College Biology Quiz: Non Mendelian Genetics
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Non Mendelian GeneticsQuestion 1 of 18

In a cross between two individuals with blood types AB and O, what percentage of offspring would be expected to have type A blood?

0%
25%
50%
75%
100%
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College Biology Quiz

College Biology Quiz: Non Mendelian Genetics

Practice Non Mendelian Genetics in College Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Non Mendelian Genetics, giving you a quick way to practice the rules, question types, and explanations that matter most for College Biology.

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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.

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Question 1

In a cross between two individuals with blood types AB and O, what percentage of offspring would be expected to have type A blood?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 75%
  5. 100%
Explanation: This question tests your understanding of ABO blood type inheritance, which follows codominant and recessive patterns. When you see blood type genetics problems, remember that A and B alleles are codominant (both expressed when present), while O is recessive. Let's work through this cross systematically. An individual with AB blood type has genotype IAIBI^A I^B, while someone with type O blood has genotype iiii. When we cross IAIB×iiI^A I^B × ii, we can use a Punnett square to find the offspring ratios. The AB parent can contribute either IAI^A or IBI^B gametes, while the O parent can only contribute ii gametes. This gives us two possible offspring genotypes: IAiI^A i (type A blood) and IBiI^B i (type B blood), each occurring 50% of the time. Looking at the wrong answers: A) 0% incorrectly assumes no type A offspring are possible, perhaps confusing this with a different cross. B) 25% might result from incorrectly treating this as a typical dihybrid cross or misunderstanding the genotypes involved. D) 75% represents a fundamental misunderstanding of the inheritance pattern, possibly confusing dominant/recessive ratios from other genetics problems. The correct answer is C) 50%. Study tip: For ABO genetics problems, always write out the genotypes first (IAIAI^A I^A, IAIBI^A I^B, IBIBI^B I^B, IAiI^A i, IBiI^B i, iiii), then work through the Punnett square systematically. Remember that codominance means both A and B alleles are expressed equally.

Question 2

A colorblind woman (XcXc) marries a man with normal color vision (XCY). What percentage of their sons will have normal color vision?

  1. 0% (correct answer)
  2. 25%
  3. 50%
  4. 75%
  5. 100%
Explanation: When you encounter genetics problems involving traits carried on sex chromosomes, you need to carefully track which parent contributes which chromosome to sons versus daughters. Color blindness is X-linked recessive, meaning the gene is located on the X chromosome and requires two copies (in females) or one copy (in males) to express the trait. Let's work through this cross systematically. The colorblind woman has genotype XcXc (both X chromosomes carry the recessive allele), while the man with normal vision has genotype XCY (his single X chromosome carries the dominant allele). For their sons, the key insight is understanding male inheritance patterns. Sons always receive their X chromosome from their mother and their Y chromosome from their father. Since this mother can only contribute Xc (she has no XC to give), and the father contributes Y, all sons will have genotype XcY. Because males only have one X chromosome, they express whatever allele is present - in this case, the recessive colorblind allele. Therefore, 0% of sons will have normal color vision, making A correct. Answer B (25%) might result from incorrectly treating this as a simple dominant/recessive cross without considering sex-linkage. Answer C (50%) could come from confusing the offspring ratios or mixing up sons with daughters. Answer D (75%) represents a fundamental misunderstanding of the inheritance pattern. Remember: for X-linked traits, sons inherit their X chromosome exclusively from their mother, so focus on what the mother can contribute when analyzing male offspring.

Question 3

In a cross between two pink-flowered snapdragons (RW × RW), where flower color shows incomplete dominance, what ratio of phenotypes is expected in the offspring?

  1. 1 red : 2 pink : 1 white (correct answer)
  2. 1 red : 1 pink : 1 white
  3. 3 red : 1 white
  4. 3 pink : 1 white
  5. All pink offspring
Explanation: When you encounter genetics problems involving incomplete dominance, remember that neither allele is completely dominant over the other, resulting in a blended phenotype in heterozygotes. In this snapdragon cross (RW × RW), you're crossing two pink-flowered plants that are heterozygotes. The R allele produces red pigment, the W allele produces white (no pigment), and the heterozygote RW appears pink due to intermediate pigment expression. Setting up a Punnett square for RW × RW:
  • Possible gametes from each parent: R and W
  • Offspring genotypes: RR, RW, WR, RW
  • Simplified ratio: 1 RR : 2 RW : 1 WW
Since this is incomplete dominance, the genotype directly determines phenotype: RR = red flowers, RW = pink flowers, WW = white flowers. Therefore, the phenotypic ratio is 1 red : 2 pink : 1 white. Answer A (1 red : 2 pink : 1 white) correctly reflects this incomplete dominance pattern. Answer B (1 red : 1 pink : 1 white) incorrectly suggests equal proportions, ignoring that heterozygotes appear twice in the cross. Answer C (3 red : 1 white) represents a typical complete dominance ratio but misses the pink phenotype entirely. Answer D (3 pink : 1 white) incorrectly treats pink as the dominant phenotype. Study tip: For incomplete dominance problems, always remember that the heterozygote has a unique, intermediate phenotype, and the phenotypic ratio matches the genotypic ratio from a standard monohybrid cross: 1:2:1.

Question 4

In epistasis, gene A masks the expression of gene B. If an individual is homozygous recessive for gene A (aa) but heterozygous for gene B (Bb), what can be concluded about the phenotype?

  1. The phenotype will show the dominant B trait because B is heterozygous
  2. The phenotype will show the recessive b trait because a is homozygous recessive
  3. The phenotype will be determined by gene A regardless of gene B genotype
  4. The phenotype will show a blended combination of both A and B traits
  5. The phenotype cannot be determined without knowing the specific type of epistasis (correct answer)
Explanation: When you encounter epistasis problems, remember that this is a gene interaction where one gene can mask or modify the expression of another gene. The key is understanding which gene is doing the masking and under what conditions. In this scenario, gene A is epistatic to gene B, meaning gene A controls whether gene B can be expressed at all. Since the individual is homozygous recessive for gene A (aa), gene A cannot mask gene B's expression. This means gene B is free to express its phenotype. With a Bb genotype, gene B will show the dominant B trait because the individual has at least one dominant B allele. Looking at the wrong answers: Choice A correctly identifies that the dominant B trait will appear but gives an incomplete explanation about why. Choice B incorrectly assumes that being homozygous recessive for the epistatic gene A somehow forces expression of the recessive b trait - this misunderstands how epistasis works. Choice C is backwards; it would be true if the individual had at least one dominant A allele, but with aa, gene A cannot exert its epistatic effect. Choice D suggests blending, which isn't how epistasis typically works - genes either mask each other or they don't. The correct answer is A - the phenotype shows the dominant B trait. When the epistatic gene is homozygous recessive, it loses its masking ability, allowing the other gene to express normally according to dominance patterns. Study tip: In epistasis problems, always identify which gene is epistatic and determine whether it's "turned on" (able to mask) based on its genotype.

Question 5

A woman who is a carrier for hemophilia (XHXh) has children with a normal man (XHY). What is the probability that their second son will have hemophilia?

  1. 0% because the first son's genotype affects the second son
  2. 25% because this represents one of four possible offspring
  3. 50% because sons have equal chance of inheriting either X chromosome (correct answer)
  4. 75% because three of four outcomes result in hemophilia
  5. 100% because all sons of carrier mothers have hemophilia
Explanation: When you encounter X-linked inheritance problems, focus on how sons inherit their single X chromosome exclusively from their mother, making each birth an independent event. Let's work through this cross systematically. The carrier mother (XHX^H XhX^h) can contribute either X^H (normal) or X^h (hemophilia) with equal probability. The normal father (XHX^H Y) contributes either X^H or Y. For sons, who receive the Y chromosome from dad, their phenotype depends entirely on which X chromosome they inherit from mom. The possible offspring are: X^H X^H (normal daughter), X^H X^h (carrier daughter), X^H Y (normal son), and X^h Y (hemophilic son). Each son has a 50% chance of inheriting the X^h chromosome and having hemophilia, regardless of birth order. Answer A incorrectly suggests that previous births affect future outcomes—this violates the principle of independent assortment. Each gamete formation is independent. Answer B miscalculates by considering all four possible offspring types, but the question specifically asks about sons only. When you focus just on male offspring, it's a 50-50 split. Answer D not only uses the wrong denominator but also incorrectly states that three outcomes result in hemophilia, when only one of the four total outcomes (XhX^h Y) produces a hemophilic child. For X-linked problems, remember this key principle: each birth is independent, and sons have a 50% probability of inheriting either maternal X chromosome. Don't let the phrase "second son" distract you from this fundamental genetic reality.

Question 6

In incomplete dominance, a cross between a red-flowered plant (RR) and a white-flowered plant (WW) produces offspring that are:

  1. All red-flowered because red is the dominant trait
  2. All white-flowered because white is the recessive trait
  3. Pink-flowered because both alleles are partially expressed (correct answer)
  4. Half red-flowered and half white-flowered in equal proportions
  5. Red and white spotted because both alleles are fully expressed
Explanation: When you encounter genetics problems involving incomplete dominance, remember that this inheritance pattern differs fundamentally from complete dominance. In incomplete dominance, neither allele is truly "dominant" over the other—instead, both contribute to the phenotype, creating a blended appearance. In this cross between RR (red) × WW (white), all offspring will have the genotype RW. Unlike complete dominance where one allele masks another, incomplete dominance allows both the R and W alleles to be partially expressed simultaneously. This results in pink flowers—a literal blend of the red and white parental traits. Looking at why the other answers miss the mark: Answer A incorrectly assumes complete dominance, where red would completely mask white expression. Answer B makes the same error but assumes white dominance instead. Answer D describes what you'd see in codominance (where both traits appear distinctly, like AB blood type) or reflects a misunderstanding of segregation ratios—the 50:50 split refers to gamete production, not phenotype expression in the F1 generation. The key distinction is that incomplete dominance produces a new, intermediate phenotype in heterozygotes. You're literally seeing a physical blend of the parental traits. Study tip: When you see genetics problems, first identify the inheritance pattern. Complete dominance = one trait masks another; incomplete dominance = blended phenotype; codominance = both traits visible separately. The question's wording often gives clues—"incomplete dominance" in the stem immediately tells you to expect a blended intermediate phenotype in heterozygotes.

Question 7

In codominance, what distinguishes it from incomplete dominance?

  1. Codominance occurs only in sex-linked traits while incomplete dominance occurs in autosomal traits
  2. Codominance shows both parental phenotypes simultaneously while incomplete dominance shows a blended phenotype (correct answer)
  3. Codominance requires multiple genes while incomplete dominance involves only one gene
  4. Codominance follows Mendelian ratios while incomplete dominance does not follow any predictable pattern
  5. Codominance is environmentally influenced while incomplete dominance is purely genetic in origin
Explanation: When you encounter genetics problems involving dominance patterns, focus on how the heterozygous phenotype compares to the parental phenotypes. This distinction is crucial for identifying different types of inheritance. In codominance, both parental traits are expressed simultaneously and distinctly in the heterozygote. Think of human ABO blood types: if you inherit both A and B alleles, your blood type is AB, showing both A and B antigens on your red blood cells. Both parental phenotypes appear together without blending. Incomplete dominance creates a blended phenotype in heterozygotes. When red and white flowers cross, you get pink offspring - a true blend where neither parental phenotype is visible. The heterozygote shows an intermediate trait that's different from both parents. Choice A is incorrect because both codominance and incomplete dominance can occur in autosomal or sex-linked traits. The chromosome location doesn't determine the dominance pattern. Choice C misrepresents the genetic basis. Both phenomena can involve single genes with multiple alleles or single genes with two alleles - the number of genes isn't the distinguishing factor. Choice D is wrong because both patterns follow predictable Mendelian ratios. In a heterozygous cross, you'll get the same 1:2:1 genotypic ratio regardless of dominance type. Remember this key distinction: codominance shows "both traits together" while incomplete dominance shows "traits blended." Look for keywords like "simultaneously," "both present," or "mixed together" versus "intermediate," "blended," or "in-between" to identify which pattern you're dealing with.

Question 8

Multiple alleles for a single gene are found in a population. How many alleles can be present in one diploid individual?

  1. Only one allele because diploid means single copy
  2. Exactly two alleles because diploid organisms have two copies of each gene (correct answer)
  3. Three alleles because multiple means at least three different versions
  4. All alleles present in the population because diploid means double
  5. Variable number depending on the specific gene and chromosome location
Explanation: When you encounter questions about alleles and diploid organisms, focus on the fundamental relationship between chromosome pairs and gene copies. Diploid organisms inherit one set of chromosomes from each parent, meaning they have two copies of every gene - one on each homologous chromosome. Since each gene occupies a specific location (locus) on a chromosome, and diploid individuals have two homologous chromosomes, they can only carry exactly two alleles for any single gene. These two alleles might be identical (homozygous) or different (heterozygous), but the total is always two. Let's examine why the other options miss the mark. Choice A incorrectly defines diploid as having single copies - this actually describes haploid organisms like gametes. Choice C confuses population-level diversity with individual genotype; while a population might have three, four, or more different alleles for a gene, any one diploid individual is limited by having only two chromosome copies. Choice D misunderstands what "diploid" means - it refers to having paired chromosomes, not the ability to carry unlimited alleles. The key distinction here is between population genetics and individual genetics. A population can maintain multiple alleles for a single gene (like the ABO blood system with A, B, and O alleles), creating genetic diversity. However, each diploid individual can only possess two of those alleles. Remember this pattern: population = multiple possible alleles; individual diploid organism = exactly two alleles per gene. This concept appears frequently in genetics problems involving inheritance patterns and population diversity.

Question 9

Human height is controlled by multiple genes, each contributing small additive effects. If both parents are of average height, most of their children will be:

  1. Taller than both parents due to hybrid vigor
  2. Shorter than both parents due to genetic regression
  3. Of average height with some variation around the mean (correct answer)
  4. Either very tall or very short with few intermediate heights
  5. Exactly the same height as their parents
Explanation: When you encounter questions about traits controlled by multiple genes with additive effects, you're dealing with quantitative genetics and polygenic inheritance. This type of inheritance produces continuous variation rather than discrete categories, and follows predictable patterns based on the normal distribution. Since human height involves many genes each contributing small additive effects, the offspring of two average-height parents will cluster around the parental average. The key principle here is that when parents are at the population mean for a polygenic trait, their children will also average around that mean, with most falling close to it and fewer at the extremes. This creates the classic bell-shaped distribution we see for height in human populations. Choice C correctly captures this pattern - most children will have average height with normal variation distributed around that mean. Choice A incorrectly suggests hybrid vigor, which refers to increased fitness in offspring from genetically diverse parents, but doesn't apply when both parents are already at the population average. Choice B mentions "genetic regression," but regression to the mean actually works in the opposite direction - it would bring extreme parental heights back toward average in offspring, not make average parents produce shorter children. Choice D describes a bimodal distribution with mostly extreme values, which would occur with simple dominant/recessive inheritance, not polygenic traits. Remember: polygenic traits always show continuous, normal distributions. When parents are average for such traits, expect their offspring to cluster around that same average with typical bell-curve variation.

Question 10

In mice, coat color is determined by two genes. Gene A controls pigment production (A = pigment produced, a = no pigment/albino). Gene B controls pigment distribution (B = black, b = brown). The albino phenotype (aa) masks the expression of gene B.

Based on the information in the passage, a cross between two mice with genotype AaBb would produce what ratio of phenotypes in the offspring?

  1. 9 black : 3 brown : 4 albino (correct answer)
  2. 9 black : 3 brown : 3 albino : 1 white
  3. 12 pigmented : 4 albino
  4. 3 black : 1 brown
  5. 1 black : 1 brown : 2 albino
Explanation: When you encounter genetics problems involving two traits, you're dealing with a dihybrid cross that follows Mendel's principles. However, this question adds a twist: epistasis, where one gene masks the expression of another. Let's work through the AaBb × AaBb cross systematically. First, determine the genotypic ratio using a Punnett square or the multiplication rule. You'll get the classic 9:3:3:1 ratio for genotypes: 9 A_B_, 3 A_bb, 3 aaB_, and 1 aabb. Now apply the epistatic effect. Since gene A controls whether any pigment is produced, the aa genotype (albino) masks gene B entirely. This means both aaB_ and aabb individuals will be albino regardless of their B gene. Converting genotypes to phenotypes: 9 A_B_ become black, 3 A_bb become brown, and 3 aaB_ + 1 aabb = 4 albino. This gives us 9 black : 3 brown : 4 albino. Choice B (9:3:3:1) represents the basic dihybrid ratio without considering epistasis—it treats all four genotypic classes as distinct phenotypes. Choice C (12:4) only considers whether pigment is present or absent, ignoring the black versus brown distinction entirely. Choice D (3:1) appears to only account for the pigmented offspring while completely overlooking the albino individuals. Remember that epistasis problems require you to think beyond simple dominant/recessive relationships. Always identify which gene is epistatic (masks others) and apply that masking effect after determining the basic genotypic ratios.

Question 11

A gene that affects multiple, seemingly unrelated traits is demonstrating:

  1. Polygenic inheritance because multiple genes are involved
  2. Codominance because multiple traits are expressed simultaneously
  3. Pleiotropy because one gene influences multiple phenotypes (correct answer)
  4. Epistasis because genes are interacting with each other
  5. Incomplete dominance because traits show intermediate phenotypes
Explanation: When you encounter questions about genes affecting multiple traits, you're dealing with fundamental concepts of how genes influence phenotypes. The key is distinguishing between scenarios where multiple genes work together versus one gene having widespread effects. The correct answer is C because pleiotropy describes exactly this situation: a single gene that influences multiple, apparently unrelated phenotypic traits. This occurs because one gene product (like an enzyme or regulatory protein) can participate in several different biological pathways or developmental processes. For example, the gene responsible for Marfan syndrome affects connective tissue throughout the body, causing seemingly unrelated symptoms in the skeletal system, cardiovascular system, and eyes. Option A is incorrect because polygenic inheritance involves multiple genes working together to influence a single trait, like height or skin color—the opposite of what's described. Option B misrepresents codominance, which occurs when two different alleles of the same gene are both fully expressed simultaneously in a heterozygote, such as AB blood type. This doesn't involve multiple traits from one gene. Option D describes epistasis, where one gene masks or modifies the expression of another gene, requiring interaction between different genes rather than one gene affecting multiple traits. Remember this distinction: "pleio" means "many" or "multiple," so pleiotropy = one gene, many effects. If you see "one gene affects multiple traits" on biology exams, think pleiotropy first. Conversely, "multiple genes affect one trait" signals polygenic inheritance.

Question 12

A trait controlled by multiple genes where each gene contributes additively to the phenotype is called:

  1. Codominance because multiple genes are dominant simultaneously
  2. Incomplete dominance because the phenotype is intermediate between extremes
  3. Polygenic inheritance because multiple genes influence one trait (correct answer)
  4. Pleiotropy because one gene affects multiple traits
  5. Epistasis because genes interact to modify each other's expression
Explanation: When you encounter questions about traits influenced by multiple genes, focus on distinguishing between different inheritance patterns and their defining characteristics. Polygenic inheritance occurs when multiple genes each contribute small, additive effects to produce a single trait. Classic examples include height, skin color, and intelligence, where each contributing gene adds incrementally to the final phenotype. This creates a continuous distribution of traits in populations rather than discrete categories. Choice C correctly identifies this pattern. The key phrase "multiple genes...each gene contributes additively" directly describes polygenic inheritance, where the cumulative effect of several genes determines the phenotype. Choice A misunderstands codominance, which occurs when two alleles of a single gene are both fully expressed simultaneously (like AB blood type), not when multiple different genes are involved. Choice B confuses incomplete dominance, where one gene produces a blended phenotype between two parental traits (like pink flowers from red and white parents), with the additive effects of multiple genes. Choice D describes pleiotropy backwards—pleiotropy is when one gene affects multiple different traits (like Marfan syndrome affecting height, heart, and eyes), not when multiple genes affect one trait. Remember this distinction: polygenic = many genes → one trait, while pleiotropy = one gene → many traits. When you see "multiple genes" and "additive effects" together, think polygenic inheritance. These questions often test whether you can differentiate between single-gene inheritance patterns (codominance, incomplete dominance) and multi-gene patterns (polygenic inheritance, pleiotropy).

Question 13

Two genes are located 20 map units apart on the same chromosome. In a testcross, approximately what percentage of offspring would show recombinant phenotypes?

  1. 10% because crossing over occurs in only half of meioses
  2. 20% because map units directly correspond to recombination frequency (correct answer)
  3. 40% because both chromatids participate in crossing over
  4. 50% because genes assort independently when far apart
  5. 80% because parental types are more common than recombinants
Explanation: When you encounter questions about gene mapping and recombination frequency, remember that map units (also called centimorgans) were specifically designed to represent recombination percentages directly. Map units are defined based on recombination frequency: 1 map unit = 1% recombination frequency. This relationship exists because early geneticists like Thomas Hunt Morgan observed that the farther apart two genes are on a chromosome, the more likely crossing over will occur between them during meiosis. Since these two genes are 20 map units apart, you can expect 20% of the offspring to show recombinant phenotypes. Let's examine why the other options are incorrect. Choice A (10%) makes the error of assuming crossing over frequency needs to be divided by two, but this misunderstands how recombination is measured - we already account for the fact that only some meioses produce crossovers in the original calculation. Choice C (40%) incorrectly suggests that having both chromatids participate doubles the recombination frequency, but recombination frequency measures the percentage of offspring showing new combinations, regardless of the mechanism. Choice D (50%) confuses this scenario with independent assortment, which only applies to genes on different chromosomes or very far apart (>50 map units) on the same chromosome. For genetics problems involving linked genes, remember this key relationship: map units = recombination frequency percentage. This direct correspondence makes calculations straightforward and is fundamental to understanding genetic mapping. When genes are closer than 50 map units apart, they're linked and will show recombination frequencies less than 50%.

Question 14

A man with type AB blood and a woman with type O blood have a child. Which of the following blood types is NOT possible for their offspring?

  1. Type A blood with genotype IA i
  2. Type B blood with genotype IB i
  3. Type AB blood with genotype IA IB (correct answer)
  4. Type A blood with genotype IA IA
  5. Both type A and type B are equally likely
Explanation: When you encounter blood type genetics problems, remember that ABO blood groups follow codominance and simple dominance patterns. The A and B alleles (IAI^A and IBI^B) are codominant with each other but both dominant over the O allele (i). Let's work through this cross systematically. The man has type AB blood (genotype IAIBI^A I^B), so he can contribute either an IAI^A or IBI^B allele. The woman has type O blood (genotype ii), so she can only contribute an i allele to any offspring. The possible offspring from this cross are:
  • IAI^A from father + i from mother = IAiI^A i (type A blood)
  • IBI^B from father + i from mother = IBiI^B i (type B blood)
Now examining each answer choice: Choice A (type A with IAiI^A i) is possible from this cross. Choice B (type B with IBiI^B i) is also possible. Choice D (type A with IAIAI^A I^A) is impossible because the mother cannot contribute an IAI^A allele—she only has i alleles. Choice C (type AB with IAIBI^A I^B) is impossible because it would require the child to inherit both IAI^A and IBI^B alleles, but the father can only pass one of these, and the mother has neither. The correct answer is C because offspring cannot have type AB blood when one parent is type O. Study tip: In blood type genetics, always check what alleles each parent can actually contribute. A type O parent (ii) can never produce type AB offspring, regardless of the other parent's genotype.

Question 15

In a population, the ABO blood group system has three alleles: IA, IB, and i. A person with type AB blood has children with a person with type O blood. If they have 4 children, what is the most likely outcome?

  1. 2 children with type A blood and 2 children with type B blood (correct answer)
  2. 1 child each of types A, B, AB, and O blood
  3. 3 children with type A blood and 1 child with type B blood
  4. All 4 children with type AB blood
  5. 4 children with type O blood
Explanation: When you encounter ABO blood group genetics problems, remember that this system involves codominance and multiple alleles. The key is determining what gametes each parent can produce and then working out the offspring probabilities. Let's work through this cross systematically. A person with type AB blood has genotype IA IB, so they can only produce gametes carrying either IA or IB (50% each). A person with type O blood has genotype ii, so they can only produce gametes carrying i. The Punnett square shows two possible offspring: IA i (type A blood) and IB i (type B blood), each with 50% probability. With 4 children, you'd statistically expect 2 type A and 2 type B children, making choice A correct. Now let's examine why the other options are wrong. Choice B suggests getting AB and O children, but this is impossible since the AB parent must contribute either IA or IB (never both together for AB, and they don't carry i for type O). Choice C proposes a 3:1 ratio, but this pattern emerges from heterozygous crosses (like Aa × Aa), not from our IA IB × ii cross. Choice D suggests all AB children, which would require both parents to contribute dominant alleles, but the type O parent can only contribute i. Study tip: For ABO genetics, always start by writing out the genotypes, then determine possible gametes. Remember that type O individuals (ii) can only pass on recessive alleles, which often simplifies these crosses significantly.

Question 16

In fruit flies, white eyes are X-linked recessive to red eyes. A white-eyed female is crossed with a red-eyed male. What percentage of the F2 generation will be white-eyed females?

  1. 0% - no white-eyed females can be produced
  2. 25% - one quarter of all F2 offspring (correct answer)
  3. 50% - half of all F2 female offspring
  4. 75% - three quarters of F2 offspring
  5. 100% - all F2 females will be white-eyed
Explanation: When you encounter X-linked inheritance problems, remember that the trait's location on the X chromosome creates different inheritance patterns for males and females. Males only need one copy of a recessive allele to express the trait, while females need two copies. Let's work through this cross systematically. The white-eyed female has genotype XwXwX^wX^w and the red-eyed male has genotype XRYX^RY. Their F1 offspring will be XRXwX^RX^w (red-eyed females) and XwYX^wY (white-eyed males). When F1 individuals mate (XRXw×XwYX^RX^w \times X^wY), we get four equally likely F2 genotypes: XRXwX^RX^w (red-eyed female), XwXwX^wX^w (white-eyed female), XRYX^RY (red-eyed male), and XwYX^wY (white-eyed male). Each represents 25% of the total F2 generation, making white-eyed females 25% of all offspring. Answer A incorrectly assumes white-eyed females cannot be produced, but F1 carrier females can pass the recessive allele to their daughters. Answer C (50%) would be correct if the question asked for the percentage of F2 females that are white-eyed, but it asks for the percentage of all F2 offspring. Answer D (75%) doesn't correspond to any logical inheritance pattern for this cross. The correct answer is B - 25% of the F2 generation will be white-eyed females. For X-linked problems, always draw out the Punnett squares for both the P and F1 crosses, and pay careful attention to whether the question asks about all offspring or just one sex.

Question 17

A man with type A blood (genotype IAIA) and a woman with type B blood (genotype IBIB) have children. All offspring have type AB blood. This is an example of:

  1. Incomplete dominance because the offspring show a blended phenotype
  2. Codominance because both parental alleles are fully expressed (correct answer)
  3. Complete dominance because one trait masks the other
  4. Multiple alleles because more than two alleles exist in the population
  5. Sex-linked inheritance because blood type is determined by X-linked genes
Explanation: When you encounter genetics problems involving ABO blood types, you're dealing with a special inheritance pattern where multiple alleles interact. The key is understanding how different alleles express themselves in the phenotype. In this cross, the man (IAI^A IAI^A) contributes only I^A alleles, while the woman (IBI^B IBI^B) contributes only I^B alleles. All offspring receive one I^A and one I^B allele, giving them I^A I^B genotype and type AB blood. Critically, these children express both A antigens and B antigens on their red blood cells simultaneously. This is codominance - both parental alleles are fully expressed without either one masking the other. Looking at the incorrect options: A) suggests incomplete dominance, but that would produce a blended phenotype (like pink flowers from red × white). Type AB blood isn't a blend - it's the simultaneous expression of both A and B traits. C) describes complete dominance, where one allele masks another, but neither I^A nor I^B is masking the other here. D) mentions multiple alleles, and while the ABO system does involve multiple alleles in the population (IAI^A, IBI^B, i), this question specifically asks about the inheritance pattern observed in the offspring, not the allele diversity in the population. Remember this pattern: codominance means "co-expression" - both alleles contribute equally to the phenotype. In genetics problems, look for situations where you can detect both parental contributions in the offspring rather than one dominating or a true blend occurring.

Question 18

In a dihybrid cross involving two traits that are linked on the same chromosome, which outcome would be observed compared to independent assortment?

  1. Higher frequency of parental combinations and lower frequency of recombinant combinations (correct answer)
  2. Lower frequency of parental combinations and higher frequency of recombinant combinations
  3. Equal frequencies of all four phenotypic classes as predicted by Mendel
  4. Complete absence of recombinant offspring in all generations
  5. Random distribution of traits with no predictable pattern whatsoever
Explanation: When you encounter questions about linked genes, remember that linkage disrupts Mendel's principle of independent assortment. Linked genes sit close together on the same chromosome and tend to be inherited together as a unit. In a typical dihybrid cross with independent assortment, you'd expect a 9:3:3:1 phenotypic ratio. However, when genes are linked, the parental combinations (the allele combinations that were present in the original parents) appear much more frequently in offspring than the recombinant combinations (new allele combinations created by crossing over). This makes option A correct: you'll observe higher frequencies of parental combinations and lower frequencies of recombinant combinations. The closer the genes are on the chromosome, the stronger this effect becomes. Option B reverses the actual pattern - linked genes don't favor recombination; they resist it. Option C describes what happens with independent assortment, where all phenotypic classes appear in predictable Mendelian ratios. This only occurs when genes are on different chromosomes or very far apart on the same chromosome. Option D suggests no recombination at all, which would only happen if genes were so tightly linked that crossing over never occurred - this is extremely rare and wouldn't apply to typical dihybrid crosses. Remember this key pattern: linkage always favors parental combinations over recombinants. The strength of linkage is actually measured by how much the ratio deviates from independent assortment - the greater the deviation toward parental types, the tighter the linkage.