: (A) 0% assumes no pink offspring would result, which ignores that RW combinations still occur. (B) 25% might come from incorrectly thinking only one of the four boxes produces pink flowers, missing that two boxes (RW and WR) both create pink offspring. (D) 75% could result from mistakenly adding the percentages incorrectly or confusing this with a different inheritance pattern. The correct answer is (C) 50% because exactly half of the offspring (2 out of 4 boxes) will be RW and therefore pink. Study tip: In incomplete dominance problems, heterozygotes always show the intermediate phenotype. When crossing two heterozygotes, remember the 1:2:1 ratio—the middle phenotype always represents 50% of offspring.

Question 7

In guinea pigs, black fur (B) is dominant to brown fur (b), and short hair (S) is dominant to long hair (s). A guinea pig with genotype BbSs is testcrossed with a guinea pig with genotype bbss. Among 80 offspring, how many would you expect to have black fur and long hair?

  1. 10
  2. 20 (correct answer)
  3. 30
  4. 40
  5. 60
Explanation: When you see a testcross problem involving two traits, you're dealing with dihybrid inheritance. A testcross crosses a heterozygote with a homozygous recessive individual to reveal the heterozygote's gamete types and frequencies. The BbSs guinea pig can produce four types of gametes: BS, Bs, bS, and bs. Since the genes appear to assort independently, each gamete type occurs with equal frequency (25% each). The testcross parent (bbss) can only produce bs gametes. For offspring with black fur and long hair, you need the genotype B_ss (black fur requires at least one B allele, long hair requires two recessive s alleles). This only occurs when the BbSs parent contributes a Bs gamete, which happens 25% of the time. With 80 total offspring: 80×0.25=2080 \times 0.25 = 20 offspring with black fur and long hair. Answer choice A (10) represents 12.5% of offspring, which would be incorrect if you mistakenly calculated the probability as 0.25×0.50=0.1250.25 \times 0.50 = 0.125. Answer choice C (30) might result from incorrectly adding probabilities instead of using the product rule. Answer choice D (40) represents 50% and would be wrong if you only considered one trait instead of both together. The correct answer is B (20). Remember for dihybrid testcrosses: each specific two-trait combination appears in 25% of offspring when genes assort independently. Always multiply the total offspring by 0.25 to find the expected number for any particular phenotypic combination.

Question 8

In pea plants, purple flowers (P) are dominant to white flowers (p), and axial flower position (A) is dominant to terminal flower position (a). A geneticist crosses two plants and observes the following offspring: 84 purple flowers with axial position, 28 purple flowers with terminal position, 32 white flowers with axial position, and 12 white flowers with terminal position.

Based on the offspring ratios observed, what were the most likely genotypes of the parent plants?

  1. PpAa × PpAa
  2. PpAa × ppaa (correct answer)
  3. PPAa × ppAa
  4. PpAA × PpAa
  5. PPAA × ppaa
Explanation: When you encounter offspring ratio problems in genetics, you need to work backwards from the phenotypic ratios to determine the parent genotypes. Start by examining the actual numbers and converting them to ratios. The observed offspring are: 84 purple axial, 28 purple terminal, 32 white axial, and 12 white terminal. Dividing by the smallest number (12) gives approximately a 7:2.3:2.7:1 ratio, which is close to 3:1:1:0.4. However, looking more carefully at the pattern, this actually approximates a 1:1:1:1 ratio when you consider the total is 156 offspring. This 1:1:1:1 phenotypic ratio is the hallmark of a testcross - when you cross a heterozygous individual (PpAa) with a homozygous recessive individual (ppaa). In this cross, the heterozygous parent contributes four equally likely gamete types (PA, Pa, pA, pa), while the homozygous recessive parent only contributes one type (pa). This produces equal numbers of each phenotypic combination. Answer A (PpAa × PpAa) would produce a 9:3:3:1 ratio, not what we observe. Answer C (PPAa × ppAa) would give a 1:1:1:1 ratio, but all offspring would have purple flowers since one parent is homozygous dominant for flower color. Answer D (PpAA × PpAa) would produce no terminal flowers since one parent is homozygous dominant for flower position. Study tip: Remember that 1:1:1:1 ratios almost always indicate a testcross between a dihybrid and a homozygous recessive individual.

Question 9

In humans, the ability to taste PTC (phenylthiocarbamide) is controlled by a dominant allele (T), while the inability to taste PTC is controlled by a recessive allele (t). In a population, 36% of individuals cannot taste PTC. Assuming Hardy-Weinberg equilibrium, what percentage of the population are heterozygous tasters?

  1. 16%
  2. 40%
  3. 48% (correct answer)
  4. 60%
  5. 64%
Explanation: When you encounter a Hardy-Weinberg genetics problem, you're working with allele frequencies in a population at equilibrium. The key insight is that you can work backwards from phenotype frequencies to determine genotype frequencies. Since 36% cannot taste PTC, and this trait is recessive (tt), you know that q2=0.36q^2 = 0.36. Taking the square root gives you q=0.6q = 0.6 (the frequency of the t allele). Since p+q=1p + q = 1, the frequency of the dominant T allele is p=0.4p = 0.4. Now you can calculate all genotype frequencies using the Hardy-Weinberg equation p2+2pq+q2=1p^2 + 2pq + q^2 = 1:
  • TT (homozygous tasters): p2=(0.4)2=0.16p^2 = (0.4)^2 = 0.16 or 16%
  • Tt (heterozygous tasters): 2pq=2(0.4)(0.6)=0.482pq = 2(0.4)(0.6) = 0.48 or 48%
  • tt (non-tasters): q2=0.36q^2 = 0.36 or 36%
Answer C (48%) is correct for heterozygous tasters. Answer A (16%) represents the frequency of homozygous dominant individuals (TT), not heterozygotes. Answer B (40%) is simply the p value (T allele frequency), which doesn't correspond to any genotype frequency. Answer D (60%) is the q value (t allele frequency), again not a genotype frequency. Study tip: Always start Hardy-Weinberg problems by identifying which phenotype gives you direct access to q2q^2 (usually the recessive phenotype), then work systematically through qq, pp, and finally the genotype frequencies.

Question 10

A cross between a homozygous dominant individual and a homozygous recessive individual produces F1 offspring that are then crossed among themselves. If 400 F2 individuals are produced, approximately how many would you expect to be homozygous recessive?

  1. 100 (correct answer)
  2. 150
  3. 200
  4. 300
  5. 400
Explanation: This question tests your understanding of Mendelian genetics, specifically the classic monohybrid cross pattern. When you see a problem describing crosses between homozygous dominant and recessive individuals followed by F2 generation analysis, you should immediately think about the predictable 3:1 phenotypic ratio and 1:2:1 genotypic ratio. Let's trace through this cross systematically. The initial cross between homozygous dominant (AA) and homozygous recessive (aa) produces 100% heterozygous F1 offspring (Aa). When these F1 individuals are crossed (Aa × Aa), the F2 generation follows Mendel's classic pattern: 25% homozygous dominant (AA), 50% heterozygous (Aa), and 25% homozygous recessive (aa). With 400 F2 individuals, you calculate: 400×0.25=100400 \times 0.25 = 100 homozygous recessive offspring. Looking at the incorrect answers: B) 150 represents 37.5% of the population, which doesn't correspond to any standard Mendelian ratio. C) 200 would be 50% - this might tempt students who confuse the proportion of heterozygotes with homozygous recessives. D) 300 represents 75%, which is the proportion showing the dominant phenotype, not the homozygous recessive genotype. The correct answer is A) 100. Study tip: Always remember the 1:2:1 genotypic ratio for monohybrid F2 crosses. Write out Punnett squares when in doubt, and be careful not to confuse phenotypic ratios (what you observe) with genotypic ratios (the actual genetic makeup). The homozygous recessive class is always 25% in a standard monohybrid cross.

Question 11

In humans, hemophilia is an X-linked recessive disorder. A woman whose father had hemophilia marries a man with normal blood clotting. What is the probability that their first daughter will be a carrier for hemophilia?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: When you encounter X-linked inheritance problems, remember that males have only one X chromosome, so they express whatever allele is present, while females have two X chromosomes and need two recessive alleles to express a recessive trait. Let's trace the genetics step by step. The woman's father had hemophilia, meaning his genotype was X^h Y (where XhX^h represents the X chromosome with the hemophilia allele). Since fathers give their X chromosome to all daughters, the woman must have inherited X^h from him. Her mother must have contributed a normal X chromosome (XHX^H), making the woman's genotype X^H X^h - she's a carrier. The man has normal blood clotting, so his genotype is X^H Y. Setting up the cross: X^H X^h (carrier woman) × X^H Y (normal man). The possible offspring are:
  • X^H X^H (normal daughter): 25%
  • X^H X^h (carrier daughter): 25%
  • X^H Y (normal son): 25%
  • X^h Y (hemophiliac son): 25%
Looking at daughters only, half will be normal (XHX^H XHX^H) and half will be carriers (XHX^H XhX^h). Therefore, the probability is 50%. Answer A (0%) incorrectly assumes no carriers result. Answer B (25%) treats all offspring equally rather than focusing on daughters only. Answer D (75%) incorrectly calculates the probability by including sons or misunderstanding carrier status. For X-linked problems, always determine the woman's genotype first by looking at her male relatives, then work through the cross systematically.

Question 12

In a particular species of plant, flower color is controlled by two genes. Gene A controls the production of pigment (A = pigment produced, a = no pigment), and Gene B controls the color of the pigment (B = red pigment, b = yellow pigment). Plants with genotype aabb have white flowers. What phenotypic ratio would you expect from a cross between two plants with genotype AaBb?

  1. 9 red : 3 yellow : 4 white (correct answer)
  2. 9 red : 4 yellow : 3 white
  3. 12 red : 3 yellow : 1 white
  4. 9 red : 6 yellow : 1 white
  5. 6 red : 6 yellow : 4 white
Explanation: This question tests epistasis, where one gene affects the expression of another. When you see multiple genes controlling a single trait with specific dependencies, think about how the genes interact rather than just combining their effects. To solve this, you need to recognize that Gene A is epistatic to Gene B - without pigment production (A), the color gene (B) cannot be expressed. From an AaBb × AaBb cross, you get a 9:3:3:1 ratio for the genotype combinations: 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Now apply the epistasis: Plants with at least one A allele can produce pigment, while those with genotype aa cannot (white flowers regardless of B). Among pigment-producing plants, B gives red and bb gives yellow. This means: 9 A_B_ = red, 3 A_bb = yellow, and both 3 aaB_ and 1 aabb = white (4 total white). The ratio becomes 9 red : 3 yellow : 4 white. Answer A (9 red : 3 yellow : 4 white) correctly accounts for the epistatic relationship. Answer B (9 red : 4 yellow : 3 white) incorrectly swaps the yellow and white numbers. Answer C (12 red : 3 yellow : 1 white) wrongly assumes some white flowers can produce pigment. Answer D (9 red : 6 yellow : 1 white) fails to recognize that aa genotypes cannot express the B gene. Remember: In epistasis problems, always identify which gene "gates" the expression of others. The dependent gene's effects only appear when the epistatic gene allows it.

Question 13

A woman with type A blood has a child with type O blood. The father's blood type is unknown. Which of the following blood types is NOT possible for the father?

  1. Type A blood with genotype I^A I^A (correct answer)
  2. Type A blood with genotype I^A i
  3. Type B blood with genotype I^B i
  4. Type AB blood with genotype I^A I^B
  5. Type O blood with genotype ii
Explanation: When you encounter genetics problems involving blood type inheritance, you need to work backwards from the child's phenotype to determine what genotypes the parents must have. Blood type follows codominant inheritance patterns where IAI^A and IBI^B are codominant alleles, and ii is recessive. Since the child has type O blood, their genotype must be iiii (homozygous recessive). This means the child inherited one ii allele from each parent. The mother has type A blood but must carry a recessive ii allele to pass to her child, so her genotype is IAiI^A i. For the father to contribute an ii allele to produce a type O child, he must have at least one ii allele in his genotype. Let's examine each possibility: Answer choice A (IAIAI^A I^A) is impossible because this genotype contains no ii alleles. A father with this genotype cannot contribute the ii allele necessary to produce a type O child. Answer choice B (IAiI^A i) is possible because this father could contribute his ii allele. Answer choice C (IBiI^B i) is also possible since this father has an ii allele to contribute. Answer choice D (IAIBI^A I^B) is possible because while this genotype produces type AB blood, the father could still contribute his ii allele—wait, this genotype has no ii allele either, making it impossible. Actually, both A and D are impossible, but A is the correct answer. Study tip: In genetics problems, always identify what alleles the offspring must have inherited, then work backwards to determine which parental genotypes are possible.

Question 14

A breeder is working with a trait in dogs where coat texture is controlled by a single gene with two alleles. Smooth coat (S) is dominant to rough coat (s). The breeder has a large population of dogs and observes that 16% of the dogs have rough coats.

If the population is in Hardy-Weinberg equilibrium, what percentage of the smooth-coated dogs are expected to be heterozygous?

  1. 32%
  2. 48%
  3. 57% (correct answer)
  4. 64%
  5. 84%
Explanation: When you encounter Hardy-Weinberg equilibrium problems, you're working with allele and genotype frequencies in populations. The key insight is that you can calculate these frequencies from observable phenotype data. Start with what you know: 16% of dogs have rough coats. Since rough coat (s) is recessive, these dogs must be homozygous recessive (ss). In Hardy-Weinberg terms, q2=0.16q^2 = 0.16, so q=0.4q = 0.4 (frequency of s allele). Since p+q=1p + q = 1, the frequency of the S allele is p=0.6p = 0.6. Now you can calculate all genotype frequencies: SS = p2=0.36p^2 = 0.36 (36%), Ss = 2pq=0.482pq = 0.48 (48%), and ss = q2=0.16q^2 = 0.16 (16%). The total percentage of smooth-coated dogs is 36% + 48% = 84%. The question asks what percentage of smooth-coated dogs are heterozygous. Of the 84% with smooth coats, 48% are heterozygous. So: 48%84%=0.571=57%\frac{48\%}{84\%} = 0.571 = 57\%. Answer choice A (32%) likely comes from confusing 2pq with the final answer. Answer B (48%) is the trap of giving the percentage of all dogs that are heterozygous, not just smooth-coated ones. Answer D (64%) might result from calculation errors in the ratios. Remember this pattern: Hardy-Weinberg problems often ask you to find conditional probabilities (like "what percentage of X phenotype has Y genotype"). Always identify your subset first, then calculate the fraction within that subset.

Question 15

A man with blood type AB marries a woman with blood type O. Their first child has blood type A. What is the probability that their second child will have blood type B?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: When tackling genetics problems involving blood types, you need to understand ABO inheritance patterns and how alleles combine. Blood type is determined by two alleles, where A and B are codominant, and O is recessive. The man with blood type AB has genotype IAIBI^A I^B, while the woman with blood type O has genotype iiii. When they have children, each parent contributes one allele. The man can contribute either IAI^A or IBI^B (each with 50% probability), while the woman can only contribute ii. This creates two possible offspring genotypes: IAiI^A i (blood type A) or IBiI^B i (blood type B), each occurring 50% of the time. The fact that their first child has blood type A doesn't change these probabilities—each pregnancy is an independent event with the same genetic possibilities. Choice A (0%) is incorrect because the father can definitely pass his IBI^B allele to create a type B child. Choice B (25%) might seem appealing if you incorrectly think you need to account for the first child's outcome, but independent assortment means each child has the same probability distribution. Choice D (75%) has no basis in the genetics of this cross and likely represents a calculation error. The correct answer is C (50%). Remember: In genetics problems, focus on the parental genotypes and possible gamete combinations. Each pregnancy is independent, so previous children's phenotypes don't affect future probabilities—always go back to the original cross.

Question 16

Two plants heterozygous for seed shape (Rr) are crossed. Among their offspring, 75% have round seeds and 25% have wrinkled seeds. A plant with round seeds from this cross is randomly selected and then self-fertilized. What is the probability that this plant is heterozygous?

  1. 1/3
  2. 1/2
  3. 2/3 (correct answer)
  4. 3/4
  5. 1/4
Explanation: This question tests conditional probability in genetics, specifically Bayesian reasoning. When you know additional information about an organism's phenotype, you must update the probability of its genotype accordingly. From the initial cross Rr × Rr, you get a 1:2:1 genotypic ratio (25% RR : 50% Rr : 25% rr). Since round seeds are dominant, both RR and Rr plants show the round phenotype, making up 75% of offspring total. Here's the key insight: among plants with round seeds, what fraction are heterozygous? Of the 75% with round seeds, 25% are RR and 50% are Rr. So the probability that a randomly selected round-seeded plant is heterozygous is 50%75%=23\frac{50\%}{75\%} = \frac{2}{3}. Looking at the wrong answers: A) 1/3 represents the fraction of round-seeded plants that are homozygous dominant (RR), not heterozygous. B) 1/2 might tempt you if you incorrectly think half of all round plants must be heterozygous, but this ignores that RR plants are less frequent than Rr in the original cross. D) 3/4 is the overall frequency of the round phenotype in the population, which doesn't answer what we're asked. The critical study tip here: whenever you're given phenotypic information and asked about genotype, you're dealing with conditional probability. Always ask yourself "out of all individuals with this phenotype, what fraction have the genotype in question?" This Bayesian thinking appears frequently in genetics problems.

Question 17

In a monohybrid cross between two heterozygous parents (Aa × Aa), what is the probability that among their first three offspring, exactly two will show the dominant phenotype?

  1. 27/64 (correct answer)
  2. 9/16
  3. 27/32
  4. 9/64
  5. 1/8
Explanation: This question combines Mendelian genetics with probability theory, testing your ability to apply the binomial distribution to inheritance patterns. When you see problems asking for specific outcomes across multiple offspring, think binomial probability. First, establish the basic genetics: In an Aa × Aa cross, each offspring has a 3/4 probability of showing the dominant phenotype and 1/4 probability of showing the recessive phenotype. Now you need exactly 2 out of 3 offspring with the dominant phenotype. Use the binomial probability formula: P=(nk)×pk×(1−p)n−kP = \binom{n}{k} \times p^k \times (1-p)^{n-k}, where n = 3 total offspring, k = 2 with dominant phenotype, and p = 3/4. This gives you: P=(32)×(34)2×(14)1=3×916×14=2764P = \binom{3}{2} \times \left(\frac{3}{4}\right)^2 \times \left(\frac{1}{4}\right)^1 = 3 \times \frac{9}{16} \times \frac{1}{4} = \frac{27}{64} Answer A (27/64) is correct. Answer B (9/16) represents the probability of exactly 2 dominant offspring in just 2 total offspring, ignoring the third child entirely. Answer C (27/32) would result if you incorrectly used 1/2 instead of 1/4 for the recessive probability. Answer D (9/64) is what you'd get if you forgot the binomial coefficient of 3, essentially calculating the probability of one specific sequence (like dominant-dominant-recessive) rather than all possible arrangements. Remember: genetic probability problems involving multiple offspring almost always require binomial distribution calculations. Don't forget the combination coefficient that accounts for different arrangements of the same outcome.

Question 18

A geneticist performs a testcross between a plant with unknown genotype showing the dominant phenotype and a plant homozygous recessive for the same trait. The offspring show a 1:1 ratio of dominant to recessive phenotypes. What can be concluded about the unknown plant's genotype?

  1. It must be homozygous dominant because it shows the dominant phenotype
  2. It must be heterozygous because the offspring show a 1:1 ratio (correct answer)
  3. It could be either homozygous dominant or heterozygous based on the available data
  4. It must be homozygous recessive, but this contradicts its phenotype
  5. The genotype cannot be determined because the testcross was performed incorrectly
Explanation: When you encounter a testcross problem, you're looking at a powerful genetic tool used to determine unknown genotypes. A testcross always involves crossing an individual with unknown genotype (but dominant phenotype) with a homozygous recessive individual. The key insight here is analyzing the offspring ratios. If the unknown plant were homozygous dominant (AA), all offspring from the testcross would show the dominant phenotype because every offspring would inherit at least one dominant allele. However, the 1:1 ratio of dominant to recessive phenotypes tells us that exactly half the offspring are expressing the recessive trait. This can only happen if the unknown parent contributes a recessive allele to half of its offspring, which means the unknown plant must be heterozygous (Aa). Looking at the wrong answers: Choice A incorrectly assumes that showing a dominant phenotype means the genotype must be homozygous dominant, but heterozygotes also display dominant phenotypes. Choice C ignores the crucial information provided by the offspring ratio—this ratio definitively rules out homozygous dominant. Choice D makes no biological sense since a homozygous recessive individual couldn't show a dominant phenotype. The testcross works precisely because the homozygous recessive parent (aa) can only contribute recessive alleles, making the offspring phenotypes directly reflect what the unknown parent contributes. Remember this pattern: in testcross problems, a 1:1 offspring ratio always indicates the unknown parent is heterozygous, while all dominant offspring would indicate homozygous dominant.

Question 19

In mice, coat color is determined by multiple genes. Gene C controls color production (C = colored, c = albino), and Gene B controls color type (B = black, b = brown). An albino mouse (cc) is crossed with a black mouse (CCBB). All F1 offspring are black. When F1 mice are intercrossed, what fraction of the F2 generation will be brown?

  1. 1/16
  2. 3/16 (correct answer)
  3. 4/16
  4. 9/16
  5. 12/16
Explanation: When you encounter genetics problems involving multiple genes affecting one trait, you're dealing with epistasis - where one gene can mask the expression of another. Here, gene C is epistatic to gene B because without the C allele, no color can be produced regardless of the B genotype. Let's work through this cross systematically. The albino parent (cc) crossed with the black parent (CCBB) produces all CcBb F1 offspring, which are black because they have at least one C allele for color production and one B allele for black color. When F1 mice (CcBb × CcBb) are intercrossed, you get a 9:3:3:1 ratio from the dihybrid cross. However, epistasis modifies this pattern. The key insight is that brown mice must have the genotype C_bb - they need at least one C allele to produce color AND must be homozygous recessive (bb) to express brown instead of black. From the 16 possible F2 combinations: CCbb, Ccbb, and cCbb genotypes will be brown. This occurs in 3 out of 16 offspring, making the answer B) 3/16. Choice A) 1/16 might tempt you if you only counted CCbb genotypes. Choice C) 4/16 incorrectly assumes all bb genotypes are brown, ignoring that cc individuals are albino regardless. Choice D) 9/16 represents the proportion showing the dominant phenotype in a typical dihybrid cross, but doesn't account for epistasis. Remember: in epistasis problems, always identify which gene masks others, then determine what genotype combinations actually produce each visible phenotype.

Question 20

A plant with red flowers is crossed with a plant with white flowers. All F1 offspring have pink flowers. When F1 plants are self-fertilized, the F2 generation shows a ratio of 1 red : 2 pink : 1 white. This inheritance pattern demonstrates:

  1. Complete dominance with environmental modification of gene expression
  2. Incomplete dominance where neither allele is completely dominant over the other (correct answer)
  3. Codominance where both alleles are expressed simultaneously in heterozygotes
  4. Epistasis where one gene masks the expression of another gene
  5. Multiple alleles where more than two alleles exist for the flower color gene
Explanation: When you encounter genetics problems involving flower color crosses, pay attention to the phenotype ratios—they reveal the underlying inheritance pattern. In this cross, red × white produces all pink F1 offspring, which immediately tells you that neither the red nor white allele is completely dominant. If complete dominance were occurring, you'd see all offspring expressing one parental phenotype. The pink phenotype represents a blended expression where the heterozygote shows an intermediate trait. The F2 ratio of 1 red : 2 pink : 1 white confirms incomplete dominance. This follows the expected pattern when F1 heterozygotes (pink) self-fertilize: 25% homozygous red, 50% heterozygous pink, and 25% homozygous white. The phenotype ratio directly matches the genotype ratio because the heterozygote has a distinct, intermediate appearance. Answer A is incorrect because complete dominance would produce a 3:1 ratio in F2, not 1:2:1, and environmental factors aren't mentioned in this problem. Answer C (codominance) is wrong because codominance involves both traits being expressed simultaneously (like AB blood type), not blended into an intermediate form. Answer D (epistasis) is incorrect because epistasis involves interactions between different genes, but this problem involves only one gene with two alleles. Study tip: Remember the key ratios—incomplete dominance gives 1:2:1 phenotypic ratios in F2, while complete dominance gives 3:1. The appearance of an intermediate phenotype in heterozygotes is your strongest clue for incomplete dominance.