Genetics Quiz: Genetic Linkage And Recombination

What this quiz covers

This quiz focuses on Genetic Linkage And Recombination, 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. 40%
2. 50% (correct answer)
3. 80%
4. 100%

Explanation: Recombination frequency is used to estimate genetic distance in map units (or centiMorgans), but the relationship is not linear over long distances. Due to the occurrence of multiple crossover events between distant genes, the observed recombination frequency never exceeds 50%. An RF of 50% is characteristic of independent assortment, which occurs for genes on different chromosomes or for genes that are very far apart on the same chromosome. Even though the map distance is 80 cM, the observed frequency of recombinant gametes will plateau at 50%.

Question 2

1. 1 wild-type : 1 black body : 1 vestigial wings : 1 black, vestigial
2. 1 wild-type : 1 black, vestigial (correct answer)
3. 1 black body : 1 vestigial wings
4. 3 wild-type : 1 black, vestigial

Explanation: Complete linkage means no crossing over occurs between the two genes. Therefore, the heterozygous parent (b+ vg+ / b vg) can only produce two types of gametes: the parental gametes b+ vg+ and b vg. The test-cross parent (b vg / b vg) produces only b vg gametes. The resulting progeny will have genotypes b+ vg+ / b vg (wild-type phenotype) and b vg / b vg (black, vestigial phenotype) in a 1:1 ratio. The other options represent outcomes for independent assortment (A), only recombinant phenotypes (C), or a different type of cross (D).

Question 3

1. 0.09
2. 0.18
3. 0.41
4. 0.82 (correct answer)

Explanation: The heterozygous parent has the genotype D e / d E (repulsion phase). This means the parental gametes are D e and d E. The phenotypes corresponding to these gametes in a test cross are dominant-D, recessive-e and recessive-d, dominant-E. The recombination frequency is 18%, which is the proportion of recombinant offspring. The proportion of parental offspring is therefore 100% - 18% = 82%. The probability of the first offspring having a parental phenotype is equal to this proportion, which is 0.82.

Question 4

1. The genes are assorting independently, as the p-value is low.
2. The recombination frequency between the two genes is exactly 1%.
3. The genes are genetically linked, as the observed results are significantly different from the expected results for independent assortment. (correct answer)
4. The experimental data is inconclusive, and the test should be repeated with a larger sample size.

Explanation: In hypothesis testing, the p-value represents the probability of obtaining the observed results (or more extreme results) if the null hypothesis were true. A low p-value (typically < 0.05) indicates that the observed data are unlikely to have occurred by chance under the null hypothesis. Therefore, we reject the null hypothesis. In this case, rejecting the null hypothesis of independent assortment provides strong statistical evidence for the alternative hypothesis, which is that the genes are genetically linked.

Question 5

1. 80
2. 160
3. 420 (correct answer)
4. 840

Explanation: The parent genotype is R T / r t, meaning the parental gametes are R T and r t. The recombinant gametes are R t and r T. The total recombination frequency is 16%, so the total frequency of parental gametes is 100% - 16% = 84%. This is split equally between the two parental gamete types, so the frequency of the R T gamete is 84% / 2 = 42%. The question asks for the number of individuals with the dominant eye color and dominant wing shape phenotype (RrTt). This phenotype results from the R T gamete from the heterozygous parent fertilizing the r t gamete from the test-cross parent. The expected number is the frequency of the R T gamete multiplied by the total number of progeny: 0.42 * 1000 = 420.

Question 6

1. The genes are assorting independently, as the p-value is low.
2. The recombination frequency between the two genes is exactly 1%.
3. The genes are genetically linked, as the observed results are significantly different from the expected results for independent assortment. (correct answer)
4. The experimental data is inconclusive, and the test should be repeated with a larger sample size.

Explanation: In hypothesis testing, the p-value represents the probability of obtaining the observed results (or more extreme results) if the null hypothesis were true. A low p-value (typically < 0.05) indicates that the observed data are unlikely to have occurred by chance under the null hypothesis. Therefore, we reject the null hypothesis. In this case, rejecting the null hypothesis of independent assortment provides strong statistical evidence for the alternative hypothesis, which is that the genes are genetically linked.

Question 7

1. 9 : 3 : 3 : 1
2. 1 : 1 : 1 : 1
3. 12 : 3 : 1
4. 13 : 2 : 2 : 3 (correct answer)

Explanation: Independent assortment in a dihybrid self-cross yields a 9:3:3:1 phenotypic ratio. Genetic linkage causes a deviation from this ratio. When genes are linked in coupling phase (AABB x aabb), the parental phenotypes (A_B_ and aabb) are overrepresented in the F2 generation, and the recombinant phenotypes (A_bb and aaB_) are underrepresented. The ratio 13:2:2:3 shows a clear excess of A_B_ (13 > 9) and aabb (3 > 1) individuals and a deficit of the recombinant types (2 < 3), which is the classic signature of linkage in coupling phase.

Question 8

1. The F2 generation exhibits four distinct phenotypes, but with an excess of the original parental combinations. (correct answer)
2. The F2 generation exhibits a phenotypic ratio that is a modification of 9:3:3:1, such as 9:7.
3. The F2 generation exhibits a novel phenotype not seen in the P or F1 generations.
4. The F2 generation exhibits only two distinct phenotypes instead of the expected four.

Explanation: When analyzing F2 generation patterns, you need to distinguish between genetic linkage (genes located close together on the same chromosome) and gene interactions like epistasis (where one gene affects the expression of another gene). Genetic linkage creates a specific, recognizable pattern: genes that are linked tend to be inherited together more often than expected by chance. When you cross heterozygous F1 individuals, linked genes don't assort independently, so you see an excess of parental combinations in the F2 generation. Answer A correctly describes this phenomenon - four phenotypes appear (indicating the genes are separate), but the original parental combinations occur more frequently than the 9:3:3:1 ratio would predict. Answer B describes epistasis, not linkage. A 9:7 ratio is a classic epistatic pattern where one gene masks the expression of another, modifying the expected Mendelian ratio. Answer C also suggests epistasis - novel phenotypes often result when gene interactions create new phenotypic expressions not seen in the parents. Answer D indicates either epistasis (where gene interaction reduces phenotypic classes) or complete linkage, but doesn't specifically point to the partial linkage that creates detectable recombination patterns. The key distinction is that linkage preserves parental combinations while still allowing some recombination, whereas epistasis fundamentally alters how genes are expressed together. Remember: when you see "excess of parental types" in a genetics problem, think linkage. When you see modified ratios or novel phenotypes, think gene interaction.

Question 9

1. Sh c and sh C (correct answer)
2. Sh C and sh c
3. Sh C and Sh c
4. sh c and sh C

Explanation: When you encounter genetics problems involving two traits, you're dealing with linkage and recombination. The key is identifying which alleles were originally together in the parental generation, then determining which gamete combinations represent crossovers. Let's trace this cross step by step. The original parents were Sh Sh C C (smooth, colored) × sh sh c c (shrunken, colorless). This means the F1 plant has genotype Sh sh C c, where the Sh and C alleles came from one parent on the same chromosome, while sh and c came from the other parent on the homologous chromosome. During meiosis in the F1 plant, if no crossing over occurs, you get parental-type gametes: Sh C and sh c (keeping the original combinations intact). However, if crossing over occurs between these loci, you get recombinant gametes: Sh c and sh C, where alleles have been shuffled into new combinations not seen in either original parent. Looking at the answer choices: A) Sh c and sh C correctly identifies the recombinant types. B) Sh C and sh c represents the parental combinations, not recombinants. C) Sh C and Sh c incorrectly includes one parental type (Sh C) with one recombinant type. D) sh c and sh C similarly mixes one parental type (sh c) with one recombinant. The answer is A. Study tip: Always identify the original parental combinations first, then look for gametes that shuffle these alleles into new pairings—those are your recombinants.

Question 10

1. It would increase to 50% because the inversion disrupts normal pairing.
2. It would decrease significantly, approaching 0%, because crossover products are non-viable. (correct answer)
3. It would remain at 20% because the physical distance between the genes is unchanged.
4. It would be unaffected, but the linkage map order of genes within the inversion would be reversed.

Explanation: In an individual heterozygous for a paracentric inversion, a single crossover event within the inverted region leads to the formation of dicentric and acentric chromatids. Gametes receiving these chromatids are typically non-viable. As a result, viable progeny are almost exclusively derived from non-crossover events or double crossovers within the loop. This effectively suppresses the recovery of recombinant offspring, causing the observed recombination frequency to decrease dramatically, often approaching zero.

Question 11

1. 20% (correct answer)
2. 25%
3. 40%
4. 80%

Explanation: First, identify the parental and recombinant progeny. The parental classes are the most numerous: Ppqq (208) and ppQq (192). This indicates the dihybrid parent had the genotype Pq/pQ. The recombinant classes are the least numerous: PpQq (42) and ppqq (58). The recombination frequency (RF) is calculated as the number of recombinant progeny divided by the total number of progeny, multiplied by 100. Total recombinant progeny = 42 + 58 = 100. Total progeny = 42 + 208 + 192 + 58 = 500. RF = (100 / 500) * 100% = 20%.

Question 12

1. The homozygous genotypes Ab/Ab and aB/aB will be present in the offspring.
2. Offspring with recombinant phenotypes (e.g., A_bb) will be completely absent.
3. The phenotypic ratio will follow a 9:3:3:1 pattern since both parents are dihybrids.
4. Every viable offspring must inherit either an AB or an ab chromosome from its father. (correct answer)

Explanation: When you encounter Drosophila genetics problems, pay close attention to sex-specific differences in crossing over. This question tests your understanding of how the absence of male recombination affects offspring possibilities. Since crossing over doesn't occur in males, the father can only produce two types of gametes: AB or ab. He cannot produce recombinant gametes (Ab or aB). The female, however, can undergo crossing over and potentially produce all four gamete types: AB, Ab, aB, and ab. This means every offspring must receive either an AB chromosome or an ab chromosome from the father - making answer D correct. The father's genetic contribution is limited to these two parental combinations only. Let's examine why the other options are wrong. Answer A is incorrect because for Ab/Ab to exist, the offspring would need Ab from both parents, but the father cannot produce Ab gametes due to no crossing over. Answer B is wrong because recombinant phenotypes can still appear - if the female produces Ab gametes and crosses with the father's ab gametes, you'd get Ab/ab offspring showing the A_bb phenotype. Answer C incorrectly assumes independent assortment applies here, but the 9:3:3:1 ratio only occurs when genes assort independently, which they don't when linkage is involved. Remember this key principle: in Drosophila problems, always check whether crossing over occurs in both sexes. Male Drosophila never undergo crossing over, which significantly constrains the possible gamete types and affects inheritance patterns. This sex-specific difference is a favorite topic on genetics exams.

Question 13

1. 32%
2. 50% (correct answer)
3. 80%
4. 160%

Explanation: Each chiasma represents a crossover event involving two of the four chromatids in a bivalent, producing 50% recombinant products from that event. While for short distances, recombination frequency is half the average number of chiasmata, this relationship breaks down as the number of chiasmata increases. An average number of chiasmata greater than 1.0 indicates that multiple crossovers are common. These multiple crossovers (especially double crossovers involving all four strands) cause the observed recombination frequency to approach a limit of 50%. With an average of 1.6 chiasmata, the genes are far enough apart that they will assort independently, yielding the maximum observable recombination frequency of 50%.

Question 14

1. The two genes must be located on different, non-homologous chromosomes.
2. The two genes are located very close to each other on the same chromosome.
3. The two genes undergo independent assortment during meiosis. (correct answer)
4. A crossover event occurs between the two genes in exactly half of all meiotic divisions.

Explanation: A recombination frequency of 50% is the hallmark of independent assortment. This means that parental and recombinant gametes are produced in equal proportions. Independent assortment occurs either when genes are on different chromosomes (A) or when they are located very far apart on the same chromosome. Since we cannot distinguish between these two physical possibilities based solely on the RF value, the most accurate and general conclusion is that the genes assort independently (C). Close linkage (B) would result in an RF close to 0%. (D) is incorrect because a crossover in every meiosis would still produce an RF of 50%, but an RF of 50% can also be achieved with fewer but more distant crossovers.

Question 15

1. 20% (correct answer)
2. 25%
3. 40%
4. 80%

Explanation: First, identify the parental and recombinant progeny. The parental classes are the most numerous: Ppqq (208) and ppQq (192). This indicates the dihybrid parent had the genotype Pq/pQ. The recombinant classes are the least numerous: PpQq (42) and ppqq (58). The recombination frequency (RF) is calculated as the number of recombinant progeny divided by the total number of progeny, multiplied by 100. Total recombinant progeny = 42 + 58 = 100. Total progeny = 42 + 208 + 192 + 58 = 500. RF = (100 / 500) * 100% = 20%.

Question 16

1. 40%
2. 50% (correct answer)
3. 80%
4. 100%

Explanation: Recombination frequency is used to estimate genetic distance in map units (or centiMorgans), but the relationship is not linear over long distances. Due to the occurrence of multiple crossover events between distant genes, the observed recombination frequency never exceeds 50%. An RF of 50% is characteristic of independent assortment, which occurs for genes on different chromosomes or for genes that are very far apart on the same chromosome. Even though the map distance is 80 cM, the observed frequency of recombinant gametes will plateau at 50%.

Question 17

1. Independent alignment of bivalents at the metaphase I plate.
2. Formation of chiasmata between sister chromatids of a single chromosome.
3. Physical exchange of DNA segments between non-sister chromatids during prophase I. (correct answer)
4. Segregation of homologous chromosomes into different daughter cells at anaphase I.

Explanation: Genetic recombination is the physical process of crossing over, where segments of DNA are exchanged between homologous chromosomes. This event occurs during the pachytene stage of prophase I of meiosis and involves non-sister chromatids. The visible manifestation of this exchange is a chiasma. Independent alignment at metaphase I (A) explains assortment of unlinked genes. Crossing over occurs between non-sister chromatids, not sister chromatids (B). Segregation (D) is the separation of chromosomes, which happens after recombination has already occurred.

Question 18

1. 1 wild-type : 1 black body : 1 vestigial wings : 1 black, vestigial
2. 1 wild-type : 1 black, vestigial (correct answer)
3. 1 black body : 1 vestigial wings
4. 3 wild-type : 1 black, vestigial

Explanation: Complete linkage means no crossing over occurs between the two genes. Therefore, the heterozygous parent (b+ vg+ / b vg) can only produce two types of gametes: the parental gametes b+ vg+ and b vg. The test-cross parent (b vg / b vg) produces only b vg gametes. The resulting progeny will have genotypes b+ vg+ / b vg (wild-type phenotype) and b vg / b vg (black, vestigial phenotype) in a 1:1 ratio. The other options represent outcomes for independent assortment (A), only recombinant phenotypes (C), or a different type of cross (D).

Question 19

1. It would increase to 50% because the inversion disrupts normal pairing.
2. It would decrease significantly, approaching 0%, because crossover products are non-viable. (correct answer)
3. It would remain at 20% because the physical distance between the genes is unchanged.
4. It would be unaffected, but the linkage map order of genes within the inversion would be reversed.

Explanation: In an individual heterozygous for a paracentric inversion, a single crossover event within the inverted region leads to the formation of dicentric and acentric chromatids. Gametes receiving these chromatids are typically non-viable. As a result, viable progeny are almost exclusively derived from non-crossover events or double crossovers within the loop. This effectively suppresses the recovery of recombinant offspring, causing the observed recombination frequency to decrease dramatically, often approaching zero.

Question 20

1. 1.0%
2. 4.0% (correct answer)
3. 5.0%
4. 8.0%

Explanation: The progeny genotype A b c / a b c arises from an A b c gamete from the trihybrid parent. This gamete results from a single crossover (SCO) between genes A and B, with no crossover between B and C. The probability of a crossover between A and B is equal to the map distance, P(CO A-B) = 0.10. The probability of no crossover between B and C is 1 - P(CO B-C) = 1 - 0.20 = 0.80. The probability of this specific SCO event is the product of these probabilities: 0.10 * 0.80 = 0.08. A single crossover event produces two recombinant chromatids out of four, so the frequency of each specific recombinant gamete (A b c and a B C) is half the event frequency. Therefore, the frequency of the A b c gamete is 0.08 / 2 = 0.04, or 4%.