IB Biology Quiz: Apply Inheritance
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Apply InheritanceQuestion 1 of 20

In pea plants, tall (T) is dominant to dwarf (t), and purple flowers (P) are dominant to white (p). A plant with genotype TtPp is self-pollinated. What is the probability that an offspring will be phenotypically tall with white flowers?

9/16
3/16
1/4
1/16
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IB Biology Quiz

IB Biology Quiz: Apply Inheritance

Practice Apply Inheritance in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Inheritance, giving you a quick way to practice the rules, question types, and explanations that matter most for IB 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 pea plants, tall (T) is dominant to dwarf (t), and purple flowers (P) are dominant to white (p). A plant with genotype TtPp is self-pollinated. What is the probability that an offspring will be phenotypically tall with white flowers?

  1. 9/16
  2. 3/16 (correct answer)
  3. 1/4
  4. 1/16
Explanation: This is a dihybrid cross. We can consider each trait independently. For height, the cross is Tt x Tt, and the probability of a tall phenotype (TT or Tt) is 3/4. For flower colour, the cross is Pp x Pp, and the probability of a white phenotype (pp) is 1/4. To find the probability of both events occurring together, we use the product rule: P(tall and white) = P(tall) × P(white) = (3/4) × (1/4) = 3/16.

Question 2

Red-green colour blindness is a human X-linked recessive trait. A female who is a carrier for colour blindness mates with a male who is colour-blind. What is the expected phenotypic ratio among their daughters only?

  1. 1 daughter with normal vision : 1 daughter who is colour-blind. (correct answer)
  2. All daughters will be carriers with normal vision.
  3. All daughters will be colour-blind.
  4. 3 daughters with normal vision : 1 daughter who is colour-blind.
Explanation: Let X^B be the allele for normal vision and X^b be the allele for colour blindness. The carrier female's genotype is X^B X^b. The colour-blind male's genotype is X^b Y. The cross is X^B X^b x X^b Y. The possible genotypes for their daughters are X^B X^b (carrier with normal vision) and X^b X^b (colour-blind). These two genotypes have an equal probability of occurring. Therefore, the expected phenotypic ratio among daughters is 1 with normal vision to 1 who is colour-blind.

Question 3

[HL] In fruit flies, grey body (G) is dominant to ebony body (g), and normal wings (W) are dominant to vestigial wings (w). A fly heterozygous for both traits is test-crossed with a fly that has an ebony body and vestigial wings. If 800 offspring are produced, what is the expected number of flies with a grey body and vestigial wings?

  1. 100
  2. 200 (correct answer)
  3. 400
  4. 450
Explanation: This is a dihybrid test cross: GgWw x ggww. The heterozygous parent (GgWw) produces four types of gametes in equal proportion (assuming independent assortment): GW, Gw, gW, and gw. The homozygous recessive parent (ggww) produces only one type of gamete: gw. The resulting offspring genotypes will be GgWw, Ggww, ggWw, and ggww, each with a probability of 1/4. The phenotype 'grey body and vestigial wings' corresponds to the genotype Ggww. Therefore, the expected frequency is 1/4. The expected number of such flies is (1/4) * 800 = 200.

Question 4

[HL] Two genes, P and Q, are linked. An individual with genotype PpQq, where alleles P and Q are on one homologous chromosome and p and q are on the other, is test-crossed. Which phenotypic classes among the offspring are evidence that crossing over occurred between the two genes?

  1. Only phenotypes corresponding to the genotypes PpQq and ppqq.
  2. Only phenotypes corresponding to the genotypes Ppqq and ppQq. (correct answer)
  3. Phenotypes corresponding to PpQq and Ppqq.
  4. All four phenotypes (P-Q-, P-qq, ppQ-, ppqq) in equal numbers.
Explanation: The parent's chromosomes have the arrangement (PQ)/(pq). Without crossing over, it produces parental gametes PQ and pq. A test cross would yield offspring with parental phenotypes (P-Q- and ppqq). Crossing over between the genes produces recombinant gametes Pq and pQ. A test cross with these gametes would yield offspring with recombinant phenotypes (P-qq and ppQ-). The presence of these recombinant phenotypes is the evidence that crossing over occurred.

Question 5

In Shorthorn cattle, the alleles for red coat (R) and white coat (W) are codominant, producing roan (RW) offspring. In snapdragons, alleles for red flowers (CRC^R) and white flowers (CWC^W) show incomplete dominance, producing pink (CRC^R CWC^W) offspring. What is the fundamental difference in the expression of the heterozygous genotypes?

  1. In cattle, both alleles are expressed distinctly, while in snapdragons, the heterozygote has a blended phenotype. (correct answer)
  2. In cattle, the heterozygote has a blended phenotype, while in snapdragons, both alleles are expressed distinctly.
  3. The F2 generation phenotypic ratio is 1:2:1 for cattle but 3:1 for snapdragons.
  4. The F2 generation phenotypic ratio is 3:1 for cattle but 1:2:1 for snapdragons.
Explanation: Codominance means both alleles are simultaneously and fully expressed in the heterozygote, resulting in a phenotype with distinct characteristics of both homozygotes (e.g., patches of red and white hair in roan cattle). Incomplete dominance means the heterozygote exhibits a phenotype that is an intermediate blend of the two homozygous phenotypes (e.g., red and white mix to make pink). The F2 phenotypic ratio for a heterozygous cross is 1:2:1 in both cases.

Question 6

A pedigree analysis for a genetic disorder reveals that the trait appears in every generation. Furthermore, all affected individuals have at least one affected parent, and affected parents can have unaffected offspring. What is the most probable mode of inheritance?

  1. Autosomal recessive
  2. Autosomal dominant (correct answer)
  3. X-linked recessive
  4. X-linked dominant
Explanation: The trait appearing in every generation strongly suggests a dominant pattern, as recessive traits can skip generations. The fact that affected individuals have an affected parent reinforces this. If it were recessive, two unaffected parents could have an affected child. The fact that affected parents can have unaffected children (e.g., Aa x aa -> aa) is consistent with a dominant trait and rules out some rare forms of inheritance. This pattern is characteristic of autosomal dominant inheritance.

Question 7

[HL] During a dihybrid test cross for two linked genes, an analysis of 800 offspring reveals 128 individuals with recombinant phenotypes. What is the calculated map distance between these two genes?

  1. 8.0 cM
  2. 16.0 cM (correct answer)
  3. 20.5 cM
  4. 84.0 cM
Explanation: Map distance is calculated from the recombination frequency. Recombination frequency = (Number of recombinant offspring / Total number of offspring) x 100. In this case, Frequency = (128 / 800) x 100. 128/800 = 0.16. 0.16 x 100 = 16%. One percent recombination frequency is equal to one map unit or centiMorgan (cM). Therefore, the map distance is 16.0 cM.

Question 8

[HL] Non-disjunction of the sex chromosomes during meiosis I in a human male results in the formation of abnormal sperm. If one of these abnormal sperm fertilizes a normal egg, which karyotype is a possible outcome for the zygote?

  1. XYY
  2. XXX
  3. XXY (correct answer)
  4. XY
Explanation: In meiosis I, homologous chromosomes separate. Non-disjunction in a male means the X and Y chromosomes fail to separate. This produces two types of abnormal sperm: one containing both X and Y (XY sperm), and one containing no sex chromosome (nullo sperm). A normal egg contains one X chromosome. Fertilization by the XY sperm results in an XXY zygote (Klinefelter syndrome). Fertilization by the nullo sperm results in an XO zygote (Turner syndrome). XYY results from non-disjunction in meiosis II.

Question 9

[HL] In a large, randomly mating population of squirrels in Hardy-Weinberg equilibrium, the frequency of the recessive allele for an albino coat is 0.2. What percentage of the population is expected to be heterozygous carriers of the albino allele?

  1. 4%
  2. 16%
  3. 32% (correct answer)
  4. 64%
Explanation: The frequency of the recessive allele (q) is 0.2. According to the Hardy-Weinberg principle, p + q = 1, where p is the frequency of the dominant allele. Therefore, p = 1 - 0.2 = 0.8. The frequency of the heterozygous genotype is given by the term 2pq. So, the frequency of heterozygotes is 2 * p * q = 2 * 0.8 * 0.2 = 2 * 0.16 = 0.32. To express this as a percentage, multiply by 100, which gives 32%.

Question 10

The fur colour of Himalayan rabbits is controlled by a gene that codes for a heat-sensitive enzyme required for melanin production. The enzyme is active only at cooler temperatures. A patch of white fur on a rabbit's back is shaved, and an ice pack is secured over the shaved area while the fur regrows. What will be the colour of the newly grown fur in that patch?

  1. Black, because the lower temperature from the ice pack allows the enzyme to be active. (correct answer)
  2. White, because the rabbit's core body temperature deactivates the enzyme.
  3. A mix of black and white, as the effect of temperature is random.
  4. White, because the fur colour genotype cannot be altered by environmental conditions.
Explanation: This is an example of gene expression being influenced by the environment. The genotype for the Himalayan pattern produces a heat-sensitive tyrosinase enzyme. The ice pack creates a local environment of low temperature on the rabbit's back. In this cold spot, the enzyme will be active, leading to the production of melanin. Consequently, the fur that regrows in this specific patch will be black, similar to the fur on the rabbit's naturally cooler extremities.

Question 11

An agricultural scientist has a pea plant that produces yellow seeds, a dominant trait. To determine if this plant is homozygous dominant (YY) or heterozygous (Yy), which cross would provide the most conclusive information in a single generation?

  1. Crossing the plant with another plant that produces yellow seeds.
  2. Self-pollinating the plant.
  3. Crossing the plant with one that is homozygous dominant (YY).
  4. Crossing the plant with one that produces green seeds (yy). (correct answer)
Explanation: This scenario requires a test cross. A test cross involves mating an individual with a dominant phenotype but unknown genotype with a homozygous recessive individual. In this case, crossing the yellow-seeded plant with a green-seeded plant (yy) is the test cross. If any offspring produce green seeds, the unknown parent must be heterozygous (Yy). If all offspring produce yellow seeds (after a large sample size), the parent is likely homozygous dominant (YY). This cross is the most direct and informative.

Question 12

In a plant species, flower colour is determined by a gene with two alleles showing incomplete dominance: R (red) and r (white), with Rr resulting in pink flowers. A plant with pink flowers is self-pollinated. If three offspring are produced from this cross, what is the probability that at least one of them will have pink flowers?

  1. 1/8
  2. 1/2
  3. 3/4
  4. 7/8 (correct answer)
Explanation: The cross is Rr x Rr, which produces offspring in the genotypic ratio 1 RR : 2 Rr : 1 rr. The corresponding phenotypic ratio is 1 red : 2 pink : 1 white. The probability of an offspring having pink flowers (Rr) is 2/4 or 1/2. The probability of an offspring not having pink flowers (being RR or rr) is 1 - 1/2 = 1/2. The probability of all three offspring not having pink flowers is (1/2) * (1/2) * (1/2) = 1/8. The probability of 'at least one' having pink flowers is the complement of 'none have pink flowers', so P(at least one pink) = 1 - P(none pink) = 1 - 1/8 = 7/8.

Question 13

[HL] In a genetic cross investigating two unlinked genes, the results are compared to the expected 9:3:3:1 ratio. A chi-squared test yields a value of 6.95. For this test, with 3 degrees of freedom, the critical value at a significance level of p=0.05 is 7.81. Which statement is a valid conclusion from this result?

  1. The null hypothesis is rejected because the calculated value is significantly high.
  2. The genes are likely linked because the deviation from the expected ratio is significant.
  3. The deviation between observed and expected results is not statistically significant. (correct answer)
  4. There is a greater than 95% probability that the deviation from the expected ratio is due to chance.
Explanation: In a chi-squared test, the null hypothesis (H₀) states that there is no significant difference between observed and expected results. The hypothesis is rejected if the calculated chi-squared value is greater than the critical value. Here, the calculated value (6.95) is less than the critical value (7.81). Therefore, we fail to reject the null hypothesis. This means the observed deviation is not statistically significant and is likely due to random chance.

Question 14

In certain mice, the allele for a yellow coat (Y) is dominant over the allele for a grey coat (y). However, the homozygous dominant genotype (YY) is lethal during embryonic development. If two yellow mice are crossed, what is the expected phenotypic ratio among their surviving offspring?

  1. 3 yellow : 1 grey
  2. 2 yellow : 1 grey (correct answer)
  3. 1 yellow : 1 grey
  4. All yellow
Explanation: Since the YY genotype is lethal, any living yellow mouse must be heterozygous (Yy). The cross is therefore Yy x Yy. The initial genotypic ratio of the zygotes is 1 YY : 2 Yy : 1 yy. However, the YY embryos do not survive. Thus, the surviving offspring are only those with genotypes Yy (yellow) and yy (grey). The ratio of these survivors is 2 Yy : 1 yy. This corresponds to a phenotypic ratio of 2 yellow : 1 grey.

Question 15

A man is a carrier for the autosomal recessive disorder cystic fibrosis. He and his partner, whose family history is unknown, have a child with cystic fibrosis. What must be true about the mother's genotype?

  1. She is homozygous dominant for the allele.
  2. She is homozygous recessive for the allele.
  3. She is heterozygous for the allele. (correct answer)
  4. Her genotype cannot be determined from the information given.
Explanation: Cystic fibrosis is an autosomal recessive disorder. A child with the disorder must have a homozygous recessive genotype (e.g., ff). This child inherited one recessive allele from each parent. We are told the father is a carrier (Ff). Therefore, the mother must also have at least one recessive allele (f) to pass to the child. Since her phenotype is not mentioned as having the disease, it is assumed she is unaffected, meaning she cannot be homozygous recessive (ff). Thus, she must be heterozygous (Ff).

Question 16

Haemophilia is an X-linked recessive disorder. A phenotypically normal woman whose father had haemophilia marries a phenotypically normal man. What is the probability that their first son will have haemophilia?

  1. 0
  2. 1/4
  3. 1/2 (correct answer)
  4. 1
Explanation: Since the woman's father had haemophilia (XhX^h Y), he must have passed his X^h allele to her. As she is phenotypically normal, her genotype is X^H X^h (a carrier). The normal man has the genotype X^H Y. The cross is X^H X^h x X^H Y. Their sons can inherit either the mother's X^H (resulting in XHX^H Y, normal) or her X^h (resulting in XhX^h Y, haemophilia). Each outcome has an equal chance. Therefore, the probability that a son born to this couple will have haemophilia is 1/2.

Question 17

Human height shows a wide range of variation and approximates a normal distribution within the population. Which genetic mechanism is the most likely explanation for this observation?

  1. A single gene with multiple alleles controlling height.
  2. Codominant alleles at a single gene locus.
  3. Polygenic inheritance where multiple genes have an additive effect. (correct answer)
  4. An X-linked gene that is expressed differently in males and females.
Explanation: Traits that show continuous variation, such as height, are typically polygenic. This means they are controlled by the cumulative, additive effects of multiple genes. Each gene contributes a small amount to the overall phenotype, leading to a wide spectrum of possible outcomes that often form a bell-shaped (normal) distribution. The other options describe patterns of discontinuous variation.

Question 18

A woman with type A blood and a man with type B blood have their first child, who has type O blood. Based on this information, what is the probability that their second child will have type B blood?

  1. 0
  2. 1/4 (correct answer)
  3. 1/2
  4. 3/4
Explanation: For a child to have type O blood (genotype ii), they must inherit an 'i' allele from each parent. Since the woman has type A blood, her genotype must be I^A i. Since the man has type B blood, his genotype must be I^B i. The cross is I^A i x I^B i. The possible offspring genotypes are I^A I^B (Type AB), I^A i (Type A), I^B i (Type B), and ii (Type O), each with a probability of 1/4. Therefore, the probability of their second child having type B blood (IBI^B i) is 1/4.

Question 19

A woman with type A blood has a child with type O blood. She claims a man with blood type AB is the child's father. Why is this claim inconsistent with genetic principles?

  1. A man with type AB blood cannot have a child with type O blood. (correct answer)
  2. A woman with type A blood cannot have a child with type O blood.
  3. A man with type AB blood can only have children with type AB blood.
  4. A child with type O blood must have at least one parent with type O blood.
Explanation: A child with type O blood has the genotype ii. This means the child must inherit one 'i' allele from each parent. A man with type AB blood has the genotype I^A I^B. He does not possess an 'i' allele to pass on to his offspring. Therefore, it is genetically impossible for him to be the father of a type O child.

Question 20

[HL] A geneticist performs a dihybrid test cross (AaBb x aabb) to determine if genes A and B are linked. Which result would provide the strongest evidence that the two genes assort independently?

  1. Offspring phenotypes are observed in an approximate 9:3:3:1 ratio.
  2. Parental phenotypes are significantly more common than recombinant phenotypes.
  3. Offspring are produced with only the two parental phenotypes.
  4. The four possible phenotypes appear in an approximate 1:1:1:1 ratio. (correct answer)
Explanation: The law of independent assortment applies to genes located on different chromosomes. For a dihybrid individual (AaBb), this leads to the formation of four different gametes (AB, Ab, aB, ab) in equal proportions. When this individual is test-crossed with a homozygous recessive individual (aabb), the phenotypic ratio of the offspring directly reflects the gametic ratio of the heterozygous parent. Therefore, observing the four possible phenotypes in a 1:1:1:1 ratio is the expected outcome for independent assortment.