Genetics Quiz
Practice Test 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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A geneticist has a male mouse displaying a dominant, autosomal phenotype, "jerker" (J_). She wants to determine if the mouse is homozygous (JJ) or heterozygous (Jj). Which of the following crosses represents the most appropriate and direct test cross?
This quiz focuses on Test 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.
A geneticist has a male mouse displaying a dominant, autosomal phenotype, "jerker" (J_). She wants to determine if the mouse is homozygous (JJ) or heterozygous (Jj). Which of the following crosses represents the most appropriate and direct test cross?
Explanation: A test cross is specifically designed to determine an unknown genotype of a dominant-phenotype individual by crossing it with an individual that is homozygous recessive for the trait. In this case, crossing the jerker male (J_) with a non-jerker female (jj) will reveal if the male carries the recessive allele. If any offspring are non-jerkers (jj), the male must be heterozygous (Jj).
A breeder is working with guinea pigs where black fur (B) is dominant to white fur (b). A black male is test-crossed, and the resulting litter contains both black and white pups. What conclusion can be drawn, and why is a test cross superior to a self-cross for this purpose?
Explanation: When you encounter genetics problems involving test crosses, you're dealing with a powerful technique to determine an unknown genotype by crossing with a homozygous recessive individual. Let's analyze this step by step. The black male could be either BB or Bb (since black is dominant). When test-crossed with a white female (bb), the offspring include both black and white pups. This is only possible if the male contributes both B and b alleles to different offspring, proving he must be Bb (heterozygous). If he were BB, all offspring would receive a B allele and appear black. Now let's examine why each answer choice succeeds or fails: Answer A correctly identifies the male as Bb and explains that any recessive offspring definitively proves heterozygosity - this is exactly what happened here. Answer B incorrectly claims the male is BB, which contradicts the evidence of white offspring appearing. The claim about producing "more offspring" is also irrelevant to the genetic analysis. Answer C correctly identifies Bb but wrongly suggests a self-cross would be superior. While self-crosses do produce 3:1 ratios, they're less efficient for determining genotype because you'd need to count many offspring to distinguish between expected ratios. Answer D makes the same BB error as B, ignoring the clear evidence from the offspring ratios. Study tip: Remember that test crosses are the gold standard for determining unknown genotypes because any recessive phenotype in the offspring immediately reveals heterozygosity in the dominant parent - it's a yes/no answer rather than requiring statistical analysis.
A trihybrid test cross is performed on a plant with genotype A_B_C_. The resulting offspring exhibit 8 different phenotypic classes in approximately equal numbers. What was the genotype of the parent plant?
Explanation: A test cross involves crossing to a fully homozygous recessive individual (aabbcc). The phenotypes of the offspring directly reflect the gametes produced by the unknown parent. To get 8 different phenotypic classes, the parent must produce 8 different types of gametes. A fully heterozygous trihybrid (AaBbCc) with independent assortment produces 2^3 = 8 gamete types (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc) in equal proportions, which would result in 8 phenotypic classes in a 1:1:1:1:1:1:1:1 ratio in a test cross.
A plant with dominant phenotypes for height (T_) and flower color (P_) is test-crossed. The results show that the recombination frequency between the T and P genes is 18%. If the parent plant's alleles are in coupling phase (TP/tp), what percentage of the test cross progeny would be expected to have a tall plant with white flowers (T_pp)?
Explanation: The progeny phenotype T_pp results from the fusion of a parental gamete (Tp) and a tester gamete (tp). Since the parent's alleles are in coupling phase (TP/tp), the gamete 'Tp' is a recombinant gamete. The total recombination frequency is 18%, which is the sum of the frequencies of both recombinant gamete types (Tp and tP). Assuming equal recombination rates, the frequency of the Tp gamete is half of the total, or 18% / 2 = 9%. Therefore, 9% of the progeny are expected to have the tall/white phenotype.
In a certain insect, two genes affect eye color and wing shape. A dihybrid test cross of an individual with dominant phenotypes for both traits yields the following offspring: 485 red eyes/normal wings, 515 red eyes/dwarf wings, 0 brown eyes/normal wings, and 0 brown eyes/dwarf wings. What is the most likely genotype of the parent with the dominant phenotypes?
Explanation: The test cross is with an rrww individual. The offspring phenotypes reveal the gametes produced by the unknown parent. Offspring are red/normal (from an RW or Rw gamete) and red/dwarf (from an RW or Rw gamete). Specifically, since the tester parent is rrww, the red/normal offspring are RrWw and the red/dwarf are Rrww. This means the parent produced RW and Rw gametes. No brown-eyed offspring were produced, so no 'r' allele was passed on, meaning the parent was homozygous RR. The presence of both normal and dwarf winged offspring indicates the parent was heterozygous for the wing gene, Ww. Therefore, the genotype is RRWw.
A breeder is trying to determine the genotype of a male dog with a dominant wire-haired coat (W_). She conducts a test cross. To be 95% confident that the male is homozygous dominant (WW), no recessive smooth-haired (ww) puppies can be born. What is the minimum number of wire-haired puppies that must be produced in the litter to achieve this level of confidence?
Explanation: The question asks for 95% confidence that the male is WW. This is equivalent to stating that the probability of the male being heterozygous (Ww) is less than 5%. If the male is heterozygous, the probability of any given puppy being wire-haired (Ww) is 1/2. The probability of having N wire-haired puppies in a row from a heterozygous father is (1/2)^N. We want this probability to be less than 5% (0.05). Let's test the values: N=3: (1/2)^3 = 1/8 = 0.125 (12.5%). N=4: (1/2)^4 = 1/16 = 0.0625 (6.25%). N=5: (1/2)^5 = 1/32 = 0.03125 (3.125%). Since 3.125% is less than 5%, a minimum of 5 wire-haired puppies is required to be at least 95% confident the father is not heterozygous.
In Labrador retrievers, coat color is controlled by two genes that exhibit epistasis. Gene B controls pigment production (B=black, b=brown). Gene E controls pigment deposition (E=deposit, e=no deposit). Dogs with genotype ee are yellow, regardless of their B/b genotype. A black lab (B_E_) is test-crossed to a brown lab (bbEE). However, this is not a standard test cross. If the black lab's genotype is BbEe, what is the expected phenotypic ratio in the offspring?
Explanation: This question tests the ability to apply the test cross concept in a non-standard scenario involving epistasis. The cross is BbEe × bbEE. We can analyze the genes separately. For the B gene, the cross is Bb × bb, which yields 1/2 Bb and 1/2 bb. For the E gene, the cross is Ee × EE, which yields 1/2 EE and 1/2 Ee. All offspring will have at least one dominant E allele, so there will be no yellow labs (ee). The phenotype is therefore determined entirely by the B/b genotype. The offspring genotypes will be 1/2 BbE_ (black) and 1/2 bbE_ (brown), resulting in a 1 black : 1 brown phenotypic ratio.
A test cross on a plant with a dominant phenotype (genotype P_) yields six offspring, all displaying the dominant phenotype. This result supports the hypothesis that the parent is homozygous (PP), but does not prove it. Which of the following subsequent crosses offers the most decisive method to determine the genotype of the original P_ parent?
Explanation: When determining an unknown genotype, you need to design crosses that will produce distinctly different offspring ratios depending on whether the parent is homozygous dominant (PP) or heterozygous (Pp). The initial test cross gave ambiguous results—six dominant offspring could occur with either parental genotype, just with different probabilities. Answer A provides the most decisive test because a backcross between F1 offspring and the original parent will yield dramatically different results depending on the parent's genotype. If the original parent is PP, then all F1 offspring are also PP, and the backcross (PP × PP) produces 100% dominant offspring. However, if the original parent is Pp, then F1 offspring are either PP or Pp, and backcrossing with the heterozygous parent can produce recessive offspring. Any recessive offspring in this backcross would definitively prove the original parent was Pp. Answer B is wrong because self-pollinating F1 offspring won't reveal the original parent's genotype—if the parent was PP, F1 self-crosses yield all dominant offspring, providing no new information. Answer C is incorrect for the same reason; crossing F1 individuals together doesn't distinguish between the two possible parental genotypes. Answer D simply repeats the original approach and suffers from the same statistical ambiguity—you'd need an impractically large sample size to distinguish between the possibilities with confidence. Remember: when genotype determination is ambiguous, design crosses that exploit the differences between possible genotypes to create distinguishable phenotypic ratios.
In a species of beetle, black body (B) is dominant to brown (b), and long antennae (L) are dominant to short (l). An individual with a black body and long antennae is test-crossed. The cross produces offspring in an approximate phenotypic ratio of 1 black/long : 1 black/short : 1 brown/long : 1 brown/short. What was the genotype of the parent beetle with the dominant phenotypes?
Explanation: A test cross for two traits involves crossing the unknown dominant individual (B_L_) with a homozygous recessive individual (bbll). To produce all four phenotypic classes, the unknown parent must carry both recessive alleles (b and l). The 1:1:1:1 ratio is characteristic of a test cross with a dihybrid individual that is heterozygous for both genes (BbLl) and where the genes assort independently.
In Drosophila, red eyes (W) are a dominant, X-linked trait, and white eyes (w) are recessive. A geneticist performs a test cross on a red-eyed female of unknown genotype. If the female is heterozygous (X^W X^w), what are the expected phenotypes of her offspring?
Explanation: A test cross for an X-linked trait in a female involves crossing her with a recessive male. The cross is X^W X^w (heterozygous female) × X^w Y (white-eyed male). For female offspring (XX), half will receive X^W from the mother and X^w from the father (X^W X^w, red-eyed), and half will receive X^w from the mother and X^w from the father (X^w X^w, white-eyed). For male offspring (XY), half will receive X^W from the mother (X^W Y, red-eyed), and half will receive X^w from the mother (X^w Y, white-eyed). Thus, both sexes show a 1:1 phenotypic ratio.
A test cross is performed on a tomato plant with red fruit (R_), a dominant trait. The cross yields 52 red-fruited and 48 yellow-fruited (rr) offspring. Two of the red-fruited F1 offspring are then randomly selected and crossed with each other. What is the probability that this second cross will produce a plant with yellow fruit?
Explanation: This is a two-step problem. First, interpret the test cross. A ratio of approximately 1:1 (52 red:48 yellow) indicates the unknown parent's genotype was heterozygous (Rr). The cross was Rr x rr. Second, determine the genotype of the red-fruited F1 offspring. In the Rr x rr cross, all red-fruited offspring must have the genotype Rr. Therefore, the second cross is between two Rr individuals (Rr x Rr). The probability of producing a yellow-fruited (rr) plant from this cross is 1/4.
In a certain insect, two genes affect eye color and wing shape. A dihybrid test cross of an individual with dominant phenotypes for both traits yields the following offspring: 485 red eyes/normal wings, 515 red eyes/dwarf wings, 0 brown eyes/normal wings, and 0 brown eyes/dwarf wings. What is the most likely genotype of the parent with the dominant phenotypes?
Explanation: The test cross is with an rrww individual. The offspring phenotypes reveal the gametes produced by the unknown parent. Offspring are red/normal (from an RW or Rw gamete) and red/dwarf (from an RW or Rw gamete). Specifically, since the tester parent is rrww, the red/normal offspring are RrWw and the red/dwarf are Rrww. This means the parent produced RW and Rw gametes. No brown-eyed offspring were produced, so no 'r' allele was passed on, meaning the parent was homozygous RR. The presence of both normal and dwarf winged offspring indicates the parent was heterozygous for the wing gene, Ww. Therefore, the genotype is RRWw.
In corn, the allele for colored kernels (C) is dominant to colorless (c), and the allele for full kernels (S) is dominant to shrunken (s). A plant grown from a colored, full kernel is test-crossed, yielding the following progeny: 144 colored, full; 156 colorless, shrunken; 26 colored, shrunken; and 24 colorless, full. Which statement best explains these results?
Explanation: The offspring do not appear in a 1:1:1:1 ratio, which rules out independent assortment. The two most numerous classes (parental types) are colored, full and colorless, shrunken. The two least numerous classes (recombinant types) are colored, shrunken and colorless, full. This indicates the genes are linked. Since the parental phenotypes match the dominant/dominant and recessive/recessive combinations, the alleles in the heterozygous parent must be in the coupling phase (CS on one chromosome and cs on the homologous chromosome).
In Drosophila, the allele for vestigial wings (vg) is recessive to the allele for normal wings (vg+). A geneticist test crosses a normal-winged male fly. The first three offspring all have normal wings. What is the probability that the fourth offspring will have vestigial wings?
Explanation: When you encounter test cross problems in genetics, remember that independent assortment means each offspring's genotype is determined independently, regardless of previous results. The key insight here is understanding what the observed data tells us about the parent's genotype and how that affects future offspring. Since vestigial wings (vg) is recessive, flies need two vg alleles to express this trait. The test cross involves mating the unknown male with a homozygous recessive female (vg/vg). If the male were homozygous dominant (vg+/vg+), ALL offspring would have normal wings. If he's heterozygous (vg+/vg), then each offspring has a 50% chance of inheriting either allele from him. The fact that three offspring all have normal wings doesn't prove the male's genotype definitively. While this outcome is more likely if he's homozygous dominant, it's still possible (probability = 81) if he's heterozygous. Assuming he could be heterozygous, each individual mating event is independent—the fourth offspring still has a 21 chance of getting the vg allele from a heterozygous father. Answer A incorrectly assumes three normal offspring prove homozygosity. Answer B wrongly suggests probabilities change based on previous outcomes. Answer C confuses this test cross scenario with a typical heterozygous × heterozygous cross, which would yield 41 recessive offspring. Answer D correctly recognizes that if the parent is heterozygous, each offspring independently has a 21 probability of being recessive. Remember: in genetics problems, each fertilization event is independent—previous offspring don't influence future probabilities.
In Drosophila, red eyes (W) are a dominant, X-linked trait, and white eyes (w) are recessive. A geneticist performs a test cross on a red-eyed female of unknown genotype. If the female is heterozygous (X^W X^w), what are the expected phenotypes of her offspring?
Explanation: A test cross for an X-linked trait in a female involves crossing her with a recessive male. The cross is X^W X^w (heterozygous female) × X^w Y (white-eyed male). For female offspring (XX), half will receive X^W from the mother and X^w from the father (X^W X^w, red-eyed), and half will receive X^w from the mother and X^w from the father (X^w X^w, white-eyed). For male offspring (XY), half will receive X^W from the mother (X^W Y, red-eyed), and half will receive X^w from the mother (X^w Y, white-eyed). Thus, both sexes show a 1:1 phenotypic ratio.
In a species of beetle, black body (B) is dominant to brown (b), and long antennae (L) are dominant to short (l). An individual with a black body and long antennae is test-crossed. The cross produces offspring in an approximate phenotypic ratio of 1 black/long : 1 black/short : 1 brown/long : 1 brown/short. What was the genotype of the parent beetle with the dominant phenotypes?
Explanation: A test cross for two traits involves crossing the unknown dominant individual (B_L_) with a homozygous recessive individual (bbll). To produce all four phenotypic classes, the unknown parent must carry both recessive alleles (b and l). The 1:1:1:1 ratio is characteristic of a test cross with a dihybrid individual that is heterozygous for both genes (BbLl) and where the genes assort independently.
A breeder is working with guinea pigs where black fur (B) is dominant to white fur (b). A black male is test-crossed, and the resulting litter contains both black and white pups. What conclusion can be drawn, and why is a test cross superior to a self-cross for this purpose?
Explanation: When you encounter genetics problems involving test crosses, you're dealing with a powerful technique to determine an unknown genotype by crossing with a homozygous recessive individual. Let's analyze this step by step. The black male could be either BB or Bb (since black is dominant). When test-crossed with a white female (bb), the offspring include both black and white pups. This is only possible if the male contributes both B and b alleles to different offspring, proving he must be Bb (heterozygous). If he were BB, all offspring would receive a B allele and appear black. Now let's examine why each answer choice succeeds or fails: Answer A correctly identifies the male as Bb and explains that any recessive offspring definitively proves heterozygosity - this is exactly what happened here. Answer B incorrectly claims the male is BB, which contradicts the evidence of white offspring appearing. The claim about producing "more offspring" is also irrelevant to the genetic analysis. Answer C correctly identifies Bb but wrongly suggests a self-cross would be superior. While self-crosses do produce 3:1 ratios, they're less efficient for determining genotype because you'd need to count many offspring to distinguish between expected ratios. Answer D makes the same BB error as B, ignoring the clear evidence from the offspring ratios. Study tip: Remember that test crosses are the gold standard for determining unknown genotypes because any recessive phenotype in the offspring immediately reveals heterozygosity in the dominant parent - it's a yes/no answer rather than requiring statistical analysis.
In Labrador retrievers, coat color is controlled by two genes that exhibit epistasis. Gene B controls pigment production (B=black, b=brown). Gene E controls pigment deposition (E=deposit, e=no deposit). Dogs with genotype ee are yellow, regardless of their B/b genotype. A black lab (B_E_) is test-crossed to a brown lab (bbEE). However, this is not a standard test cross. If the black lab's genotype is BbEe, what is the expected phenotypic ratio in the offspring?
Explanation: This question tests the ability to apply the test cross concept in a non-standard scenario involving epistasis. The cross is BbEe × bbEE. We can analyze the genes separately. For the B gene, the cross is Bb × bb, which yields 1/2 Bb and 1/2 bb. For the E gene, the cross is Ee × EE, which yields 1/2 EE and 1/2 Ee. All offspring will have at least one dominant E allele, so there will be no yellow labs (ee). The phenotype is therefore determined entirely by the B/b genotype. The offspring genotypes will be 1/2 BbE_ (black) and 1/2 bbE_ (brown), resulting in a 1 black : 1 brown phenotypic ratio.
A test cross on a plant with a dominant phenotype (genotype P_) yields six offspring, all displaying the dominant phenotype. This result supports the hypothesis that the parent is homozygous (PP), but does not prove it. Which of the following subsequent crosses offers the most decisive method to determine the genotype of the original P_ parent?
Explanation: When determining an unknown genotype, you need to design crosses that will produce distinctly different offspring ratios depending on whether the parent is homozygous dominant (PP) or heterozygous (Pp). The initial test cross gave ambiguous results—six dominant offspring could occur with either parental genotype, just with different probabilities. Answer A provides the most decisive test because a backcross between F1 offspring and the original parent will yield dramatically different results depending on the parent's genotype. If the original parent is PP, then all F1 offspring are also PP, and the backcross (PP × PP) produces 100% dominant offspring. However, if the original parent is Pp, then F1 offspring are either PP or Pp, and backcrossing with the heterozygous parent can produce recessive offspring. Any recessive offspring in this backcross would definitively prove the original parent was Pp. Answer B is wrong because self-pollinating F1 offspring won't reveal the original parent's genotype—if the parent was PP, F1 self-crosses yield all dominant offspring, providing no new information. Answer C is incorrect for the same reason; crossing F1 individuals together doesn't distinguish between the two possible parental genotypes. Answer D simply repeats the original approach and suffers from the same statistical ambiguity—you'd need an impractically large sample size to distinguish between the possibilities with confidence. Remember: when genotype determination is ambiguous, design crosses that exploit the differences between possible genotypes to create distinguishable phenotypic ratios.
A geneticist has a male mouse displaying a dominant, autosomal phenotype, "jerker" (J_). She wants to determine if the mouse is homozygous (JJ) or heterozygous (Jj). Which of the following crosses represents the most appropriate and direct test cross?
Explanation: A test cross is specifically designed to determine an unknown genotype of a dominant-phenotype individual by crossing it with an individual that is homozygous recessive for the trait. In this case, crossing the jerker male (J_) with a non-jerker female (jj) will reveal if the male carries the recessive allele. If any offspring are non-jerkers (jj), the male must be heterozygous (Jj).