TEAS: Science Quiz: Apply Genetics Principles
20 questions · exam conditions
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Apply Genetics PrinciplesQuestion 1 of 20

In a test cross, an individual with unknown genotype is crossed with a homozygous recessive individual. If 50% of offspring show the recessive phenotype, what is the unknown genotype?

Homozygous dominant
Heterozygous
Homozygous recessive
Codominant heterozygous
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TEAS: Science Quiz

TEAS: Science Quiz: Apply Genetics Principles

Practice Apply Genetics Principles in TEAS: Science 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 Apply Genetics Principles, giving you a quick way to practice the rules, question types, and explanations that matter most for TEAS: Science.

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.

All questions

Question 1

In a test cross, an individual with unknown genotype is crossed with a homozygous recessive individual. If 50% of offspring show the recessive phenotype, what is the unknown genotype?

  1. Homozygous dominant
  2. Heterozygous (correct answer)
  3. Homozygous recessive
  4. Codominant heterozygous
Explanation: In a test cross (Aa × aa), a heterozygous individual produces 50% dominant and 50% recessive offspring. This 1:1 ratio indicates the unknown parent is heterozygous. Choice A would produce 100% dominant offspring, choice C would produce 100% recessive offspring, and choice D describes codominance rather than a simple dominant-recessive relationship.

Question 2

In a cross between two heterozygous individuals (Aa × Aa), what is the probability that their first three offspring will all have the dominant phenotype?

  1. 2764\frac{27}{64} (correct answer)
  2. 916\frac{9}{16}
  3. 34\frac{3}{4}
  4. 14\frac{1}{4}
Explanation: Each offspring has a 3/4 probability of showing the dominant phenotype (AA, Aa, or Aa genotypes). For three independent events, multiply the probabilities: (3/4) × (3/4) × (3/4) = 27/64. Choice B represents the probability for two offspring, choice C represents the probability for one offspring, and choice D represents the probability of the recessive phenotype for one offspring.

Question 3

A color-blind man marries a woman who is a carrier for color blindness. Color blindness is an X-linked recessive trait. What is the probability that their son will be color-blind?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 100%
Explanation: The man is XcY (color-blind) and the woman is XCXc (carrier). Sons inherit their X chromosome from the mother and Y from the father. The mother can pass either XC (normal) or Xc (color-blind allele), each with 50% probability. Choice A is impossible since the mother carries the recessive allele, choice B would apply if considering all children, and choice D would only occur if the mother were homozygous recessive.

Question 4

Two genes are located on the same chromosome with a recombination frequency of 20%. What is the distance between these genes?

  1. 10 map units
  2. 20 map units (correct answer)
  3. 40 map units
  4. 50 map units
Explanation: Recombination frequency directly equals map units (centimorgans). A 20% recombination frequency means the genes are 20 map units apart and are linked. Choice A incorrectly halves the distance, choice C incorrectly doubles it, and choice D represents unlinked genes that would show 50% recombination due to independent assortment.

Question 5

In humans, widow's peak (W) is dominant over straight hairline (w), and free earlobes (F) are dominant over attached earlobes (f). A man with genotype WwFf marries a woman with genotype wwff. What fraction of their children will have widow's peak and attached earlobes?

  1. 18\frac{1}{8} due to independent assortment of traits
  2. 14\frac{1}{4} following dihybrid cross ratios (correct answer)
  3. 12\frac{1}{2} based on single trait dominance
  4. 34\frac{3}{4} representing combined dominant expression
Explanation: This is a test cross: WwFf × wwff. The man can produce WF, Wf, wF, or wf gametes with equal probability. Widow's peak and attached earlobes (W_ff) occurs when he contributes Wf, which happens 1/4 of the time. Choice A represents a different probability calculation, choice C represents single trait inheritance, and choice D represents an incorrect application of ratios.

Question 6

In mice, agouti coat color (A) is dominant over non-agouti (a), and the dominant allele for color (C) is required for pigment production. Mice with cc are albino regardless of other genes. What is the expected ratio from AaCc × AaCc?

  1. 9 agouti : 3 non-agouti : 4 albino showing recessive epistatic interaction (correct answer)
  2. 12 agouti : 3 non-agouti : 1 albino following dominant epistatic patterns
  3. 9 agouti : 4 non-agouti : 3 albino due to gene interaction
  4. 13 agouti : 2 non-agouti : 1 albino with modified dihybrid ratios
Explanation: This shows recessive epistasis where cc (albino) masks the A gene's expression. From 9:3:3:1, we get 9 A_C_ (agouti), 3 aaC_ (non-agouti), 3 A_cc (albino), and 1 aacc (albino). The albino classes combine to give 9:3:4. Choice B represents dominant epistasis, choice C uses incorrect groupings, and choice D doesn't follow standard epistatic patterns.

Question 7

In a certain plant, flower color is controlled by incomplete dominance. Red flowers (RR) crossed with white flowers (WW) produce pink flowers (RW). If pink flowers are self-fertilized, what is the expected phenotypic ratio in the offspring?

  1. 1 red : 2 pink : 1 white following incomplete dominance patterns (correct answer)
  2. 3 red : 1 white showing complete dominance reversion
  3. 2 red : 1 pink : 1 white due to allelic interaction
  4. 4 pink : 0 others maintaining parental phenotype
Explanation: Self-fertilization of RW × RW produces 1 RR (red) : 2 RW (pink) : 1 WW (white). In incomplete dominance, the F2 generation shows a 1:2:1 phenotypic ratio that matches the genotypic ratio. Choice B represents complete dominance, choice C uses incorrect proportions, and choice D incorrectly assumes all offspring remain pink.

Question 8

A trait shows epistasis where gene A masks the expression of gene B. If AaBb individuals are crossed, and gene A is epistatic to gene B when in the dominant form, what is the expected phenotypic ratio?

  1. 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb with standard dihybrid expression
  2. 12 A_ : 3 aaB_ : 1 aabb due to dominant epistatic masking (correct answer)
  3. 13 A_ : 3 aaB_ : 0 aabb with complete epistatic suppression
  4. 15 A_ : 1 aaB_ : 0 aabb showing recessive epistatic interaction
Explanation: When A is epistatic to B, the presence of dominant A masks B's expression. From a 9:3:3:1 dihybrid ratio, A_B_ and A_bb both show the same phenotype (masked B), giving 9+3=12. Only aaB_ shows B's expression (3), and aabb shows the double recessive (1). The other ratios represent different types of epistasis or incorrect calculations.

Question 9

In garden peas, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are dominant over green (y). A plant with round, green seeds is crossed with a plant with wrinkled, yellow seeds. All F1 offspring have round, yellow seeds. What are the genotypes of the parents?

  1. RRyy × rrYY producing uniform F1 hybrid offspring (correct answer)
  2. Rryy × rrYy showing partial dominance expression
  3. RrYy × RrYy following standard dihybrid cross patterns
  4. RRYY × rryy creating maximum heterozygote expression
Explanation: Since all F1 show both dominant traits, each parent must be homozygous for one dominant and one recessive allele. Round, green parent is RRyy; wrinkled, yellow parent is rrYY. F1 are all RrYy (round, yellow). Choice B includes heterozygous parents which would show segregation, choice C represents F1 × F1, and choice D would require different parental phenotypes.

Question 10

A scientist observes that a particular flower has petals that are bright red. This observable trait is an example of the flower's:

  1. genotype.
  2. karyotype.
  3. phenotype. (correct answer)
  4. allele.
Explanation: When you encounter genetics questions on the TEAS, you need to distinguish between what an organism has genetically versus what you can actually observe about that organism. The bright red petals you can see represent the flower's phenotype (C) — its observable, physical characteristics. Phenotype includes anything you can directly observe: color, shape, size, behavior, or any other measurable trait. Think of phenotype as the "physical expression" of genetic information. Now let's examine why the other options don't fit. Genotype (A) refers to the actual genetic makeup — the specific combination of genes or alleles an organism carries, which you cannot see just by looking at the flower. The red petals are the result of the genotype, not the genotype itself. Karyotype (B) specifically describes the number and appearance of chromosomes in a cell, typically viewed under a microscope — this has nothing to do with petal color. An allele (D) is a specific version of a gene, like the particular gene variant that codes for red pigment, but again, this is the underlying genetic unit, not the observable trait. Remember this distinction for the TEAS: if you can observe it with your senses (see, hear, measure), it's phenotype. If it requires looking at DNA, genes, or chromosomes, it's likely genotype, alleles, or karyotype. Physical traits you observe are always phenotype.

Question 11

In cats, coat color is determined by multiple alleles. Black (B) is dominant over brown (b), but both are recessive to orange (O). If a black cat (Bb) is crossed with a brown cat (bb), what percentage of offspring will be brown?

  1. 25%
  2. 50% (correct answer)
  3. 75%
  4. 100%
Explanation: This is a simple monohybrid test cross: Bb × bb produces 50% Bb (black) and 50% bb (brown) offspring. The orange allele is not involved in this cross. Choice A represents a different cross ratio, choice C represents the wrong proportion, and choice D incorrectly assumes all offspring would be brown.

Question 12

A woman with blood type AB marries a man with blood type O. What percentage of their children will have blood type A?

  1. 25%
  2. 50% (correct answer)
  3. 75%
  4. 100%
Explanation: The woman (AB) can contribute either IA or IB alleles, while the man (OO) can only contribute i alleles. The cross is IA IB × ii, producing 50% IA i (type A) and 50% IB i (type B) offspring. Choice A would be incorrect for this cross, choice C represents three-quarters which doesn't apply here, and choice D is impossible since half will have type B blood.

Question 13

In humans, the ability to taste PTC is controlled by a dominant allele (T). If 36% of a population cannot taste PTC, what is the frequency of the dominant allele in this population?

  1. 0.4 (correct answer)
  2. 0.6
  3. 0.64
  4. 0.36
Explanation: Using Hardy-Weinberg equilibrium: non-tasters have genotype tt, representing q² = 0.36, so q = 0.6. Since p + q = 1, then p = 0.4. The dominant allele frequency is 0.4. Choice B represents the recessive allele frequency, choice C represents the frequency of individuals with the dominant phenotype, and choice D represents the given frequency of the recessive phenotype.

Question 14

In a population of flowers, red (RR) = 36%, pink (RW) = 48%, and white (WW) = 16%. Is this population in Hardy-Weinberg equilibrium for this trait?

  1. Yes, the observed frequencies match Hardy-Weinberg predictions exactly (correct answer)
  2. No, there is an excess of heterozygotes indicating outbreeding
  3. No, there is a deficiency of homozygotes suggesting inbreeding
  4. Cannot determine without additional population genetic information
Explanation: If p = 0.6 and q = 0.4, then expected frequencies are p² = 36% (RR), 2pq = 48% (RW), and q² = 16% (WW), which match observed frequencies exactly. Choice B incorrectly identifies excess heterozygotes, choice C incorrectly suggests homozygote deficiency, and choice D is wrong because we have sufficient information to calculate equilibrium expectations.

Question 15

In humans, polydactyly (extra fingers) is a dominant trait. A woman with polydactyly whose mother was normal marries a normal man. What is the probability their child will have polydactyly?

  1. 0% because the trait skips generations randomly
  2. 25% due to autosomal recessive inheritance pattern
  3. 50% since the woman must be heterozygous (correct answer)
  4. 100% because polydactyly is completely dominant
Explanation: The woman has polydactyly but her mother was normal (pp), so the woman must be Pp (heterozygous). Crossing Pp × pp gives 50% Pp (polydactyly) and 50% pp (normal). Choice A incorrectly suggests skipping generations, choice B incorrectly treats it as recessive, and choice D ignores the heterozygous nature of the affected parent.

Question 16

In corn, kernel color is controlled by two genes. Purple requires dominant alleles at both loci (A_B_), while white results from any other combination. What phenotypic ratio results from AaBb × AaBb?

  1. 9 purple : 7 white due to complementary gene interaction (correct answer)
  2. 12 purple : 4 white following dominant epistatic patterns
  3. 15 purple : 1 white showing additive gene effects
  4. 3 purple : 1 white representing simple dominant inheritance
Explanation: This is complementary gene interaction where both dominant alleles are required for purple color. From the standard 9:3:3:1 ratio, only A_B_ (9/16) produces purple, while A_bb, aaB_, and aabb (3+3+1=7/16) all produce white. Choice B represents dominant epistasis, choice C represents a different interaction, and choice D represents single gene inheritance.

Question 17

Two genes are 40 map units apart on the same chromosome. In a test cross involving these genes, what percentage of offspring will show recombinant phenotypes?

  1. 20% due to crossing over frequency limitations
  2. 40% reflecting the map distance directly (correct answer)
  3. 60% representing parental type combinations
  4. 80% showing majority recombination events
Explanation: Map units directly correspond to recombination frequency. Genes 40 map units apart show 40% recombination, meaning 40% of offspring will have recombinant phenotypes and 60% will have parental phenotypes. Choice A incorrectly halves the frequency, choice C gives the parental type frequency, and choice D incorrectly doubles the recombination frequency.

Question 18

A woman who is a carrier for sickle cell anemia marries a man with sickle cell anemia. What is the probability that their child will have normal hemoglobin (not be a carrier and not have the disease)?

  1. 0% due to parental genotype combination
  2. 25% following autosomal recessive inheritance (correct answer)
  3. 50% based on carrier parent contribution
  4. 75% representing the dominant phenotype frequency
Explanation: The cross is HbAHbS × HbSHbS, producing 25% HbAHbA (normal), 50% HbAHbS (carriers), and 25% HbSHbS (sickle cell anemia). Only 25% will have completely normal hemoglobin. Choice A is incorrect because the woman can contribute the normal allele, choice C represents the carrier frequency, and choice D doesn't apply to this specific cross.

Question 19

In pea plants, the allele for tall stems (T) is dominant to the allele for short stems (t).

If two heterozygous tall pea plants (Tt) are crossed, what is the probability that their offspring will also be heterozygous?

  1. 25%
  2. 50% (correct answer)
  3. 75%
  4. 100%
Explanation: This question tests your understanding of Mendelian genetics and Punnett squares. When you see a cross between two organisms with known genotypes, you need to determine all possible offspring combinations and their probabilities. To solve this, set up a Punnett square for the cross Tt × Tt. Each parent can contribute either a T allele or a t allele to their offspring. The four possible combinations are:
  • TT (homozygous dominant): 1 out of 4 = 25%
  • Tt (heterozygous): 2 out of 4 = 50%
  • tt (homozygous recessive): 1 out of 4 = 25%
Since the question asks specifically for heterozygous offspring (Tt), the probability is 50%. Looking at the wrong answers: Choice A (25%) represents the probability of getting either TT or tt offspring individually, but not the heterozygous combination. Choice C (75%) would be the probability of getting tall offspring (TT + Tt combined), since both genotypes produce the tall phenotype due to T being dominant. Choice D (100%) incorrectly assumes all offspring will be heterozygous, which ignores the fact that homozygous combinations are also possible in this cross. For TEAS genetics questions, always draw out the Punnett square rather than trying to guess. Remember that when two heterozygotes cross, you'll consistently see a 1:2:1 genotypic ratio (25% homozygous dominant : 50% heterozygous : 25% homozygous recessive). This pattern appears frequently on the exam.

Question 20

In a certain species of mammal, brown fur (B) is dominant to white fur (b).

A homozygous dominant male (BB) is mated with a heterozygous female (Bb). What percentage of their offspring is expected to have brown fur?

  1. 25%
  2. 50%
  3. 75%
  4. 100% (correct answer)
Explanation: When you encounter genetics problems involving dominant and heterozygous traits, start by setting up a Punnett square to visualize all possible offspring combinations. Here, you're crossing a homozygous dominant male (BB) with a heterozygous female (Bb). The male can only contribute B alleles, while the female can contribute either B or b alleles. Setting up the cross: 
    B    B
B  BB   BB
b  Bb   Bb
This gives you four possible offspring: two BB (homozygous dominant) and two Bb (heterozygous). Since brown fur (B) is dominant over white fur (b), any offspring with at least one B allele will express brown fur. All four offspring have at least one B allele, so 100% will have brown fur. Choice A (25%) would be incorrect because that represents the typical ratio for recessive traits in a heterozygous × heterozygous cross. Choice B (50%) reflects the proportion that would be heterozygous, but the question asks about phenotype (fur color), not genotype. Choice C (75%) would be the brown fur percentage if both parents were heterozygous (Bb × Bb), where you'd get a 3:1 dominant to recessive ratio. The key insight is that when one parent is homozygous dominant, it's impossible for any offspring to be homozygous recessive, since they must inherit at least one dominant allele from that parent. For TEAS genetics questions, always distinguish between genotype (genetic makeup) and phenotype (observable traits), and remember that one homozygous dominant parent guarantees the dominant phenotype in all offspring.