Epistasis & Gene Interactions - Genetics

1. In a hypothetical insect, eye color is determined by a two-gene pathway where a colorless precursor is converted to a red pigment by the product of gene R, and the red pigment is converted to a purple pigment by the product of gene P. Alleles R and P are dominant and produce functional enzymes. A dihybrid insect (RrPp) is test-crossed. What is the expected phenotypic ratio in the progeny?
   Answer: 1 purple : 1 red : 2 colorless First, determine the phenotype for each genotype class. Pathway: Colorless --(R)--> Red --(P)--> Purple. So, R_P_ is purple, R_pp is red (stuck at intermediate), and rr__ is colorless (can't start the pathway). A test cross involves mating RrPp with rrpp. The expected genotypic ratio of the offspring is 1 RrPp : 1 Rrpp : 1 rrPp : 1 rrpp. Now, let's assign phenotypes to these genotypes: - RrPp: Has functional R and P enzymes -> Purple. - Rrpp: Has functional R but not P -> Red. - rrPp: Lacks functional R -> Colorless. - rrpp: Lacks functional R -> Colorless. Combining the phenotypes, we get a ratio of 1 purple : 1 red : 2 colorless.
2. Seed shape in the Shepherd's purse plant is controlled by two independently assorting genes, A and B. The presence of at least one dominant allele at either locus (A_ or B_) results in a triangular seed shape. The double homozygous recessive genotype (aabb) results in an ovoid seed shape. If a plant with genotype AABb is crossed with a plant of genotype Aabb, what fraction of the progeny is expected to have ovoid seeds?
   Answer: 0 This is a case of duplicate dominant epistasis (15:1 ratio in a standard dihybrid cross). The ovoid phenotype requires the genotype aabb. We need to determine the probability of getting this genotype from the cross AABb x Aabb. We can analyze each gene locus independently. For the A locus, the cross is AA x Aa. The possible offspring genotypes are AA and Aa. The probability of getting an aa genotype from this cross is 0. For the B locus, the cross is Bb x bb. The probability of getting a bb genotype is 1/2. To find the probability of the aabb genotype, we multiply the probabilities for each locus: P(aabb) = P(aa) * P(bb) = 0 * (1/2) = 0. Therefore, no offspring from this cross are expected to have ovoid seeds.
3. In a species of flowering plant, a biochemical pathway synthesizes petal pigment: a colorless precursor is converted to a yellow intermediate by Enzyme A (encoded by gene A), which is then converted to a red pigment by Enzyme B (encoded by gene B). Recessive alleles a and b produce non-functional enzymes. A plant of genotype AaBb is crossed to a plant with yellow petals. The resulting progeny display a phenotypic ratio of 3 red : 3 yellow : 2 colorless. What was the genotype of the yellow-flowered parent?
   Answer: Aabb The problem requires working backward from the progeny ratio to deduce the parental genotype. The phenotypes correspond to genotypes as follows: Red = A_B_, Yellow = A_bb, Colorless = aa__. The cross is AaBb x (Yellow Parent). A yellow parent must have the genotype A_bb. The specific genotype could be AAbb or Aabb. Let's test both possibilities. Case 1: Cross is AaBb x AAbb. Progeny genotypes: 1/2 AABb (Red), 1/2 AAbb (Yellow). Progeny phenotypes: 1 Red : 1 Yellow. This does not match the 3:3:2 ratio. Case 2: Cross is AaBb x Aabb. Aa x Aa -> 3/4 A_, 1/4 aa. Bb x bb -> 1/2 Bb, 1/2 bb. Progeny genotypes: - A_Bb (Red): (3/4) * (1/2) = 3/8 - A_bb (Yellow): (3/4) * (1/2) = 3/8 - aaBb (Colorless): (1/4) * (1/2) = 1/8 - aabb (Colorless): (1/4) * (1/2) = 1/8 Combining the colorless phenotypes gives 1/8 + 1/8 = 2/8. The resulting phenotypic ratio is 3 Red : 3 Yellow : 2 Colorless. This matches the observed data. Therefore, the yellow-flowered parent had the genotype Aabb.
4. A cross is performed and the F2 phenotypic ratio is approximately 13:3. This ratio is most likely the result of which type of gene interaction?
   Answer: A dominant allele at one locus and the recessive homozygous genotype at a second locus produce the same phenotype. When you encounter unusual F2 ratios like 13:3, you're dealing with gene interactions where two loci work together to produce phenotypes in ways that deviate from the classic 9:3:3:1 Mendelian ratio. To understand the 13:3 ratio, start with a standard dihybrid cross that normally gives 9:3:3:1. The 13:3 pattern emerges when certain genotype classes produce the same phenotype. Specifically, this happens when a dominant allele at one locus (let's call it A_B_) produces the same phenotype as the double recessive at the second locus combined with either genotype at the first (A_bb and aabb). When you combine these classes: 9 (A_B_) + 3 (A_bb) + 1 (aabb) = 13, leaving 3 (aaB_) as the contrasting phenotype. Answer choice D correctly describes this interaction: a dominant allele at one locus and the recessive homozygous genotype at a second locus produce the same phenotype. Choice A describes complementary gene action, which typically gives a 9:7 ratio. Choice B describes epistasis where one dominant allele masks another, usually producing a 12:3:1 ratio. Choice C describes recessive epistasis, which creates a 9:3:4 ratio when the recessive homozygote masks expression at the second locus. Study tip: Memorize the common epistatic ratios and their corresponding interactions: 9:7 (complementary), 12:3:1 (dominant epistasis), 9:3:4 (recessive epistasis), and 13:3 (duplicate interaction). When you see an unusual F2 ratio, immediately think gene interaction and match the ratio to the mechanism.
5. In a certain plant, height is controlled by two genes, T and H. The genotype tt is epistatic and results in a dwarf phenotype regardless of the H locus. For plants with at least one T allele, the H locus determines height, with H_ resulting in tall plants and hh resulting in medium plants. A cross between two plants of genotype TtHh is made. What is the probability that an offspring will be the same height as its parents?
   Answer: 9/16 When you encounter epistasis problems, remember that one gene can mask the expression of another gene. Here, the tt genotype creates a dwarf phenotype regardless of what happens at the H locus, while plants with at least one T allele follow normal height determination based on their H genotype. Since both parents are TtHh (and we need to determine their height first), they have at least one T allele, so the H locus determines their height. With Hh genotypes, both parents are tall (since H is dominant over h). For a TtHh × TtHh cross, you need to find all offspring that are also tall. Using a 16-square Punnett square, the genotype frequencies are: - T_H_ (tall): 9/16 - T_hh (medium): 3/16 - ttH_ (dwarf): 3/16 - tthh (dwarf): 1/16 Only the T_H_ offspring (9/16) will be tall like their parents. Answer choice (A) 3/4 represents the probability of having at least one T allele, ignoring the H locus entirely. Answer choice (B) 1/4 is the probability of being hh at the H locus, which would give you medium height, not tall. Answer choice (C) 3/16 represents either the medium (T_hh) or dwarf (ttH_) categories individually. The correct answer is (D) 9/16. Study tip: In epistasis problems, always determine the parents' phenotypes first, then systematically work through all possible offspring genotypes. Don't forget that epistatic genes override other gene effects completely.
6. In summer squash, fruit color is determined by two genes. A dominant allele W at one locus results in white fruit, regardless of the alleles at the second locus (Y/y). In the absence of a dominant W allele (ww), the second locus determines color, with Y_ resulting in yellow fruit and yy resulting in green fruit. A cross is performed between a plant of genotype WwYy and a plant of genotype wwYy. What is the expected phenotypic ratio of the offspring?
   Answer: 4 white : 3 yellow : 1 green This is a case of dominant epistasis, but the cross is not a standard dihybrid self-cross. We must analyze the specific cross WwYy x wwYy. First, analyze the W locus: Ww x ww -> 1/2 Ww (white), 1/2 ww (not white). Next, analyze the Y locus: Yy x Yy -> 3/4 Y_ (yellow potential), 1/4 yy (green potential). Now, combine the probabilities for the phenotypes: - White fruit: Any offspring with a W allele will be white. The probability of this is P(Ww) = 1/2. So, 1/2 or 4/8 of the offspring are white. - Yellow fruit: These must have the genotype wwY_. P(ww) = 1/2 and P(Y_) = 3/4. The combined probability is (1/2) * (3/4) = 3/8. - Green fruit: These must have the genotype wwyy. P(ww) = 1/2 and P(yy) = 1/4. The combined probability is (1/2) * (1/4) = 1/8. The resulting phenotypic ratio is 4/8 white : 3/8 yellow : 1/8 green, which simplifies to 4:3:1.
7. Feather color in a certain chicken breed is controlled by two genes. A dominant allele I is an inhibitor of color expression, resulting in white feathers. In the absence of inhibition (ii), the C locus determines color, with C_ resulting in colored feathers and cc in white feathers. A true-breeding colored bird is crossed with a true-breeding white bird of genotype IICC. All F1 birds are white. If these F1 birds are test-crossed, what is the expected phenotypic ratio in the progeny?
   Answer: 1 colored : 3 white First, determine the genotypes of the parental and F1 generations. The true-breeding colored bird must have functional color (C_) and no inhibitor (ii), so its genotype is iiCC. The true-breeding white bird is given as IICC. The cross is iiCC x IICC, producing an F1 generation with the genotype IiCc. These F1 birds are all white because the dominant I allele inhibits color expression. A test cross involves mating the F1 individual (IiCc) to a homozygous recessive individual, which in this case is iicc. The progeny of the cross IiCc x iicc will have four equally likely genotypes: IiCc, Iicc, iiCc, and iicc. Next, we determine the phenotype for each genotype: - IiCc: Has the dominant inhibitor I. Phenotype is white. - Iicc: Has the dominant inhibitor I. Phenotype is white. - iiCc: Lacks the inhibitor (ii) and has the color allele (C_). Phenotype is colored. - iicc: Lacks the inhibitor (ii) but is homozygous recessive for color (cc). Phenotype is white. Summing the phenotypes, we find 1 colored progeny (iiCc) and 3 white progeny (IiCc, Iicc, iicc). Therefore, the expected phenotypic ratio is 1 colored : 3 white.
8. In the onion, two independently assorting genes are involved in bulb color. Gene C is required for any color to be produced; cc bulbs are white. At the second locus, the dominant allele R produces a red bulb, while the recessive allele r produces a yellow bulb. A pure-line red onion is crossed with a white onion, and the F1 are all red. The F1 are selfed to produce the F2. What proportion of the white F2 onions are expected to be homozygous for the red allele (RR)?
   Answer: 1/4 When you encounter genetics problems involving multiple genes affecting one trait, you're dealing with gene interaction. Here, gene C acts as an epistatic gene—it controls whether any color can be expressed at all, while gene R determines which color appears. Let's work through this cross systematically. The pure-line red onion must be CCRR (needs C for color, RR for red). The white onion could be ccRR, ccRr, or *ccrr*—since it's white, it must be cc, but we need to determine its R genotype. Since all F1 are red, they must be CcR_ (have at least one C and one R). For this to happen from a CCRR × cc?? cross, the white parent must contribute an R allele, making it ccRR. So the cross is CCRR × ccRR, producing F1 that are all CcRR. When F1 (CcRR) self-fertilize, the F2 will be: $\frac{1}{4}CCRR : \frac{1}{2}CcRR : \frac{1}{4}ccRR$ The white onions are those with cc genotype—that's $\frac{1}{4}$ of all F2. All of these white onions have genotype ccRR, meaning they're all homozygous for the red allele. Answer choice A (0) incorrectly assumes white onions can't carry color alleles. Choice B (1/3) and choice C (1/7) likely come from incorrectly assuming the white parent was ccrr and miscalculating the resulting ratios. Study tip: In epistasis problems, always determine the complete genotypes of the parents first by working backwards from the F1 phenotypes—this prevents calculation errors in later generations.
9. A mutation in the purple gene (pr) in Drosophila results in purple eyes instead of wild-type red. A separate, unlinked mutation in the suppressor of purple gene (su(pr)) restores red eyes in flies homozygous for pr. This suppressor allele is recessive. A cross is made between a purple-eyed fly (pr/pr ; su(pr)+/su(pr)) and a red-eyed fly (pr+/pr ; su(pr)/su(pr)). What proportion of the offspring is expected to have purple eyes?
   Answer: 1/4 Purple eyes require the genotype pr/pr combined with at least one dominant suppressor allele (su(pr)+/_). The cross is pr/pr ; su(pr)+/su(pr) × pr+/pr ; su(pr)/su(pr). Analyzing each locus independently: For the pr locus, pr/pr × pr+/pr gives 1/2 pr/pr offspring. For the su(pr) locus, su(pr)+/su(pr) × su(pr)/su(pr) gives 1/2 su(pr)+/su(pr) and 1/2 su(pr)/su(pr) offspring. Purple eyes require both pr/pr AND su(pr)+/_ (not suppressed). The probability is P(pr/pr) × P(su(pr)+/_) = (1/2) × (1/2) = 1/4.
10. A recessive allele a is lethal in the embryonic stage when homozygous. A second, unlinked gene B determines tail length, with B_ for long tail and bb for short tail. A cross is made between two long-tailed parents with the genotype AaBb. What is the phenotypic ratio of tail length among the live-born progeny?
    Answer: 3 long : 1 short First, consider the standard dihybrid cross AaBb x AaBb, which yields a 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb genotypic ratio. The problem states that the aa genotype is lethal. This means all progeny with genotypes aaB_ and aabb will not survive. The surviving genotypes are A_B_ and A_bb. The ratio of these two classes in the initial cross is 9:3. This ratio simplifies to 3:1. The phenotype for A_B_ is long tail (since B_ is present and the organism is alive), and the phenotype for A_bb is short tail. Therefore, the phenotypic ratio among the surviving progeny is 3 long tail : 1 short tail. The lethality at the A locus removes 1/4 of the total offspring, but it does not alter the 3:1 segregation ratio at the unlinked B locus among the survivors.
11. In foxgloves, two genes interact to determine flower color. The first gene controls pigment production (D_ = pigment, dd = no pigment/albino). The second gene controls pigment deposition (W_ = purple, ww = pink). A dihybrid plant with purple flowers is crossed with an albino plant of genotype ddWw. What fraction of the offspring is expected to have pink flowers?
    Answer: 1/8 First, identify the genotypes of the parents. The dihybrid plant with purple flowers has the genotype DdWw. The albino plant is given as ddWw. The cross is DdWw x ddWw. The pink flower phenotype requires pigment production (D_) and the pink deposition allele (ww). So, we need to find the probability of the genotype D_ww. Let's analyze the loci from the cross DdWw x ddWw separately: - For the D locus: Dd x dd -> 1/2 Dd (D_), 1/2 dd. So, P(D_) = 1/2. - For the W locus: Ww x Ww -> 1/4 WW, 1/2 Ww, 1/4 ww. So, P(ww) = 1/4. To find the probability of pink offspring (D_ww), we multiply the probabilities of the required genotypes at each locus: P(D_ww) = P(D_) * P(ww) = (1/2) * (1/4) = 1/8.
12. In corn, the synthesis of the purple pigment anthocyanin requires the dominant alleles of two genes, A and B. Any other genotype results in a colorless (white) kernel. A plant with genotype AaBb is test-crossed. The kernels on the resulting ear of corn are counted. Out of 200 total kernels, how many are expected to be purple?
    Answer: 50 This is a case of complementary gene action (duplicate recessive epistasis), where the purple phenotype requires at least one dominant allele at both loci (A_B_). A test cross involves mating the dihybrid AaBb with the homozygous recessive parent aabb. Let's determine the genotypes of the progeny: 1/4 AaBb, 1/4 Aabb, 1/4 aaBb, 1/4 aabb. Now, let's assign the phenotypes: - AaBb: Has both dominant alleles, so it is purple. - Aabb: Lacks a dominant B allele, so it is white. - aaBb: Lacks a dominant A allele, so it is white. - aabb: Lacks both dominant alleles, so it is white. Only 1/4 of the progeny are expected to be purple. To find the expected number out of 200 kernels, we calculate (1/4) * 200 = 50. Therefore, 50 kernels are expected to be purple.
13. Epistasis is distinct from dominance because:
    Answer: epistasis describes interactions between alleles at different loci, whereas dominance describes interactions between alleles at the same locus. The core distinction between epistasis and dominance lies in the location of the interacting alleles. Dominance refers to the relationship between two alleles of the same gene (at the same locus), where one dominant allele masks the expression of a recessive allele. Epistasis refers to a phenomenon where an allele of one gene masks the expression of alleles of a different gene (at a different locus). Choice B accurately captures this fundamental difference. Choice A is incorrect; pleiotropy is when one gene affects multiple traits. Choice C is incorrect as dominance is a concept for diploid organisms with pairs of alleles. Choice D is misleading; dominance is the reason for the 3:1 ratio, and epistasis modifies this and other ratios, but the statement doesn't correctly define the distinction.
14. Two true-breeding strains of a plant with white flowers are crossed, and all F1 offspring have purple flowers. When the F1 is self-pollinated, the F2 generation exhibits a ratio of 9 purple : 7 white. If a purple F2 plant is chosen at random and self-pollinated, what is the probability that its offspring will all have purple flowers?
    Answer: 1/9 The 9:7 ratio indicates complementary gene action (duplicate recessive epistasis). Let the purple phenotype be A_B_. The white phenotypes are A_bb, aaB_, and aabb. The F1 must be AaBb. The purple F2 plants have the genotype A_B_. We want the probability that a randomly selected purple F2 plant, when selfed, produces only purple offspring. This is only possible if the plant is homozygous dominant for both genes, i.e., AABB. We need to find the proportion of AABB genotypes among the purple F2 plants. The genotypes of the purple F2 plants are: 1 AABB, 2 AABb, 2 AaBB, and 4 AaBb. There are a total of 1+2+2+4 = 9 parts representing the purple phenotype. Out of these 9 parts, only 1 part is the AABB genotype. Therefore, the probability of selecting an AABB plant from the purple F2 population is 1/9.
15. In wheat, kernel color is controlled by two genes (A and B) with duplicate dominant epistasis, where a dominant allele at either locus results in a red kernel. The intensity of the color is additive, with each dominant allele contributing equally to the redness, on top of a baseline white color for the aabb genotype. A cross between a dark red plant (AABB) and a white plant (aabb) produces an F1 generation with an intermediate red color. If this F1 is self-crossed, what fraction of the F2 offspring will be phenotypically identical to the F1 parent?
    Answer: 6/16 This question combines duplicate dominant epistasis with a quantitative (additive) component. The F1 parent from the cross AABB x aabb has the genotype AaBb. Its phenotype, 'intermediate red', is due to the presence of two dominant alleles. The question asks for the fraction of F2 offspring that will also have exactly two dominant alleles. The F2 generation is produced by self-crossing the F1 (AaBb x AaBb). We need to find all the genotypes in the F2 that have exactly two dominant alleles: - AAbb: P(AA) = 1/4, P(bb) = 1/4. Total P = (1/4)*(1/4) = 1/16. - aaBB: P(aa) = 1/4, P(BB) = 1/4. Total P = (1/4)*(1/4) = 1/16. - AaBb: P(Aa) = 1/2, P(Bb) = 1/2. Total P = (1/2)*(1/2) = 4/16. The total fraction of offspring with two dominant alleles is the sum of the probabilities of these genotypes: 1/16 (AAbb) + 1/16 (aaBB) + 4/16 (AaBb) = 6/16. This simplifies to 3/8. Choice C is 6/16.
16. In Labrador retrievers, coat color is determined by two genes. The B locus determines pigment type (B_ for black, bb for brown). The E locus controls pigment deposition (E_ allows deposition, ee prevents deposition, resulting in a yellow coat regardless of the B locus genotype). A black lab (B_E_) and a yellow lab (__ee) are mated and produce a litter containing black, brown, and yellow puppies. Which of the following must be the genotypes of the parents?
    Answer: BbEe (black) and Bb ee (yellow) To solve this, we must determine what parental genotypes could produce all three phenotypes: black (B_E_), brown (bbE_), and yellow (__ee). The presence of brown puppies (bbE_) means both parents must carry a b allele. The presence of yellow puppies (__ee) means both parents must carry an e allele. The black parent has genotype B_E_. Since it must carry b and e, its genotype must be BbEe. The yellow parent has genotype __ee. Since it must carry a b allele to produce brown offspring, its genotype must be Bbee or bbee. Let's test the given options. Option B is BbEe x Bbee. The offspring from this cross include BbEe (black), bbEe (brown), and B_ee (yellow). This cross can produce all three phenotypes and fits the parental phenotypes. Option A (BBEe x bbee) cannot produce brown puppies. Option C (BbEE x bbee) cannot produce yellow puppies. Option D (BbEe x bbee) can produce all three phenotypes, but the yellow parent genotype bbee is not one of the choices given for that cross. The only choice that fits all constraints is B.
17. Based on the complementation data provided in the table, what is the minimum number of distinct genes responsible for the Trp- phenotype among these five strains?
    Answer: 3 A complementation test determines if two mutations are in the same or different genes. If two mutant strains, when crossed, produce a wild-type phenotype (growth, in this case), they complement each other, meaning their mutations are in different genes. If they produce a mutant phenotype (no growth), they fail to complement, and their mutations are in the same gene. We need to group the strains that fail to complement. - Strain 1 fails to complement with Strain 4 (1 and 4 are in the same gene). Group 1: {1, 4}. - Strain 2 fails to complement with Strain 5 (2 and 5 are in the same gene). Group 2: {2, 5}. - Strain 3 complements with all other strains (1, 2, 4, 5). This means its mutation is in a different gene from the others. Group 3: {3}. We have identified three distinct complementation groups: {1, 4}, {2, 5}, and {3}. Each group represents a single gene. Therefore, a minimum of 3 distinct genes are responsible for the Trp- phenotype.

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