Mendelian Genetics

Mendelian Genetics is topic 5.3 of AP Biology, inside Heredity. This page works through three real practice questions on it, with the full reasoning behind each credited answer.

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22 questionsStatistical Tests and Data Analysis73% with a figure
Worked examples

Three real mendelian genetics questions

From the practice pool, not the mock papers — each with the reasoning that produces the answer.

73% of them come with a figure. Reading the graph or diagram correctly is most of the work here before any content knowledge applies.

77% test a single AP skill: Statistical Tests and Data Analysis. That makes this topic unusually predictable to prepare for.

Statistical Tests and Data Analysis · with figure

A dihybrid cross is expected to produce a 9:3:3:1 phenotype ratio. The observed counts for 160 offspring are shown. What is the approximate chi-square value?

  1. A1.20; only the third class contribution
  2. B0.18; only the first class contribution approximated
  3. C3.84; a common critical value, not this statistic
  4. D4.98; the sum of all four chi-square termscorrect
Why D is correct

Expected counts are 90, 30, 30, and 10. Chi-square = 4/90 + 4/30 + 36/30 + 36/10 = approximately 4.98.

The stem asks for a numerical or quantitative judgment: "A dihybrid cross is expected to produce a 9:3:3:1 phenotype ratio. The observed counts for 160 offspring are...". The safe route is to identify the counted event, use the full denominator, and then interpret the number biologically. Here the worked reasoning is Expected counts are 90, 30, 30, and 10, so choices based on 1.20; only the third class contribution or 0.18; only the first class contribution approximated lose the required setup.

Statistical Tests and Data Analysis

A recessive disorder is caused by genotype aa. Two parents are both heterozygous carriers. What is the probability that their child will have the disorder?

  1. A1/4; one of four expected genotypes is aacorrect
  2. B1/2; two of four expected genotypes are carriers
  3. C3/4; three of four offspring are not aa
  4. D0; carriers cannot have affected children
Why A is correct

An Aa x Aa cross produces AA, Aa, Aa, and aa genotypes in a 1:2:1 ratio. The probability of aa is 1/4.

The stem asks for a numerical or quantitative judgment: "A recessive disorder is caused by genotype aa. Two parents are both heterozygous carriers. What is the probability...". The safe route is to identify the counted event, use the full denominator, and then interpret the number biologically. Here the worked reasoning is The probability of aa is 1/4, so choices based on 1/2; two of four expected genotypes are carriers or 3/4; three of four offspring are not aa lose the required setup.

Statistical Tests and Data Analysis

A monohybrid cross is expected to produce a 3:1 dominant-to-recessive phenotype ratio. In 60 offspring, 48 show the dominant phenotype and 12 show the recessive phenotype. What is the chi-square value for these data?

  1. A(48-45)^2/45 = 0.20
  2. B(12-15)^2/15 = 0.60
  3. C|48-45| + |12-15| = 6.0
  4. D0.20 + 0.60 = 0.80correct
Why D is correct

The expected counts are 45 dominant and 15 recessive. The chi-square value is (48-45)^2/45 + (12-15)^2/15 = 9/45 + 9/15 = 0.20 + 0.60 = 0.80.

The stem asks for a numerical or quantitative judgment: "A monohybrid cross is expected to produce a 3:1 dominant-to-recessive phenotype ratio. In 60 offspring, 48 show...". The safe route is to identify the counted event, use the full denominator, and then interpret the number biologically. Here the worked reasoning is The expected counts are 45 dominant and 15 recessive, so choices based on (48-45)^2/45 = 0.20 or (12-15)^2/15 = 0.60 lose the required setup.

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Last reviewed 2026-08-28. Topic and unit names follow the College Board course framework. Question counts describe the PrepScore practice bank, not the exam.
Same unit
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Work mendelian genetics until the reasoning is automatic.

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