Genetics Quiz: Designing Genetic Crosses

What this quiz covers

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

How to use this quiz

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

Question 1

1. Cross the mutant (m/m) to a wild-type strain and test cross the F1 to each deletion strain, looking for linkage.
2. Cross different deletion strains to each other and screen for the production of non-viable progeny.
3. Cross each deletion strain to a wild-type strain and look for any F1 progeny that display the mutant phenotype.
4. Cross the mutant (m/m) to each deletion strain and look for F1 progeny that display the mutant phenotype. (correct answer)

Explanation: When you encounter deletion mapping problems, remember that deletions act as null alleles—they completely remove genetic material from a chromosome. This creates a powerful tool for quickly localizing mutations. The correct approach (D) exploits a key principle: if you cross a recessive mutant (m/m) with a deletion strain that has deleted the chromosomal region containing the m gene, the F1 offspring will be m/deletion. Since the deletion provides no functional copy of the gene, these offspring will express the recessive mutant phenotype despite being technically heterozygous. When the deletion doesn't include the m locus, F1 offspring will be m/+ and show the wild-type phenotype. Option A is unnecessarily complex and time-consuming. You'd first need to generate F1 progeny, then perform test crosses with each deletion strain, then analyze linkage patterns—a multi-generation approach when a single cross suffices. Option B makes no biological sense. Crossing deletion strains to each other won't tell you anything about where your specific mutation m is located, and the viability of progeny depends on which essential genes are deleted, not on the location of m. Option C has the logic backwards. Crossing deletion strains to wild-type produces F1 that are heterozygous (deletion/+), which will show wild-type phenotypes regardless of which chromosomal region is deleted, since they retain one functional copy of all genes. Remember: deletion mapping works because deletions unmask recessive mutations when both are present in the same individual. Look for crosses that directly combine your mutation with each deletion.

Question 2

1. Perform the reciprocal cross: a dd female with a DD male, and observe that all F1 progeny are sinistral.
2. Test cross the Dd F1 progeny with a dd parent, and observe that all F2 progeny are dextral.
3. Intercross the Dd F1 progeny, and observe that all F2 progeny are dextral despite segregation of their own genotypes.
4. Intercross the Dd F1 progeny, allow the F2 to self-fertilize, and observe a 3 dextral : 1 sinistral phenotypic ratio in the F3. (correct answer)

Explanation: The key to confirming a maternal effect is to show that an individual's phenotype does not match its genotype but instead matches its mother's genotype. After intercrossing the Dd F1s, all the F2 snails are dextral because their mothers were Dd. However, the F2 snails have genotypes DD, Dd, and dd in a 1:2:1 ratio. When these F2 snails reproduce, the dd mothers will produce all sinistral offspring (in the F3 generation). Observing this 3:1 segregation in the F3 generation, produced by an F2 generation with a uniform phenotype, is the most powerful confirmation of the maternal effect mechanism.

Question 3

1. She must be from a stock where recombination is known to be elevated to maximize double crossovers.
2. Her parental cross must establish a known linkage phase (e.g., she was produced from an ABC/ABC x abc/abc cross). (correct answer)
3. She must be old enough to ensure that age-related increases in recombination do not affect the results.
4. Her phenotype must be dominant for all three traits to ensure complete penetrance of the alleles.

Explanation: To interpret the results of a three-point cross, one must identify the parental and double-crossover classes among the progeny. This is only possible if the original arrangement (linkage phase) of alleles on the homologous chromosomes of the heterozygous parent is known (i.e., whether alleles are in cis, ABC/abc, or trans, AbC/aBc). Without this information, it is impossible to distinguish parental from recombinant gametes. Elevated recombination (A) is not necessary. Age effects (C) are a variable to control, not a prerequisite for design. The phenotype (D) is irrelevant to the transmission genetics being studied.

Question 4

1. Cross each mutant strain to the wild-type strain and analyze the F2 segregation ratios to confirm single-gene inheritance.
2. Perform systematic, pairwise crosses between all 12 mutant strains and observe the phenotype of the resulting diploid. (correct answer)
3. Sequence the genomes of all 12 mutant strains and the wild-type strain to identify the locations of the mutations.
4. Measure the accumulation of arginine precursor molecules in each strain to determine the specific enzymatic block.

Explanation: This is a description of a complementation test. By crossing two recessive mutants, if the resulting diploid is wild-type (Arg+), the mutations complement each other and are therefore in different genes. If the diploid is mutant (Arg-), the mutations fail to complement and are likely alleles of the same gene. Grouping the mutants into complementation groups reveals the number of genes involved. Crossing to wild-type (A) doesn't compare the mutants to each other. Sequencing (C) and biochemical analysis (D) are valid but are molecular/biochemical approaches, not a genetic crossing experiment to determine functional gene groups (complementation groups).

Question 5

1. Cross the F1 progeny with the true-breeding purple-eyed, short-winged parental strain. (correct answer)
2. Intercross the F1 progeny with each other.
3. Cross the F1 progeny with the true-breeding wild-type parental strain.
4. Cross the F1 progeny with a different true-breeding strain that has white eyes and curled wings.

Explanation: This describes a test cross (dihybrid F1 crossed to a homozygous recessive individual). The phenotypes of the offspring from this cross directly reveal the frequencies of the different gamete types (parental and recombinant) produced by the F1 parent, allowing for a direct calculation of recombination frequency and a powerful test for linkage. Intercrossing the F1 (B) would result in a 9:3:3:1 ratio if unlinked, and a distorted ratio if linked, but calculating recombination frequency from this is more complex. A cross to the wild-type parent (C) would mask the recessive alleles. A cross to an unrelated mutant strain (D) would unnecessarily complicate the experiment.

Question 6

1. Cross l/+ flies to CyO/+ flies and select the Curly-winged F1 progeny to interbreed. (correct answer)
2. Cross l/+ flies to each other for several generations, selecting for wild-type flies each time.
3. Cross l/+ flies to homozygous wild-type flies, then interbreed the F1 progeny.
4. Create homozygous l/l flies through inbreeding and then cross them to the CyO/+ stock.

Explanation: To create a balanced stock, the mutation (l) and the balancer (CyO) must be on homologous chromosomes in the same fly. First, cross l/+ to CyO/+ flies. The F1 progeny will include the desired l/CyO genotype (phenotypically Curly). Selecting these Curly flies and interbreeding them creates the stable stock. In this l/CyO x l/CyO cross, the l/l and CyO/CyO progeny are lethal, so only l/CyO progeny survive, maintaining the stock without selection. Option D is impossible as l/l is lethal. Options B and C do not create a balanced system and risk losing the mutation.

Question 7

1. Cross a/a ; +/+ individuals with +/+ ; b/b individuals, and then intercross the resulting F1 progeny. (correct answer)
2. Cross a/+ ; +/+ individuals with +/+ ; b/+ individuals and screen the progeny for the double-mutant phenotype.
3. Perform a test cross by mating an a/+ ; b/+ individual with an a/a ; b/b individual.
4. Chemically mutagenize an a/a single-mutant stock and screen for individuals that also exhibit the b/b phenotype.

Explanation: The standard and most efficient method is a two-step process. First, cross the two single-mutant homozygous lines (a/a and b/b) to create an F1 generation that is uniformly dihybrid (a/+ ; b/+). Second, intercross these F1 individuals. In the resulting F2 generation, the desired a/a ; b/b double mutants will appear in the expected 1/16 proportion. Crossing heterozygotes directly (B) also works but is less controlled if the initial stocks are not already dihybrid. A test cross (C) is used for mapping or determining genotype, not for efficiently generating double homozygotes. Mutagenesis (D) is for creating new mutations, not combining existing ones.

Question 8

1. Perform the reciprocal cross: a dd female with a DD male, and observe that all F1 progeny are sinistral.
2. Test cross the Dd F1 progeny with a dd parent, and observe that all F2 progeny are dextral.
3. Intercross the Dd F1 progeny, and observe that all F2 progeny are dextral despite segregation of their own genotypes.
4. Intercross the Dd F1 progeny, allow the F2 to self-fertilize, and observe a 3 dextral : 1 sinistral phenotypic ratio in the F3. (correct answer)

Explanation: The key to confirming a maternal effect is to show that an individual's phenotype does not match its genotype but instead matches its mother's genotype. After intercrossing the Dd F1s, all the F2 snails are dextral because their mothers were Dd. However, the F2 snails have genotypes DD, Dd, and dd in a 1:2:1 ratio. When these F2 snails reproduce, the dd mothers will produce all sinistral offspring (in the F3 generation). Observing this 3:1 segregation in the F3 generation, produced by an F2 generation with a uniform phenotype, is the most powerful confirmation of the maternal effect mechanism.

Question 9

1. Backcross the F1 to the large-fruited parent and measure the fruit weight of the progeny.
2. Intercross the F1 generation and analyze the distribution of fruit weights in the F2. (correct answer)
3. Backcross the F1 to the small-fruited parent and measure the fruit weight of the progeny.
4. Grow the F1 in different environmental conditions to measure phenotypic plasticity.

Explanation: Intercrossing the F1 is the most informative cross. If the trait is controlled by a single gene with incomplete dominance, the F2 generation will show three discrete phenotypic classes (small, intermediate, large) in a 1:2:1 ratio. If the trait is polygenic, the F2 generation will exhibit a wide, continuous distribution of fruit weights, including the parental extremes but also many new intermediate values. The backcrosses (A, C) are less informative because they don't reveal the full range of segregation. Environmental testing (D) assesses heritability but doesn't distinguish between the number of genes involved.

Question 10

1. Cross two heterozygous (+/Igf2-) mice and observe that the ratio of normal to dwarf mice is 1:1, not 3:1.
2. Analyze DNA methylation patterns at the Igf2 locus in dwarf vs. normal mice to identify epigenetic differences.
3. Perform reciprocal crosses: (+/+ female x +/Igf2- male) and (+/Igf2- female x +/+ male) and compare the offspring. (correct answer)
4. Create a homozygous mutant (Igf2-/ Igf2-) and observe that it is always dwarf regardless of parental origin.

Explanation: Genomic imprinting is defined by the differential expression of an allele depending on its parent of origin. The definitive test is therefore a set of reciprocal crosses. By crossing a mutant male to a wild-type female and a mutant female to a wild-type male, the experiment directly compares the phenotypic outcomes when the mutant allele is passed through the sperm versus the egg. In this case, the first cross would produce dwarf offspring while the second would produce normal offspring, confirming the proposed imprinting pattern. Option A is an expected result but is less direct than the reciprocal cross design. Option B tests the molecular mechanism, not the genetic inheritance pattern. Option D tests the homozygous phenotype, not the parent-of-origin effect on heterozygotes.

Question 11

1. An F1 x F1 cross; data needed is the mean and variance of the F2 population.
2. An F1 x F1 cross; data needed is the fraction of F2 individuals with an extreme parental phenotype (8 cm or 24 cm). (correct answer)
3. A test cross of F1 to the 8 cm parent; data needed is the total number of progeny in the F2 generation.
4. A backcross of F1 to the 24 cm parent; data needed is the correlation between F1 and F2 ear lengths.

Explanation: The Castle-Wright estimator is a classic method for estimating the number of genes (n) contributing to a quantitative trait. It relies on the principle that the proportion of the F2 generation that resembles one of the original homozygous parental lines is equal to (1/4)^n. Therefore, the essential experiment is an F1 x F1 cross to generate the F2, and the critical data is the proportion of F2 individuals that fall into one of the extreme phenotypic classes. Mean and variance (A) can be used in other estimation formulas but are more sensitive to environmental effects. The other options (C, D) describe crosses and data used for different types of quantitative genetic analysis, like heritability.

Question 12

1. Backcross the F1 to the large-fruited parent and measure the fruit weight of the progeny.
2. Intercross the F1 generation and analyze the distribution of fruit weights in the F2. (correct answer)
3. Backcross the F1 to the small-fruited parent and measure the fruit weight of the progeny.
4. Grow the F1 in different environmental conditions to measure phenotypic plasticity.

Explanation: Intercrossing the F1 is the most informative cross. If the trait is controlled by a single gene with incomplete dominance, the F2 generation will show three discrete phenotypic classes (small, intermediate, large) in a 1:2:1 ratio. If the trait is polygenic, the F2 generation will exhibit a wide, continuous distribution of fruit weights, including the parental extremes but also many new intermediate values. The backcrosses (A, C) are less informative because they don't reveal the full range of segregation. Environmental testing (D) assesses heritability but doesn't distinguish between the number of genes involved.

Question 13

1. Irradiate the pollen from the variegated plant before using it in a cross to induce mutations.
2. Analyze the DNA sequence of chloroplast genes from both the variegated and green plants.
3. Self-pollinate the F1 progeny from both reciprocal crosses and analyze the F2 segregation ratios.
4. Perform repeated backcrosses of F1 females to the paternal green parent for several generations. (correct answer)

Explanation: When you encounter reciprocal crosses with different outcomes, you're dealing with either cytoplasmic inheritance or maternal effects. The key challenge is distinguishing between these mechanisms since both show maternal-only transmission in the F1 generation. In cytoplasmic inheritance, organelles like chloroplasts are permanently inherited maternally and persist indefinitely. With maternal effects, a nuclear gene in the mother affects offspring phenotype, but the underlying nuclear alleles still follow Mendelian inheritance patterns in subsequent generations. Option D is correct because repeated backcrossing over several generations will reveal the true mechanism. If it's cytoplasmic inheritance, the variegated phenotype will persist indefinitely since chloroplasts don't segregate like nuclear genes. However, if it's a maternal effect, the phenotype will eventually disappear as the maternal gene products are diluted out over generations, and normal Mendelian ratios will emerge. Option A is incorrect because irradiating pollen might create new mutations but won't distinguish between the two inheritance mechanisms. Option B fails because while DNA sequencing might reveal chloroplast differences, it won't prove whether those differences actually control the phenotype or are just coincidental. Option C won't work because F2 self-pollination from both crosses would likely show similar patterns initially, regardless of the mechanism, since both involve maternal transmission in the F1. Remember: when distinguishing cytoplasmic inheritance from maternal effects, think long-term. Cytoplasmic traits persist indefinitely, while maternal effects fade over generations as nuclear genes reassert Mendelian patterns.

Question 14

1. Cross a/a ; +/+ individuals with +/+ ; b/b individuals, and then intercross the resulting F1 progeny. (correct answer)
2. Cross a/+ ; +/+ individuals with +/+ ; b/+ individuals and screen the progeny for the double-mutant phenotype.
3. Perform a test cross by mating an a/+ ; b/+ individual with an a/a ; b/b individual.
4. Chemically mutagenize an a/a single-mutant stock and screen for individuals that also exhibit the b/b phenotype.

Explanation: The standard and most efficient method is a two-step process. First, cross the two single-mutant homozygous lines (a/a and b/b) to create an F1 generation that is uniformly dihybrid (a/+ ; b/+). Second, intercross these F1 individuals. In the resulting F2 generation, the desired a/a ; b/b double mutants will appear in the expected 1/16 proportion. Crossing heterozygotes directly (B) also works but is less controlled if the initial stocks are not already dihybrid. A test cross (C) is used for mapping or determining genotype, not for efficiently generating double homozygotes. Mutagenesis (D) is for creating new mutations, not combining existing ones.

Question 15

1. Cross an F1 blue-winged female with a wild-type gray-winged male. (correct answer)
2. Cross an F1 gray-winged male with one of his F1 blue-winged sisters.
3. Cross an F1 blue-winged female with her blue-winged father.
4. Cross F1 gray-winged males with true-breeding wild-type females.

Explanation: The initial cross suggests X-linked dominance. The definitive test for sex linkage is a reciprocal cross. The initial cross was mutant male × wild-type female. The reciprocal cross is wild-type male × mutant female (an F1 blue-winged female). If the trait is X-linked dominant, this cross (A) would produce blue and gray offspring of both sexes, confirming the pattern of X-linkage. Cross (B) is an F1 intercross, which is also informative but less direct than the reciprocal cross. Cross (C) is a backcross that is less clean for interpretation. Cross (D) would produce all wild-type offspring and provide no information.

Question 16

1. Cross the mutant (m/m) to a wild-type strain and test cross the F1 to each deletion strain, looking for linkage.
2. Cross different deletion strains to each other and screen for the production of non-viable progeny.
3. Cross each deletion strain to a wild-type strain and look for any F1 progeny that display the mutant phenotype.
4. Cross the mutant (m/m) to each deletion strain and look for F1 progeny that display the mutant phenotype. (correct answer)

Explanation: When you encounter deletion mapping problems, remember that deletions act as null alleles—they completely remove genetic material from a chromosome. This creates a powerful tool for quickly localizing mutations. The correct approach (D) exploits a key principle: if you cross a recessive mutant (m/m) with a deletion strain that has deleted the chromosomal region containing the m gene, the F1 offspring will be m/deletion. Since the deletion provides no functional copy of the gene, these offspring will express the recessive mutant phenotype despite being technically heterozygous. When the deletion doesn't include the m locus, F1 offspring will be m/+ and show the wild-type phenotype. Option A is unnecessarily complex and time-consuming. You'd first need to generate F1 progeny, then perform test crosses with each deletion strain, then analyze linkage patterns—a multi-generation approach when a single cross suffices. Option B makes no biological sense. Crossing deletion strains to each other won't tell you anything about where your specific mutation m is located, and the viability of progeny depends on which essential genes are deleted, not on the location of m. Option C has the logic backwards. Crossing deletion strains to wild-type produces F1 that are heterozygous (deletion/+), which will show wild-type phenotypes regardless of which chromosomal region is deleted, since they retain one functional copy of all genes. Remember: deletion mapping works because deletions unmask recessive mutations when both are present in the same individual. Look for crosses that directly combine your mutation with each deletion.

Question 17

1. Cross an F1 blue-winged female with a wild-type gray-winged male. (correct answer)
2. Cross an F1 gray-winged male with one of his F1 blue-winged sisters.
3. Cross an F1 blue-winged female with her blue-winged father.
4. Cross F1 gray-winged males with true-breeding wild-type females.

Explanation: The initial cross suggests X-linked dominance. The definitive test for sex linkage is a reciprocal cross. The initial cross was mutant male × wild-type female. The reciprocal cross is wild-type male × mutant female (an F1 blue-winged female). If the trait is X-linked dominant, this cross (A) would produce blue and gray offspring of both sexes, confirming the pattern of X-linkage. Cross (B) is an F1 intercross, which is also informative but less direct than the reciprocal cross. Cross (C) is a backcross that is less clean for interpretation. Cross (D) would produce all wild-type offspring and provide no information.

Question 18

1. Cross the F1 progeny with the true-breeding purple-eyed, short-winged parental strain. (correct answer)
2. Intercross the F1 progeny with each other.
3. Cross the F1 progeny with the true-breeding wild-type parental strain.
4. Cross the F1 progeny with a different true-breeding strain that has white eyes and curled wings.

Explanation: This describes a test cross (dihybrid F1 crossed to a homozygous recessive individual). The phenotypes of the offspring from this cross directly reveal the frequencies of the different gamete types (parental and recombinant) produced by the F1 parent, allowing for a direct calculation of recombination frequency and a powerful test for linkage. Intercrossing the F1 (B) would result in a 9:3:3:1 ratio if unlinked, and a distorted ratio if linked, but calculating recombination frequency from this is more complex. A cross to the wild-type parent (C) would mask the recessive alleles. A cross to an unrelated mutant strain (D) would unnecessarily complicate the experiment.

Question 19

1. Mutagenize the his3 strain, plate on media lacking histidine, and select for any colonies that grow. (correct answer)
2. Mutagenize a wild-type (HIS3) strain and screen for mutants that cannot grow without histidine.
3. Mutagenize the his3 strain, mate it with a wild-type strain, and look for F1 diploids that cannot grow without histidine.
4. Mutagenize the his3 strain, mate it to a non-mutagenized his3 strain, and screen the diploid progeny for growth on media lacking histidine.

Explanation: A suppressor screen starts with the mutant phenotype (his3, unable to grow) and looks for a reversion to the wild-type phenotype (able to grow). The most direct method is to mutagenize the his3 strain to create new mutations and then apply a strong selection (plating on media without histidine) that allows only the suppressed individuals to survive. This is a selection, not just a screen. The other options describe different screens: (B) screens for new histidine auxotrophs, (C) is a screen for dominant negative mutations, and (D) is overly complex and screens for dominant suppressors.

Question 20

1. Cross l/+ flies to CyO/+ flies and select the Curly-winged F1 progeny to interbreed. (correct answer)
2. Cross l/+ flies to each other for several generations, selecting for wild-type flies each time.
3. Cross l/+ flies to homozygous wild-type flies, then interbreed the F1 progeny.
4. Create homozygous l/l flies through inbreeding and then cross them to the CyO/+ stock.

Explanation: To create a balanced stock, the mutation (l) and the balancer (CyO) must be on homologous chromosomes in the same fly. First, cross l/+ to CyO/+ flies. The F1 progeny will include the desired l/CyO genotype (phenotypically Curly). Selecting these Curly flies and interbreeding them creates the stable stock. In this l/CyO x l/CyO cross, the l/l and CyO/CyO progeny are lethal, so only l/CyO progeny survive, maintaining the stock without selection. Option D is impossible as l/l is lethal. Options B and C do not create a balanced system and risk losing the mutation.