Genetics Quiz: Parental And Recombinant Classes

What this quiz covers

This quiz focuses on Parental And Recombinant Classes, 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 genes are assorting independently, and deviations from a 1:1:1:1 ratio are due to random chance.
2. The genes are linked, and the parental gametes from the dihybrid were Cs and cS.
3. The genes are linked, and the parental gametes from the dihybrid were CS and cs. (correct answer)
4. The experiment is invalid because the two most frequent classes are not reciprocal.

Explanation: In this testcross, the progeny phenotypes reflect the gametes from the CcSs parent. The frequencies are: cs (275), CS (270), Cs (230), and cS (225). While close to a 1:1:1:1 ratio, there is a clear pattern where two classes are more frequent than the other two. The most frequent gametes are cs and CS. These represent the parental gametes. The less frequent gametes, Cs and cS, are the recombinants. The deviation from a 250:250:250:250 ratio is statistically significant (χ² > 7.81 for 3 df), indicating linkage. Because CS and cs are the parental gametes, the parent was in coupling phase.

Question 2

1. Phenotypes AC and ac
2. Phenotypes Ac and aC (correct answer)
3. Phenotypes ABC and abc
4. Phenotypes Abc and aBC

Explanation: To analyze the linkage between A and C, we must ignore gene B and sum the counts for the four possible A-C phenotypic combinations. AC = ABC (350) + ABc (4) = 354 ac = abC (48) + abc (342) = 390 Ac = AbC (98) + Abc (52) = 150 aC = aBC (50) + aBc (106) = 156 The most frequent combinations are AC (354) and ac (390), making them the parental classes for this pair of genes. The least frequent combinations are Ac (150) and aC (156), which are therefore the recombinant classes.

Question 3

1. 8%
2. 16% (correct answer)
3. 42%
4. 84%

Explanation: The parent is in repulsion phase: dpy + / + unc. With 16% recombination, parental gametes (dpy + and + unc) each have frequency 42%, while recombinant gametes (dpy unc and + +) each have frequency 8%. Progeny with the Dumpy phenotype come from two sources: dpy + gametes (42%) giving Dumpy-wild type, and dpy unc gametes (8%) giving Dumpy-Unc. Total Dumpy progeny = 42% + 8% = 50%. Among Dumpy progeny, those that are also Unc = 8%/(42%+8%) = 8%/50% = 16%.

Question 4

1. Pl and pL (correct answer)
2. PL and pl
3. Pp and Ll
4. PL, Pl, pL, and pl in equal amounts

Explanation: When you encounter linked genes that deviate from the expected 9:3:3:1 ratio, you're dealing with genes that don't assort independently because they're located close together on the same chromosome. The key insight is identifying which allele combinations were together in the original parents by looking at the most frequent F2 classes. Since the F1 dihybrid produces mostly purple flowers, round pollen (P_ll) and red flowers, long pollen (ppL_) in the F2, these represent the parental combinations that were linked together in the F1. For these to be the most common classes, the F1 must have received Pl from one parent and pL from the other parent. During meiosis, the F1 produces mostly Pl and pL gametes (parental types) because crossing over between linked genes is relatively rare. Looking at the answer choices: A) Pl and pL correctly identifies the two parental gamete types that would produce the observed pattern. B) PL and pl would generate purple long and red round as the most common F2 classes, opposite of what's observed. C) Pp and Ll incorrectly suggests these are separate, unlinked genes rather than gamete combinations. D) PL, Pl, pL, and pl in equal amounts describes independent assortment, which would give the standard 9:3:3:1 ratio we're told doesn't occur. Study tip: When analyzing linkage problems, always work backward from the most frequent F2 phenotypes to determine the original parental combinations. The rare classes represent recombinants from crossing over.

Question 5

1. 1 long, spots : 1 short, no spots
2. 1 long, no spots : 1 short, spots (correct answer)
3. 1 long, no spots : 1 long, spots : 1 short, no spots : 1 short, spots
4. 9 long, spots : 3 long, no spots : 3 short, spots : 1 no spots

Explanation: The P generation cross is AAbb (long, no spots) × aaBB (short, spots). The resulting F1 generation is heterozygous with alleles in repulsion: Ab/aB. Because the genes are completely linked, no crossing over occurs. The F1 individual can only produce two types of gametes, the parental gametes Ab and aB. When this F1 is testcrossed to an aabb individual, the progeny will have genotypes Ab/ab and aB/ab, resulting in phenotypes of long, no spots and short, spots in a 1:1 ratio. The recombinant phenotypes (long, spots and short, no spots) will not be produced.

Question 6

1. 3% each
2. 6% each
3. 94% each
4. 47% each (correct answer)

Explanation: When you encounter testcross problems involving linked genes, remember that recombination frequency directly tells you about crossover events, and the remaining gametes represent the parental combinations. In this F1 individual with genotype $Xy/xY$, the two chromosomes carry different combinations of alleles. During meiosis, most gametes will maintain these original parental combinations ($Xy$ and $xY$), while fewer will result from crossing over between the linked genes. Since 6% of progeny show recombinant phenotypes, this means 6% of the F1's gametes were recombinant types. In a testcross, the recombinant gametes would be $XY$ and $xy$ (3% each). The remaining 94% must be the parental types: $Xy$ and $xY$. Since these two parental combinations are produced equally during meiosis, each represents $94\% ÷ 2 = 47\%$ of total gametes. Answer choice A (3% each) represents the frequency of each recombinant gamete type, not the parental types. Answer choice B (6% each) incorrectly assumes the 6% recombination frequency applies to each parental type rather than total recombinants. Answer choice C (94% each) would give you 188% total frequency, which is impossible—this represents the total frequency of all parental types combined. The correct answer is D (47% each). Study tip: In linkage problems, always remember that recombination frequency + parental frequency = 100%, and both recombinant types are equally frequent, as are both parental types.

Question 7

1. The two parental classes together represent approximately 12.5% of the progeny.
2. The two recombinant classes together represent approximately 87.5% of the progeny.
3. The genes are assorting independently, and the ratio is an anomaly.
4. The two parental classes are about 7 times more frequent than the two recombinant classes. (correct answer)

Explanation: The 7:1:1:7 ratio clearly indicates linkage, as it deviates from the 1:1:1:1 ratio expected for independent assortment. The two '7' categories are the parental classes, and the two '1' categories are the recombinant classes. The total ratio parts are 7+1+1+7=16. The parental classes make up (7+7)/16 = 14/16 of the total, while recombinant classes make up (1+1)/16 = 2/16. The ratio of parental frequency to recombinant frequency is 14/2 = 7. Therefore, parental classes are 7 times more frequent than recombinant classes. Recombination frequency is 2/16 = 12.5%.

Question 8

1. Only progeny with genotypes PqR/pqr and pQr/pqr will be observed.
2. Progeny resulting from double-crossover events will not be observed. (correct answer)
3. Parental and recombinant progeny will be observed in equal frequencies.
4. The two single-crossover classes will be absent from the progeny.

Explanation: Interference (I) is calculated as 1 minus the coefficient of coincidence (c). If I = 1.0, then c = 0. The coefficient of coincidence is the ratio of observed double crossovers (DCO) to expected DCO. If c = 0, the observed number of DCOs is zero. This means that a crossover in one region completely prevents a crossover in the adjacent region. Therefore, no progeny resulting from double-crossover events will be observed. Parental and single-crossover classes will still be produced.

Question 9

1. leu pan⁺
2. leu⁺ pan
3. leu pan (correct answer)
4. leu⁺ pan⁺ / leu pan

Explanation: The parent strain has the genotype leu pan⁺ / leu⁺ pan. This means the alleles leu and pan⁺ are on one chromosome, and leu⁺ and pan are on the homologous chromosome. These represent the parental combinations. Recombinant genotypes are produced by a crossover event between the two genes, which would result in spores with genotypes leu pan (double mutant) and leu⁺ pan⁺ (wild type). Among the choices, leu pan is a recombinant class. leu pan⁺ and leu⁺ pan are parental classes. The genotype in choice D is diploid, not a haploid spore.

Question 10

1. Dg/dG (correct answer)
2. DG/dg
3. DD/gg
4. Dd/Gg

Explanation: This question tests your understanding of genetic linkage and recombination frequency. When you see a test cross with unequal phenotype classes, you're dealing with linked genes where recombination creates predictable patterns. In the original cross (DG/dg × dg/dg), the F1 offspring could have genotypes DG/dg or dg/dg. Since we're told an F1 plant was selected for test crossing, it must be DG/dg (the heterozygote). Now, when this F1 plant undergoes meiosis, it can produce four types of gametes: DG, dg (parental types) and Dg, dG (recombinant types). The key insight is that the most abundant progeny classes are D-gg and ddG-. These phenotypes come from Dg and dG gametes respectively - the recombinant types! This means the F1 plant produced more recombinant gametes than parental gametes, which happens when crossing over occurs more than 50% of the time. For this to be true, the F1 must have been Dg/dG, making the recombinant classes (DG and dg) less frequent. Looking at the choices: A) Dg/dG correctly explains why Dg and dG gametes are most abundant. B) DG/dg would make DG and dg the most frequent classes, not Dg and dG. C) DD/gg is impossible from the original cross since both parents carry recessive alleles. D) Dd/Gg uses incorrect notation for linked genes. Remember: in linkage problems, identify which gamete types produce the most abundant offspring classes, then work backward to determine the parental genotype arrangement.

Question 11

1. The gametes that produce Class R progeny result from meiosis with no crossover between the two genes.
2. The dihybrid parent must have its alleles in coupling phase (e.g., AB/ab).
3. The gametes that produce Class P progeny result from meiosis where no crossover occurred between the two genes. (correct answer)
4. Meiosis in the dihybrid parent produces Class P and Class R gametes in equal proportions.

Explanation: Class P represents the parental progeny classes, which are most frequent (totaling 42%+42%=84%). Class R represents the recombinant progeny, which are least frequent (totaling 8%+8%=16%). Parental progeny arise from parental gametes. Parental gametes are the direct products of meiosis where the chromosome segment containing the two genes did not undergo a crossover event. Recombinant gametes (leading to Class R) are the products of meiosis where a crossover did occur between the genes.

Question 12

1. PQR/pqr (correct answer)
2. PQr/pqR
3. PqR/pQr
4. Pqr/pQR

Explanation: The parental gametes are always the most frequent classes produced by the heterozygous parent. In this case, the most frequent progeny genotypes are pqr and PQR, which correspond to gametes pqr and PQR. Therefore, the heterozygous parent must have had one chromosome with the alleles PQR and its homolog with pqr. Its genotype is written as PQR/pqr.

Question 13

1. Parental Ditype (PD)
2. A result of failed meiosis.
3. Tetratype (T)
4. Non-Parental Ditype (NPD) (correct answer)

Explanation: When analyzing yeast crosses and asci, you need to classify the resulting spores based on how the alleles segregated during meiosis. The key is identifying whether the four spores show parental or recombinant combinations of the original traits. In this cross (his4 arg6 × + +), the parental genotypes are his4 arg6 (both mutant alleles) and + + (both wild-type alleles). The four spores produced are: his4 +, his4 +, + arg6, and + arg6. Notice that each spore type appears twice, and both represent new combinations that weren't present in either parent - these are recombinant genotypes. This makes the ascus a Non-Parental Ditype (NPD), answer D. An NPD contains only recombinant genotypes, with no spores matching either parental combination. Answer A is incorrect because a Parental Ditype (PD) would contain only the original parental combinations: his4 arg6 and + +. Answer B is wrong because this represents normal meiotic segregation - failed meiosis would produce abnormal spore numbers or non-viable gametes. Answer C is incorrect because a Tetratype (T) would contain both parental types plus two recombinant types, giving you four different genotypes rather than two types appearing twice each. Remember the pattern: PD = only parental types, NPD = only recombinant types, T = mix of both parental and recombinant types. NPDs are relatively rare and indicate that the two genes are linked and underwent a four-strand double crossover.

Question 14

1. The genes are assorting independently on non-homologous chromosomes.
2. The alleles are in repulsion phase on homologous chromosomes (Ro/rO).
3. The alleles are in coupling phase on homologous chromosomes (RO/ro). (correct answer)
4. The data are insufficient to determine linkage phase with any confidence.

Explanation: In a testcross, the phenotypes of the progeny directly reflect the gametes produced by the heterozygous parent. The most frequent progeny phenotypes are red, round (from an RO gamete) and yellow, ovate (from an ro gamete). These represent the parental classes. The less frequent phenotypes, red, ovate (Ro) and yellow, round (rO), are the recombinant classes. Since the parental gametes are RO and ro, the alleles in the dihybrid parent must be in the coupling configuration: RO/ro.

Question 15

1. PQR/pqr (correct answer)
2. PQr/pqR
3. PqR/pQr
4. Pqr/pQR

Explanation: The parental gametes are always the most frequent classes produced by the heterozygous parent. In this case, the most frequent progeny genotypes are pqr and PQR, which correspond to gametes pqr and PQR. Therefore, the heterozygous parent must have had one chromosome with the alleles PQR and its homolog with pqr. Its genotype is written as PQR/pqr.

Question 16

1. The two parental classes together represent approximately 12.5% of the progeny.
2. The two recombinant classes together represent approximately 87.5% of the progeny.
3. The genes are assorting independently, and the ratio is an anomaly.
4. The two parental classes are about 7 times more frequent than the two recombinant classes. (correct answer)

Explanation: The 7:1:1:7 ratio clearly indicates linkage, as it deviates from the 1:1:1:1 ratio expected for independent assortment. The two '7' categories are the parental classes, and the two '1' categories are the recombinant classes. The total ratio parts are 7+1+1+7=16. The parental classes make up (7+7)/16 = 14/16 of the total, while recombinant classes make up (1+1)/16 = 2/16. The ratio of parental frequency to recombinant frequency is 14/2 = 7. Therefore, parental classes are 7 times more frequent than recombinant classes. Recombination frequency is 2/16 = 12.5%.

Question 17

1. A-B-D- and aabbdd
2. A-bbD- and aaB-dd
3. A-B-dd and aabbD- (correct answer)
4. A-bbD- and aaB-D-

Explanation: First, determine the genotype of the F1 fly. The cross is ab/ab; D/D × AB/AB; d/d. The F1 will receive an ab D gamete from the first parent and an AB d gamete from the second. Thus, the F1 genotype is abD/ABd. This means genes A and B are in coupling, but D is linked to A and B in a specific arrangement. The parental gametes from this F1 are abD and ABd. When testcrossed (to ab/ab; d/d), these gametes will produce the most abundant progeny, with genotypes abD/abd and ABd/abd. The phenotypes of these progeny are aabbD- and A-B-dd, respectively.

Question 18

1. Only progeny with genotypes PqR/pqr and pQr/pqr will be observed.
2. Progeny resulting from double-crossover events will not be observed. (correct answer)
3. Parental and recombinant progeny will be observed in equal frequencies.
4. The two single-crossover classes will be absent from the progeny.

Explanation: Interference (I) is calculated as 1 minus the coefficient of coincidence (c). If I = 1.0, then c = 0. The coefficient of coincidence is the ratio of observed double crossovers (DCO) to expected DCO. If c = 0, the observed number of DCOs is zero. This means that a crossover in one region completely prevents a crossover in the adjacent region. Therefore, no progeny resulting from double-crossover events will be observed. Parental and single-crossover classes will still be produced.

Question 19

1. 8%
2. 16% (correct answer)
3. 42%
4. 84%

Explanation: The parent is in repulsion phase: dpy + / + unc. With 16% recombination, parental gametes (dpy + and + unc) each have frequency 42%, while recombinant gametes (dpy unc and + +) each have frequency 8%. Progeny with the Dumpy phenotype come from two sources: dpy + gametes (42%) giving Dumpy-wild type, and dpy unc gametes (8%) giving Dumpy-Unc. Total Dumpy progeny = 42% + 8% = 50%. Among Dumpy progeny, those that are also Unc = 8%/(42%+8%) = 8%/50% = 16%.

Question 20

1. 3% each
2. 6% each
3. 94% each
4. 47% each (correct answer)

Explanation: When you encounter testcross problems involving linked genes, remember that recombination frequency directly tells you about crossover events, and the remaining gametes represent the parental combinations. In this F1 individual with genotype $Xy/xY$, the two chromosomes carry different combinations of alleles. During meiosis, most gametes will maintain these original parental combinations ($Xy$ and $xY$), while fewer will result from crossing over between the linked genes. Since 6% of progeny show recombinant phenotypes, this means 6% of the F1's gametes were recombinant types. In a testcross, the recombinant gametes would be $XY$ and $xy$ (3% each). The remaining 94% must be the parental types: $Xy$ and $xY$. Since these two parental combinations are produced equally during meiosis, each represents $94\% ÷ 2 = 47\%$ of total gametes. Answer choice A (3% each) represents the frequency of each recombinant gamete type, not the parental types. Answer choice B (6% each) incorrectly assumes the 6% recombination frequency applies to each parental type rather than total recombinants. Answer choice C (94% each) would give you 188% total frequency, which is impossible—this represents the total frequency of all parental types combined. The correct answer is D (47% each). Study tip: In linkage problems, always remember that recombination frequency + parental frequency = 100%, and both recombinant types are equally frequent, as are both parental types.