Genetics Quiz
Practice Two Point Gene Maps 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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Two genes, A and B, are 24 map units apart on the same autosome. An individual with genotype A B / a b is crossed with an individual with genotype a b / a b. If 1,000 progeny are produced, what is the expected number of progeny with the genotype A b / a b?
This quiz focuses on Two Point Gene Maps, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.
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Two genes, A and B, are 24 map units apart on the same autosome. An individual with genotype A B / a b is crossed with an individual with genotype a b / a b. If 1,000 progeny are produced, what is the expected number of progeny with the genotype A b / a b?
Explanation: The map distance of 24 cM corresponds to a recombination frequency of 24%. This is the total frequency of both recombinant types. The parental gametes are AB and ab, and the recombinant gametes are Ab and aB. The two recombinant gamete types are produced in approximately equal frequencies. Therefore, the frequency of the Ab gamete is half of the total recombination frequency, which is 24% / 2 = 12%. The expected number of progeny with the A b / a b genotype is 12% of the total, or 0.12 × 1,000 = 120.
If a single crossover event occurs between two linked genes in 28% of the meioses of a dihybrid individual, what is the expected map distance between these two genes?
Explanation: Map distance is defined as the percentage of recombinant offspring. A single crossover event involves only two of the four chromatids in a bivalent. These two chromatids become recombinant, while the other two remain parental. Therefore, if a crossover occurs in a certain percentage of meioses, the resulting frequency of recombinant gametes will be half of that percentage. In this case, the frequency of recombinant gametes is 28% / 2 = 14%. Since 1% recombination frequency equals 1 map unit (cM), the map distance is 14 cM.
In a dihybrid test cross, the resulting progeny exhibit a recombination frequency of 8%. If this cross were repeated and allowed to produce a total of 400 offspring, what is the expected number of offspring with a parental phenotype?
Explanation: A recombination frequency of 8% means that 8% of the offspring are of the recombinant types. The remaining offspring must be of the parental types. The percentage of parental offspring is 100% - 8% = 92%. To find the expected number of parental offspring out of a total of 400, we calculate 92% of 400: 0.92 × 400 = 368.
Two yeast genes, TRP1 and HIS4, are linked. A diploid yeast strain of genotype TRP1 HIS4 / trp1 his4 is sporulated. If the map distance between the genes is 22 cM, what is the approximate probability that a random spore will have the genotype trp1 HIS4?
Explanation: The diploid parent is in coupling phase (TRP1 HIS4 / trp1 his4). The map distance of 22 cM means the total recombination frequency is 22% (0.22). Meiosis produces four spores (a tetrad), which represent the four chromatids. The gametes (spores in this case) will be a mix of parental and recombinant types. The parental genotypes are TRP1 HIS4 and trp1 his4. The recombinant genotypes are TRP1 his4 and trp1 HIS4. The total frequency of recombinant spores is 22%. Since the two recombinant types are produced in equal numbers, the frequency of the trp1 HIS4 spore is half of the total recombination frequency: 22% / 2 = 11%, or 0.11.
A two-point test cross is conducted to map two linked genes, P and Q. The resulting data show 450 parental progeny and 50 recombinant progeny. A second, independent two-point cross for genes Q and R shows 880 parental progeny and 120 recombinant progeny. Based on these data, what is the arrangement of these three genes?
Explanation: First, calculate the map distances. For P-Q: Total progeny = 450 + 50 = 500. Recombination frequency = (50/500) × 100 = 10%. Distance = 10 cM. For Q-R: Total progeny = 880 + 120 = 1000. Recombination frequency = (120/1000) × 100 = 12%. Distance = 12 cM. Two-point crosses can establish the distance between pairs of genes, but they cannot determine the order of genes on a chromosome. For example, the order could be P-Q-R or R-P-Q. To determine the correct order, a three-point cross involving all three genes simultaneously is required. Therefore, while we know the distances between the pairs, the gene order remains ambiguous.
A test cross for two linked genes, D and E, yields 17% recombinant progeny. What is the expected frequency of parental gametes produced by the heterozygous parent?
Explanation: The frequency of recombinant progeny directly reflects the frequency of recombinant gametes produced by the heterozygous parent. If the total recombination frequency is 17% (or 0.17), then the remaining gametes must be of the parental type. The total frequency of all gametes is 100% (or 1.0). Therefore, the frequency of parental gametes is 100% - 17% = 83%, or 1.0 - 0.17 = 0.83.
In maize, the genes for colored aleurone (C) and shrunken endosperm (sh) are linked. A dihybrid plant with the genotype C sh / c Sh is test-crossed to a plant with genotype c sh / c sh. The resulting progeny consist of 149 colored, shrunken; 152 colorless, non-shrunken; 6 colored, non-shrunken; and 4 colorless, shrunken kernels. What is the approximate map distance between the C and sh genes?
Explanation: The dihybrid parent is in repulsion phase (C sh / c Sh), so the parental progeny are colored, shrunken (C sh) and colorless, non-shrunken (c Sh). The recombinant progeny are colored, non-shrunken (C Sh) and colorless, shrunken (c sh). Total progeny = 149 + 152 + 6 + 4 = 311. Recombinant progeny = 6 + 4 = 10. Recombination frequency = (Number of recombinants / Total progeny) × 100 = (10 / 311) × 100 ≈ 3.2%. Therefore, the map distance is approximately 3.2 cM.
A geneticist performs a test cross for two genes, F and G, and observes that the four resulting phenotypic classes appear in a 1:1:1:1 ratio, within the bounds of random statistical fluctuation. A chi-square test yields a p-value of 0.85. Which conclusion is most strongly supported by these results?
Explanation: A 1:1:1:1 phenotypic ratio is the classic expectation for a test cross involving two unlinked genes that assort independently. A high p-value (typically > 0.05) from a chi-square test indicates that there is no statistically significant difference between the observed data and the expected results under the null hypothesis. In this case, the null hypothesis is independent assortment. Therefore, the data strongly support the conclusion that the genes are unlinked (or are so far apart on the same chromosome that they assort independently), and the observed ratio is consistent with this hypothesis.
In a certain species of plant, genes G and H are linked. A plant of genotype G H / g h is crossed with a plant of genotype G h / g H. If the map distance between the two genes is 10 cM, what proportion of the progeny is expected to have the genotype g h / g h?
Explanation: When you encounter linked genes with map distances, you're dealing with recombination frequency and need to identify which offspring represent recombinants versus parentals. First, identify the parental types from each parent. The cross is G H / g h × G h / g H. Parent 1 produces GH and gh gametes (these are parental), while Parent 2 produces Gh and gH gametes (also parental). With 10 cM map distance, recombination frequency is 10%, so recombinant gametes (Gh and gH from Parent 1; GH and gh from Parent 2) each occur at 5% frequency, while parental gametes each occur at 45% frequency. To get g h / g h, you need a gh gamete from each parent. From Parent 1, gh is parental (45% frequency). From Parent 2, gh is recombinant (5% frequency). The probability is 0.45×0.05=0.0225. Looking at the wrong answers: A) 0.0225 appears twice, which is actually the correct answer, making this a duplicate. B) 0.0025 would result from incorrectly treating gh as recombinant from both parents (0.05×0.05). C) 0.2025 comes from mistakenly using parental frequencies from both parents (0.45×0.45), ignoring that gh has different classifications in each parent. Study tip: Always determine whether each required gamete type is parental or recombinant for each specific parent — the same allele combination can have different classifications depending on how the chromosomes were originally arranged.
Two genes are determined to be 70 map units apart on the same chromosome based on a three-point cross. What would be the expected recombination frequency observed in a standard two-point test cross for these same two genes?
Explanation: While map distances can exceed 50 cM (representing the sum of recombination frequencies between intervening genes), the maximum observable recombination frequency between any two points in a single cross is 50%. This occurs because for genes far apart, multiple crossover events are common. The probability of an odd number of crossovers (which produces recombinant gametes) becomes equal to the probability of an even number of crossovers (which produces parental gametes). As a result, the genes behave as if they are unlinked, assorting independently and producing 50% recombinant progeny.
A test cross for two genes in coupling phase (A B / a b) produces 18% recombinant offspring. If the same cross was performed but the heterozygous parent was in repulsion phase (A b / a B), what would be the expected frequency of progeny with the genotype a b / a b?
Explanation: The map distance between the genes is 18 cM, as determined by the 18% recombination frequency. This distance is a property of the genes and does not change with the linkage phase. For a parent in repulsion phase (A b / a B), the parental gametes are Ab and aB. The recombinant gametes are AB and ab. The total frequency of recombinant gametes is 18% (0.18). The frequency of each specific recombinant gamete (AB or ab) is half of the total, so f(ab) = 18% / 2 = 9% (or 0.09). In a test cross, the progeny genotype directly reflects the gamete from the heterozygous parent. Therefore, the frequency of a b / a b progeny will be equal to the frequency of the ab gamete, which is 0.09.
A two-point test cross is conducted to map two linked genes, P and Q. The resulting data show 450 parental progeny and 50 recombinant progeny. A second, independent two-point cross for genes Q and R shows 880 parental progeny and 120 recombinant progeny. Based on these data, what is the arrangement of these three genes?
Explanation: First, calculate the map distances. For P-Q: Total progeny = 450 + 50 = 500. Recombination frequency = (50/500) × 100 = 10%. Distance = 10 cM. For Q-R: Total progeny = 880 + 120 = 1000. Recombination frequency = (120/1000) × 100 = 12%. Distance = 12 cM. Two-point crosses can establish the distance between pairs of genes, but they cannot determine the order of genes on a chromosome. For example, the order could be P-Q-R or R-P-Q. To determine the correct order, a three-point cross involving all three genes simultaneously is required. Therefore, while we know the distances between the pairs, the gene order remains ambiguous.
In Drosophila, the genes for cut wings (ct) and sable body (s) are X-linked and 15 cM apart. A female fly with genotype ct s+ / ct+ s is crossed with a wild-type male (ct+ s+ / Y). What proportion of the male offspring is expected to have cut wings and a sable body?
Explanation: When you encounter X-linked genes with map distances in Drosophila, you're dealing with recombination frequency and need to track which gametes the female can produce. The female has genotype ct s+ / ct+ s, meaning one X chromosome carries cut and wild-type sable, while the other carries wild-type cut and sable. Since these genes are 15 cM apart, recombination occurs 15% of the time. This means 85% of her gametes will be parental types (ct s+ and ct+ s), while 15% will be recombinant types (ct s and ct+ s+). Each recombinant class represents half of the 15%, so ct s gametes make up 7.5% of her total gametes. The male contributes either ct+ s+ (X chromosome) or Y chromosome. Male offspring receive the Y chromosome, so their phenotype depends entirely on which X chromosome they inherit from their mother. To have cut wings AND sable body, males need the ct s X chromosome from mom. Therefore, 7.5% or 0.075 of male offspring will have both cut wings and sable body, making D correct. A (0.000) assumes these alleles never appear together, ignoring recombination. B (0.150) incorrectly uses the total recombination frequency instead of recognizing that ct s represents only one of two recombinant classes. C (0.425) likely represents a parental class frequency (42.5%), confusing recombinant with parental types. Strategy tip: Always identify which gamete type produces the desired phenotype, then calculate that specific gamete's frequency using recombination data.
A plant breeder crosses a dihybrid tall plant with purple flowers to a dwarf plant with white flowers, and obtains the following progeny: 122 tall, purple; 128 dwarf, white; 27 tall, white; and 23 dwarf, purple. A new researcher analyzes only the recombinant classes and concludes the map distance is (27+23)/(27+23) = 100 cM. What is the primary flaw in the new researcher's calculation?
Explanation: The researcher correctly identified the recombinant classes (tall, white and dwarf, purple; 27+23=50). However, the formula for map distance is (Number of Recombinants / Total Number of Progeny) × 100. The researcher divided the number of recombinants by itself instead of the total. The correct calculation is: Total progeny = 122 + 128 + 27 + 23 = 300. Map distance = (50 / 300) × 100 ≈ 16.7 cM. The primary flaw described is the incorrect denominator in the frequency calculation.
In maize, the genes for colored aleurone (C) and shrunken endosperm (sh) are linked. A dihybrid plant with the genotype C sh / c Sh is test-crossed to a plant with genotype c sh / c sh. The resulting progeny consist of 149 colored, shrunken; 152 colorless, non-shrunken; 6 colored, non-shrunken; and 4 colorless, shrunken kernels. What is the approximate map distance between the C and sh genes?
Explanation: The dihybrid parent is in repulsion phase (C sh / c Sh), so the parental progeny are colored, shrunken (C sh) and colorless, non-shrunken (c Sh). The recombinant progeny are colored, non-shrunken (C Sh) and colorless, shrunken (c sh). Total progeny = 149 + 152 + 6 + 4 = 311. Recombinant progeny = 6 + 4 = 10. Recombination frequency = (Number of recombinants / Total progeny) × 100 = (10 / 311) × 100 ≈ 3.2%. Therefore, the map distance is approximately 3.2 cM.
If a single crossover event occurs between two linked genes in 28% of the meioses of a dihybrid individual, what is the expected map distance between these two genes?
Explanation: Map distance is defined as the percentage of recombinant offspring. A single crossover event involves only two of the four chromatids in a bivalent. These two chromatids become recombinant, while the other two remain parental. Therefore, if a crossover occurs in a certain percentage of meioses, the resulting frequency of recombinant gametes will be half of that percentage. In this case, the frequency of recombinant gametes is 28% / 2 = 14%. Since 1% recombination frequency equals 1 map unit (cM), the map distance is 14 cM.
A test cross for two linked genes, D and E, yields 17% recombinant progeny. What is the expected frequency of parental gametes produced by the heterozygous parent?
Explanation: The frequency of recombinant progeny directly reflects the frequency of recombinant gametes produced by the heterozygous parent. If the total recombination frequency is 17% (or 0.17), then the remaining gametes must be of the parental type. The total frequency of all gametes is 100% (or 1.0). Therefore, the frequency of parental gametes is 100% - 17% = 83%, or 1.0 - 0.17 = 0.83.
In a dihybrid test cross, the resulting progeny exhibit a recombination frequency of 8%. If this cross were repeated and allowed to produce a total of 400 offspring, what is the expected number of offspring with a parental phenotype?
Explanation: A recombination frequency of 8% means that 8% of the offspring are of the recombinant types. The remaining offspring must be of the parental types. The percentage of parental offspring is 100% - 8% = 92%. To find the expected number of parental offspring out of a total of 400, we calculate 92% of 400: 0.92 × 400 = 368.
A plant breeder crosses a dihybrid tall plant with purple flowers to a dwarf plant with white flowers, and obtains the following progeny: 122 tall, purple; 128 dwarf, white; 27 tall, white; and 23 dwarf, purple. A new researcher analyzes only the recombinant classes and concludes the map distance is (27+23)/(27+23) = 100 cM. What is the primary flaw in the new researcher's calculation?
Explanation: The researcher correctly identified the recombinant classes (tall, white and dwarf, purple; 27+23=50). However, the formula for map distance is (Number of Recombinants / Total Number of Progeny) × 100. The researcher divided the number of recombinants by itself instead of the total. The correct calculation is: Total progeny = 122 + 128 + 27 + 23 = 300. Map distance = (50 / 300) × 100 ≈ 16.7 cM. The primary flaw described is the incorrect denominator in the frequency calculation.
A geneticist performs a test cross for two genes, F and G, and observes that the four resulting phenotypic classes appear in a 1:1:1:1 ratio, within the bounds of random statistical fluctuation. A chi-square test yields a p-value of 0.85. Which conclusion is most strongly supported by these results?
Explanation: A 1:1:1:1 phenotypic ratio is the classic expectation for a test cross involving two unlinked genes that assort independently. A high p-value (typically > 0.05) from a chi-square test indicates that there is no statistically significant difference between the observed data and the expected results under the null hypothesis. In this case, the null hypothesis is independent assortment. Therefore, the data strongly support the conclusion that the genes are unlinked (or are so far apart on the same chromosome that they assort independently), and the observed ratio is consistent with this hypothesis.