Genetics Quiz
Practice Monohybrid Crosses in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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In snapdragons, flower color is controlled by a single gene with incomplete dominance. The (C^R) allele produces red pigment, and the (C^W) allele produces no pigment. (C^RC^R) plants have red flowers, (C^WC^W) have white flowers, and (C^RC^W) have pink flowers. A pink-flowered plant is backcrossed to its red-flowered parent. What proportion of the offspring is expected to have the pink-flowered phenotype?
This quiz focuses on Monohybrid Crosses, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.
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.
In snapdragons, flower color is controlled by a single gene with incomplete dominance. The (C^R) allele produces red pigment, and the (C^W) allele produces no pigment. (C^RC^R) plants have red flowers, (C^WC^W) have white flowers, and (C^RC^W) have pink flowers. A pink-flowered plant is backcrossed to its red-flowered parent. What proportion of the offspring is expected to have the pink-flowered phenotype?
Explanation: A backcross involves crossing an offspring with one of its parents. The pink-flowered plant has the genotype (C^RC^W). Its red-flowered parent must have the genotype (C^RC^R). The cross is (C^RC^W \times C^RC^R). A Punnett square for this cross shows that the possible offspring genotypes are (C^RC^R) and (C^RC^W). The expected genotypic ratio is 1 (C^RC^R) : 1 (C^RC^W). The phenotype for (C^RC^W) is pink. Therefore, the proportion of offspring expected to have pink flowers is 1/2.
A geneticist crosses two fruit flies with long wings. The cross produces 243 offspring with long wings and 81 with short (vestigial) wings. Based on this result, which of the following crosses most likely represents the genotypes of the parent flies? (L = long wings, l = short wings)
Explanation: The problem requires deducing parental genotypes from offspring phenotypic ratios. The observed ratio is 243 long wings to 81 short wings, which is approximately a 3:1 ratio (243/81 = 3). A 3:1 phenotypic ratio is the characteristic result of a monohybrid cross between two heterozygous parents. Both parents displayed the dominant long-wing phenotype, which is consistent with them being heterozygous (Ll). Therefore, the most likely cross was Ll x Ll.
Tay-Sachs disease is an autosomal recessive disorder. A man and a woman are both carriers of the Tay-Sachs allele. If they have three children, what is the probability that at least one child will have the disease?
Explanation: Since both parents are carriers, the cross is Tt x Tt. The probability of a child having Tay-Sachs (tt) is 1/4. The probability of a child being unaffected is 3/4. The probability of 'at least one' affected child is most easily calculated as 1 minus the probability of the complementary event, which is 'no children are affected'. The probability that all three children are unaffected is the product of their individual probabilities: (3/4) × (3/4) × (3/4) = 27/64. Therefore, the probability of at least one child being affected is 1 - 27/64 = 37/64.
A gene in canaries affects feather color. The allele (Y) for yellow feathers is dominant over (y) for white feathers. The homozygous dominant genotype ((YY)) is lethal, with the embryo failing to develop. A breeder crosses two yellow canaries. What is the probability that a surviving yellow chick from this cross is a purebred?
Explanation: The two yellow canaries must be heterozygous (Yy) because the YY genotype is lethal. The cross is Yy x Yy, which produces genotypes in a 1 YY : 2 Yy : 1 yy ratio. The YY embryos do not survive. The surviving offspring are Yy (yellow) and yy (white) in a 2:1 ratio. The question asks for the probability that a surviving yellow chick is a purebred (homozygous). All surviving yellow chicks have the genotype Yy, which is heterozygous. Therefore, there are no purebred yellow chicks, and the probability is 0.
A plant breeder has a tall pea plant with an unknown genotype (T_). To determine the genotype, she performs a test cross with a dwarf plant (tt) and obtains 6 offspring, all of which are tall. What is the probability of obtaining this specific result if the unknown parent plant were heterozygous (Tt)?
Explanation: The question asks for the probability of a specific outcome (6 tall offspring) under the assumption that the parent is heterozygous (Tt). In a test cross between a heterozygous individual (Tt) and a homozygous recessive individual (tt), the probability of producing a tall offspring (Tt) is 1/2, and the probability of producing a dwarf offspring (tt) is 1/2. The outcome of each offspring is an independent event. To find the probability of getting 6 tall offspring in a row, we multiply the individual probabilities: P(6 tall) = (1/2) × (1/2) × (1/2) × (1/2) × (1/2) × (1/2) = (1/2)^6 = 1/64.
A student analyzes a monohybrid cross between a homozygous dominant (GG) parent and a heterozygous (Gg) parent. The student incorrectly concludes there is a 25% chance of a homozygous recessive (gg) offspring. Which conceptual error most likely led to this conclusion?
Explanation: The cross described is GG x Gg. All gametes from the GG parent are G. Gametes from the Gg parent are 1/2 G and 1/2 g. The offspring will be 1/2 GG and 1/2 Gg. There is a 0% chance of a gg offspring. A 25% (or 1/4) probability of a homozygous recessive offspring is the characteristic outcome of a cross between two heterozygous parents (Gg x Gg). Therefore, the most likely error was that the student ignored the provided parental genotypes and incorrectly assumed the standard heterozygous cross.
A purebred tall pea plant is crossed with a purebred dwarf pea plant (P generation). The resulting F1 generation is allowed to self-pollinate to produce an F2 generation. If two F2 plants showing the dominant (tall) phenotype are randomly selected and crossed, what is the probability that they will produce a dwarf offspring?
Explanation: This is a multi-step problem. First, P cross: TT x tt → F1 is all Tt. Second, F1 self-cross: Tt x Tt → F2 is 1 TT : 2 Tt : 1 tt. The F2 plants with the dominant phenotype (tall) are TT and Tt. The conditional probability of a tall F2 plant being TT is 1/3, and of being Tt is 2/3. For two randomly selected tall F2 plants to produce a dwarf (tt) offspring, both must be heterozygous (Tt). The probability of selecting two Tt plants is P(first is Tt) × P(second is Tt) = (2/3) × (2/3) = 4/9. Given that both parents are Tt, the probability of them having a tt offspring is 1/4. The final probability is the product of these two events: P(both parents are Tt) × P(tt offspring from Tt x Tt) = (4/9) × (1/4) = 4/36 = 1/9.
Huntington's disease is a rare autosomal dominant disorder. A man whose father had Huntington's disease marries a woman with no family history of the disorder. The man himself is currently asymptomatic. What is the probability their first child will be unaffected by the disease?
Explanation: This problem requires careful analysis of the man's probable genotype. His father had the disease (H_), and since it is rare, we assume his mother was homozygous recessive (hh). To have a child, the father must have passed on an allele. The man is a child of a person with Huntington's and an unaffected person. Thus, his father must have been heterozygous (Hh). The cross that produced the man was Hh x hh, meaning there was a 1/2 chance the man inherited the H allele (and is Hh) and a 1/2 chance he inherited the h allele (and is hh). The woman has no family history, so her genotype is hh. There are two possibilities for the man: 1) He is Hh (with probability 1/2). The cross would be Hh x hh, and the chance of an unaffected (hh) child is 1/2. 2) He is hh (with probability 1/2). The cross would be hh x hh, and the chance of an unaffected (hh) child is 1. The total probability of an unaffected child is the sum of the probabilities of these two mutually exclusive scenarios: P(unaffected) = P(man is Hh) * P(unaffected child) + P(man is hh) * P(unaffected child) = (1/2 * 1/2) + (1/2 * 1) = 1/4 + 1/2 = 3/4.
A couple, both carriers for cystic fibrosis (autosomal recessive), have two children. What is the probability that their first child is unaffected and their second child is affected?
Explanation: When you encounter genetics problems involving multiple offspring, you need to calculate the probability of each event separately, then multiply them together since these are independent events. Both parents are carriers (Cc), so let's first determine the probability for any single child. Using a Punnett square: CC (25%, unaffected), Cc (50%, carrier but unaffected), and cc (25%, affected). This means each child has a 3/4 chance of being unaffected and a 1/4 chance of being affected. For this specific scenario, you need the first child to be unaffected AND the second child to be affected. Since these are independent events, you multiply the probabilities: 43×41=163 Looking at the wrong answers: Choice A (1/16) would be the probability if you needed both children to be affected (41×41). Choice B (9/16) represents the probability that both children are unaffected (43×43). Choice C (1/4) is the probability for just one child being affected, ignoring the condition about the first child being unaffected. The correct answer is D (3/16). Study tip: For genetics problems involving multiple offspring, always identify the probability for each individual event first, then multiply probabilities for independent events occurring together. Remember that birth order matters when the question specifies which child has which phenotype.
In Manx cats, the allele for taillessness (M) is dominant over the allele for a tail (m). However, the homozygous dominant genotype (MM) is lethal and causes embryos to abort. If two tailless Manx cats are mated, what is the probability that a surviving kitten will have a tail?
Explanation: Since taillessness (M) is dominant and lethal when homozygous (MM), any living tailless cat must be heterozygous (Mm). The cross is therefore Mm x Mm. The initial genotypic ratio from the Punnett square is 1 MM : 2 Mm : 1 mm. However, the MM genotype is lethal, so these embryos do not survive. The surviving offspring will only have the genotypes Mm (tailless) and mm (tailed), in a ratio of 2:1. The total number of possible surviving outcomes is 2 + 1 = 3. The number of outcomes corresponding to a tailed kitten (mm) is 1. Therefore, the probability that a surviving kitten has a tail is 1/3.
A dominant allele (D) causes a genetic disorder, but it has an incomplete penetrance of 80%. A man who is heterozygous for the allele (Dd) has a child with a woman who is homozygous recessive (dd). What is the probability that their child will be affected by the disorder?
Explanation: This problem requires combining Mendelian probability with the concept of penetrance. First, determine the probability of the child inheriting the disease-causing allele. The cross is Dd x dd. The probability that the child will have the genotype Dd is 1/2 or 0.50. Second, consider the penetrance. Penetrance is the probability that an individual with a specific genotype will express the corresponding phenotype. Here, the penetrance is 80% (0.80). The overall probability of being affected is the probability of having the genotype MULTIPLIED by the probability of expressing the phenotype (penetrance). P(affected) = P(inheriting D allele) × P(expressing the trait | has D allele) = 0.50 × 0.80 = 0.40.
A seed company produces seeds from a cross between two tomato plants heterozygous for fruit color, where red (R) is dominant to yellow (r). They sell a packet of 800 seeds. If the germination rate for these seeds is 80%, approximately how many of the resulting plants are expected to have yellow fruit?
Explanation: This problem combines Mendelian probability with a practical calculation. First, determine the probability of yellow fruit. The cross is Rr x Rr, which yields a 1/4 probability of rr genotype (yellow fruit). Second, calculate the expected number of yellow-fruit seeds: 800 seeds × (1/4) = 200 seeds. Third, apply the germination rate. Not all seeds will grow into plants. The number of expected plants is the number of seeds multiplied by the germination rate: 200 yellow-fruit seeds × 80% = 200 × 0.80 = 160 plants.
In a species of ornamental fish, scale color is determined by a single gene with incomplete dominance. (B^1B^1) individuals are blue, (B^2B^2) are yellow, and (B^1B^2) are green. A breeder crosses two green fish. Blue fish sell for 3each,yellowfor3, and green for $8. If the breeder raises 200 offspring from this cross to maturity, what is the expected total revenue from selling all the fish?
Explanation: First, determine the expected phenotypic ratio. The cross is between two green fish ((B^1B^2 \times B^1B^2)). The offspring genotypic ratio is 1 (B^1B^1) : 2 (B^1B^2) : 1 (B^2B^2), which corresponds to a phenotypic ratio of 1 blue : 2 green : 1 yellow. Out of 200 offspring, we expect: (1/4) * 200 = 50 blue fish; (1/2) * 200 = 100 green fish; (1/4) * 200 = 50 yellow fish. Next, calculate the revenue for each phenotype: Revenue(blue) = 50 * 3=150; Revenue(green) = 100 * 8=800; Revenue(yellow) = 50 * 3=150. Finally, sum the revenues: 150+800 + 150=1100.
A plant with purple flowers is self-pollinated, and the offspring are produced in a ratio of approximately 3 purple to 1 white. This original purple parent plant is then crossed with a white-flowered plant. What is the expected phenotypic ratio in the offspring of this second cross?
Explanation: This is a two-step problem. First, deduce the genotype of the original purple parent. When self-pollinated, it produced a 3:1 phenotypic ratio. This is the classic signature of a monohybrid cross between two heterozygotes. Therefore, the purple parent's genotype must be heterozygous (let's use Pp). Second, determine the outcome of the new cross. This purple parent (Pp) is crossed with a white-flowered plant. Since white is recessive, its genotype must be pp. The cross is Pp x pp. A Punnett square for this cross predicts offspring genotypes of 1/2 Pp and 1/2 pp. This corresponds to a phenotypic ratio of 1 purple : 1 white.
In guinea pigs, rough coat (R) is dominant to smooth coat (r). A heterozygous rough-coated guinea pig is crossed with another heterozygous rough-coated guinea pig. What is the probability that their first offspring is homozygous dominant and their second offspring is heterozygous?
Explanation: When you encounter genetics problems involving multiple offspring, you're dealing with independent events where each birth has the same probability regardless of previous outcomes. First, determine the probabilities for a single cross between two heterozygotes (Rr × Rr). Using a Punnett square: RR appears in 1/4 of offspring (homozygous dominant), Rr appears in 2/4 or 1/2 of offspring (heterozygous), and rr appears in 1/4 of offspring (homozygous recessive). For the first offspring to be homozygous dominant (RR): probability = 1/4 For the second offspring to be heterozygous (Rr): probability = 1/2 Since these are independent events, multiply the individual probabilities: 41×21=81 Looking at the wrong answers: Choice A (1/16) represents what you'd get if you mistakenly thought both offspring needed to be homozygous dominant (1/4 × 1/4). Choice B (3/4) is the probability of getting a rough coat (dominant phenotype) in a single cross, ignoring the specific genotype requirements. Choice C (1/4) is the probability of getting homozygous dominant in just one offspring, failing to account for the second requirement. Remember that when genetics problems ask about multiple specific outcomes in sequence, you must multiply the individual probabilities together. Each birth is an independent event with the same underlying probabilities from the parental cross.
A cross is performed between two pea plants heterozygous for seed shape (Rr), where round is dominant to wrinkled. What is the probability that a randomly selected offspring will be purebred for this trait?
Explanation: This question tests your understanding of Mendelian genetics and the difference between genotype ratios and phenotype ratios in monohybrid crosses. When crossing two heterozygotes (Rr × Rr), you can use a Punnett square to determine all possible offspring genotypes. The cross produces: RR, Rr, rR, and rr, each with equal probability. Since Rr and rR represent the same genotype, the actual ratio is 1 RR : 2 Rr : 1 rr, or in probability terms: 41 RR, 21 Rr, and 41 rr. The key insight is recognizing what "purebred" means. Purebred organisms are homozygous—they have two identical alleles for a trait. In this cross, both RR (homozygous dominant) and rr (homozygous wrinkled) are purebred. Together, these represent 41+41=21 of all offspring, making D correct. Answer A (41) represents the probability of getting only one specific purebred genotype (either RR or rr alone), not both types combined. Answer B (1) would mean all offspring are purebred, which ignores the 21 that are heterozygous. Answer C (43) represents the probability of showing the dominant phenotype (round seeds), confusing phenotype with genotype. Remember: "purebred" always means homozygous, regardless of whether it's dominant or recessive. When calculating probabilities for "either/or" scenarios, add the individual probabilities together.
In Manx cats, the allele for taillessness (M) is dominant over the allele for a tail (m). However, the homozygous dominant genotype (MM) is lethal and causes embryos to abort. If two tailless Manx cats are mated, what is the probability that a surviving kitten will have a tail?
Explanation: Since taillessness (M) is dominant and lethal when homozygous (MM), any living tailless cat must be heterozygous (Mm). The cross is therefore Mm x Mm. The initial genotypic ratio from the Punnett square is 1 MM : 2 Mm : 1 mm. However, the MM genotype is lethal, so these embryos do not survive. The surviving offspring will only have the genotypes Mm (tailless) and mm (tailed), in a ratio of 2:1. The total number of possible surviving outcomes is 2 + 1 = 3. The number of outcomes corresponding to a tailed kitten (mm) is 1. Therefore, the probability that a surviving kitten has a tail is 1/3.
Phenylketonuria (PKU) is an autosomal recessive disorder. A couple are both carriers for PKU. They have a child who does not have PKU. What is the probability that this child is a carrier of the PKU allele?
Explanation: Let P be the dominant allele and p be the recessive allele for PKU. Since both parents are carriers, their genotypes are Pp. The cross is Pp x Pp. The possible offspring genotypes are 1 PP : 2 Pp : 1 pp. The child does not have PKU, which means their genotype cannot be pp. This is a conditional probability problem. We must consider only the unaffected genotypes: PP and Pp. There is 1 part PP and 2 parts Pp, for a total of 3 possible unaffected outcomes. Out of these 3 possibilities, 2 are carriers (Pp). Therefore, the probability that the unaffected child is a carrier is 2/3.
A purebred tall pea plant is crossed with a purebred dwarf pea plant (P generation). The resulting F1 generation is allowed to self-pollinate to produce an F2 generation. If two F2 plants showing the dominant (tall) phenotype are randomly selected and crossed, what is the probability that they will produce a dwarf offspring?
Explanation: This is a multi-step problem. First, P cross: TT x tt → F1 is all Tt. Second, F1 self-cross: Tt x Tt → F2 is 1 TT : 2 Tt : 1 tt. The F2 plants with the dominant phenotype (tall) are TT and Tt. The conditional probability of a tall F2 plant being TT is 1/3, and of being Tt is 2/3. For two randomly selected tall F2 plants to produce a dwarf (tt) offspring, both must be heterozygous (Tt). The probability of selecting two Tt plants is P(first is Tt) × P(second is Tt) = (2/3) × (2/3) = 4/9. Given that both parents are Tt, the probability of them having a tt offspring is 1/4. The final probability is the product of these two events: P(both parents are Tt) × P(tt offspring from Tt x Tt) = (4/9) × (1/4) = 4/36 = 1/9.
A cross is performed between two pea plants heterozygous for seed shape (Rr), where round is dominant to wrinkled. What is the probability that a randomly selected offspring will be purebred for this trait?
Explanation: This question tests your understanding of Mendelian genetics and the difference between genotype ratios and phenotype ratios in monohybrid crosses. When crossing two heterozygotes (Rr × Rr), you can use a Punnett square to determine all possible offspring genotypes. The cross produces: RR, Rr, rR, and rr, each with equal probability. Since Rr and rR represent the same genotype, the actual ratio is 1 RR : 2 Rr : 1 rr, or in probability terms: 41 RR, 21 Rr, and 41 rr. The key insight is recognizing what "purebred" means. Purebred organisms are homozygous—they have two identical alleles for a trait. In this cross, both RR (homozygous dominant) and rr (homozygous wrinkled) are purebred. Together, these represent 41+41=21 of all offspring, making D correct. Answer A (41) represents the probability of getting only one specific purebred genotype (either RR or rr alone), not both types combined. Answer B (1) would mean all offspring are purebred, which ignores the 21 that are heterozygous. Answer C (43) represents the probability of showing the dominant phenotype (round seeds), confusing phenotype with genotype. Remember: "purebred" always means homozygous, regardless of whether it's dominant or recessive. When calculating probabilities for "either/or" scenarios, add the individual probabilities together.