Genetics Quiz: Linked Vs Unlinked Genes

What this quiz covers

This quiz focuses on Linked Vs Unlinked Genes, 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. The alleles were in repulsion phase (Fg/fG).
2. The alleles were in coupling phase (FG/fg). (correct answer)
3. The genes F and G were on different chromosomes.
4. The FfGg parent was homozygous for one of the genes.

Explanation: In a test cross, the offspring phenotypic ratio directly reflects the gametic ratio of the heterozygous parent. The most frequent phenotypes are F_G_ (43%) and ffgg (42%). These correspond to the parental gametes. Therefore, the parental gametes produced by the FfGg parent were FG and fg. This means the alleles were arranged on the homologous chromosomes in the coupling (or cis) phase: FG/fg.

Question 2

1. 6.3% (correct answer)
2. 7.5%
3. 13.9%
4. 43.2%

Explanation: The heterozygous parent was YZ/yz, so the parental offspring are YyZz and yyzz. The recombinant offspring are Yyzz and yyZz. The total number of offspring is 391 + 384 + 68 + 57 = 900. The total number of recombinant offspring is 68 + 57 = 125. The total recombination frequency is (125/900) * 100 ≈ 13.9%. This total frequency is split between the two recombinant gamete types, Yz and yZ. The frequency of the yZ gamete is directly reflected by the proportion of yyZz offspring, which is 57/900 ≈ 0.0633, or approximately 6.3%.

Question 3

1. The genes are on different chromosomes.
2. There was a systematic error in classifying the offspring phenotypes. (correct answer)
3. The genes are linked and are 60 map units apart.
4. The genes are located on the X chromosome.

Explanation: The maximum possible recombination frequency between two genes is 50%, which occurs when genes are on different chromosomes or very far apart on the same chromosome. A recombination frequency greater than 50% is not biologically possible through standard meiotic crossing over, as this would imply that non-parental combinations are more likely than parental ones. Therefore, observing 60% recombinant offspring strongly suggests a misclassification of which phenotypes are parental and which are recombinant, or some other significant experimental error. If the classes were swapped, the RF would be 40%, a plausible value for linkage.

Question 4

1. 4%
2. 10%
3. 16% (correct answer)
4. 25%

Explanation: Each parent has the genotype T P / t p. The parental gametes are TP and tp, while the recombinant gametes are Tp and tP. Since the recombination frequency is 20%, the total frequency of recombinant gametes is 20%, and the total frequency of parental gametes is 80%. The frequency of each individual gamete type is: f(TP) = 40%, f(tp) = 40%, f(Tp) = 10%, f(tP) = 10%. The ttpp genotype can only arise from the fusion of a 'tp' gamete from the first parent and a 'tp' gamete from the second parent. The probability of this event is f(tp) × f(tp) = 0.40 × 0.40 = 0.16, or 16%.

Question 5

1. A cross between two true-breeding individuals: OrpOrp BluBlu × orporp blublu.
2. A cross between two F1 heterozygotes from the above cross: Orp_orp Blu_blu × Orp_orp Blu_blu.
3. A test cross between an F1 heterozygote and a homozygous recessive individual: Orp_orp Blu_blu × orporp blublu. (correct answer)
4. A monohybrid cross for one of the genes: Orp_orp × Orp_orp.

Explanation: A test cross (C) is the most effective method for determining linkage. In this cross, the homozygous recessive parent only contributes recessive alleles, so the phenotypes of the offspring directly reveal the gametic contributions of the heterozygous parent. If the genes are unlinked, the four phenotypes will appear in a 1:1:1:1 ratio. If they are linked, the parental phenotypes will be significantly more common than the recombinant phenotypes. A dihybrid cross (B) produces a 9:3:3:1 ratio, and deviations from this are more complex to interpret. The P-generation cross (A) only produces the F1, and a monohybrid cross (D) cannot test for linkage between two different genes.

Question 6

1. 90
2. 180
3. 410 (correct answer)
4. 820

Explanation: The P cross is vgb/vgb × ++/++. The F1 is vgb/++. The alleles are in coupling (cis) phase. In the test cross of the F1 female, the parental gametes are vgb and ++. The recombinant gametes are vg+ and +b. The recombination frequency is 18%, so the parental gametes make up the remaining 100% - 18% = 82% of the total. Each parental gamete type has a frequency of 82%/2 = 41%. The wild-type phenotype (long wings, gray body) results from the parental '++' gamete. Therefore, the expected number of wild-type offspring is 41% of 1000, which is 410.

Question 7

1. The null hypothesis of independent assortment is supported, as the p-value is low.
2. The null hypothesis of independent assortment is rejected, providing strong evidence that the genes are linked. (correct answer)
3. The genes are unlinked, but other factors such as reduced viability are skewing the ratios.
4. The experiment must be repeated with a larger sample size because the results are statistically inconclusive.

Explanation: The null hypothesis for this test is that the genes assort independently, which predicts a 1:1:1:1 ratio. A very low p-value (typically < 0.05) indicates that the observed deviation from the expected ratio is statistically significant and not likely due to random chance. Therefore, the null hypothesis of independent assortment is rejected. The large discrepancy between the parental classes (398, 382) and recombinant classes (55, 45) is strong evidence for gene linkage.

Question 8

1. 4%
2. 10%
3. 16% (correct answer)
4. 25%

Explanation: Each parent has the genotype T P / t p. The parental gametes are TP and tp, while the recombinant gametes are Tp and tP. Since the recombination frequency is 20%, the total frequency of recombinant gametes is 20%, and the total frequency of parental gametes is 80%. The frequency of each individual gamete type is: f(TP) = 40%, f(tp) = 40%, f(Tp) = 10%, f(tP) = 10%. The ttpp genotype can only arise from the fusion of a 'tp' gamete from the first parent and a 'tp' gamete from the second parent. The probability of this event is f(tp) × f(tp) = 0.40 × 0.40 = 0.16, or 16%.

Question 9

1. PD = NPD; T = 0
2. PD > NPD; T is variable
3. PD = NPD; T is variable (correct answer)
4. PD = T; NPD = 0

Explanation: This question requires distinguishing linked vs. unlinked genes using tetrad analysis ratios, a related concept. For unlinked genes, the orientation of non-homologous chromosomes at metaphase I is random. This leads to an equal probability of producing parental ditype (PD) tetrads (all parental spores) and non-parental ditype (NPD) tetrads (all recombinant spores) when no crossovers occur between the genes and their centromeres. Tetratype (T) tetrads (containing both parental and recombinant spores) are formed when a crossover occurs between one of the genes and its centromere. Thus, for unlinked genes, PD = NPD. In contrast, for linked genes, PD > NPD, because NPDs require a four-strand double crossover, which is rare. The ratio is PD = NPD; T is variable.

Question 10

1. The ABC and abc parental classes are the most numerous.
2. The ABc and abC classes are more numerous than the aBC and Abc classes.
3. The AbC and aBc double-crossover classes are the least numerous.
4. All eight phenotypic classes appear in approximately equal numbers. (correct answer)

Explanation: Linkage, by definition, means that parental allele combinations are inherited together more often than not, leading to an overrepresentation of parental phenotypes and an underrepresentation of recombinant phenotypes in the offspring. If all eight phenotypic classes appeared in approximately equal numbers, it would signify that the three genes are assorting independently of each other, which is the opposite of linkage. Statements A, B, and C describe expected patterns for linked genes (parentals most frequent, double crossovers least frequent, and single crossovers at intermediate frequencies).

Question 11

1. A and B only
2. C and D only
3. C and D, and E and F (correct answer)
4. A and B, C and D, and E and F

Explanation: Independent assortment occurs when genes are on different chromosomes or when they are very far apart on the same chromosome. Genes C and D are on different chromosomes, so they will assort independently. Genes E and F are 75 map units apart. A genetic distance of 50 map units or more results in a recombination frequency of 50%, which is phenotypically indistinguishable from independent assortment. Therefore, E and F will also produce ratios consistent with this law. Genes A and B are on the same chromosome, and since their distance is not specified to be 50 cM or more, they are assumed to be linked and would not assort independently.

Question 12

1. 10%
2. 20% (correct answer)
3. 40%
4. 80%

Explanation: In a test cross, the two larger phenotypic classes represent the parental types, and the two smaller classes represent the recombinant types. Here, the parental classes are in the proportion of 40 and 40, while the recombinants are 10 and 10. The total proportion is 10+40+40+10 = 100. The total proportion of recombinants is 10 + 10 = 20. The recombination frequency is the percentage of recombinant offspring, which is 20%.

Question 13

1. The calculated χ² value is high, so the genes are definitely linked.
2. The calculated χ² value is high, so the genes assort independently.
3. The calculated χ² value is low, so the deviation from the 1:1:1:1 ratio is significant, indicating linkage.
4. The calculated χ² value is low, so there is no significant evidence to reject the hypothesis of independent assortment. (correct answer)

Explanation: The null hypothesis for the chi-square test is that the genes assort independently (predicting a 1:1:1:1 ratio). A statistically significant result that rejects the null hypothesis occurs when the calculated chi-square value is greater than the critical value. In this case, the calculated value (2.1) is less than the critical value (7.81). This means the observed deviation from the expected 1:1:1:1 ratio is small and likely due to random sampling error. Therefore, we fail to reject the null hypothesis, and there is no statistical evidence for linkage.

Question 14

1. The p-value is high, so the null hypothesis of independent assortment should be rejected.
2. The p-value is high, indicating the observed results are consistent with the hypothesis of independent assortment. (correct answer)
3. The genes are weakly linked, and a larger sample size is needed to achieve a significant p-value.
4. The p-value of 0.82 represents the recombination frequency between the two genes.

Explanation: The null hypothesis in this context is that the genes assort independently, predicting a 1:1:1:1 ratio. A high p-value (conventionally > 0.05) means that there is a high probability of observing such deviations from the expected ratio by random chance alone. Therefore, there is no statistical reason to reject the null hypothesis. The data are consistent with the genes being unlinked.

Question 15

1. Codominance between alleles of one gene.
2. Independent assortment of the two genes.
3. Complete linkage with no recombination.
4. Partial linkage with recombination. (correct answer)

Explanation: This observation describes the classic outcome for linked genes. The two more frequent classes correspond to the parental allele combinations, which are inherited together most of the time. The two less frequent classes correspond to the recombinant allele combinations, which are produced by crossing over. Independent assortment (B) would result in four classes of roughly equal frequency. Complete linkage (C) would result in only two parental classes and zero recombinant classes. Codominance (A) describes allelic interactions at a single locus and does not explain the ratio pattern between two different genes.

Question 16

1. 9% (correct answer)
2. 18%
3. 41%
4. 82%

Explanation: The female fly has alleles in the trans (repulsion) configuration (De/dE). Her parental gametes are De and dE, and her recombinant gametes are DE and de. The total recombination frequency is 18%, meaning the two recombinant gametes (DE and de) together make up 18% of the total. Each recombinant gamete is expected to have a frequency of half the total, so frequency(DE) = 18% / 2 = 9%. The D_E_ phenotype in the offspring results from the fertilization of a de ovum by a DE sperm from the female parent. Therefore, the expected proportion of D_E_ offspring is 9%.

Question 17

1. The alleles were in repulsion phase (Fg/fG).
2. The alleles were in coupling phase (FG/fg). (correct answer)
3. The genes F and G were on different chromosomes.
4. The FfGg parent was homozygous for one of the genes.

Explanation: In a test cross, the offspring phenotypic ratio directly reflects the gametic ratio of the heterozygous parent. The most frequent phenotypes are F_G_ (43%) and ffgg (42%). These correspond to the parental gametes. Therefore, the parental gametes produced by the FfGg parent were FG and fg. This means the alleles were arranged on the homologous chromosomes in the coupling (or cis) phase: FG/fg.

Question 18

1. The ABC and abc parental classes are the most numerous.
2. The ABc and abC classes are more numerous than the aBC and Abc classes.
3. The AbC and aBc double-crossover classes are the least numerous.
4. All eight phenotypic classes appear in approximately equal numbers. (correct answer)

Explanation: Linkage, by definition, means that parental allele combinations are inherited together more often than not, leading to an overrepresentation of parental phenotypes and an underrepresentation of recombinant phenotypes in the offspring. If all eight phenotypic classes appeared in approximately equal numbers, it would signify that the three genes are assorting independently of each other, which is the opposite of linkage. Statements A, B, and C describe expected patterns for linked genes (parentals most frequent, double crossovers least frequent, and single crossovers at intermediate frequencies).

Question 19

1. 6.3% (correct answer)
2. 7.5%
3. 13.9%
4. 43.2%

Explanation: The heterozygous parent was YZ/yz, so the parental offspring are YyZz and yyzz. The recombinant offspring are Yyzz and yyZz. The total number of offspring is 391 + 384 + 68 + 57 = 900. The total number of recombinant offspring is 68 + 57 = 125. The total recombination frequency is (125/900) * 100 ≈ 13.9%. This total frequency is split between the two recombinant gamete types, Yz and yZ. The frequency of the yZ gamete is directly reflected by the proportion of yyZz offspring, which is 57/900 ≈ 0.0633, or approximately 6.3%.

Question 20

1. The p-value is high, so the null hypothesis of independent assortment should be rejected.
2. The p-value is high, indicating the observed results are consistent with the hypothesis of independent assortment. (correct answer)
3. The genes are weakly linked, and a larger sample size is needed to achieve a significant p-value.
4. The p-value of 0.82 represents the recombination frequency between the two genes.

Explanation: The null hypothesis in this context is that the genes assort independently, predicting a 1:1:1:1 ratio. A high p-value (conventionally > 0.05) means that there is a high probability of observing such deviations from the expected ratio by random chance alone. Therefore, there is no statistical reason to reject the null hypothesis. The data are consistent with the genes being unlinked.