Free practice final
Genetics practice final (free, with answers)
A typical genetics final covers Mendelian inheritance and probability, gene interaction, pedigree analysis, sex linkage, linkage mapping, chromosome changes, population genetics, and gene regulation. Many of these 15 questions ask you to calculate, so write out each step on paper before you open the answer. If your result does not match, use the explanation to find the step where it went wrong.
Topics
- Mendelian Inheritance
- Gene Interaction
- Pedigree Analysis
- Sex Linkage
- Linkage and Mapping
- Chromosome Changes
- Population Genetics
- Gene Regulation
- Mutation
15 questions
Try each one before you open its answer.
1. In the cross AaBbCc × AaBbCc, the three genes assort independently. What fraction of the offspring is expected to have the genotype AaBbcc?
- A 1/64
- B 9/64
- C 1/16
- D 1/8
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Answer: C. 1/16
Treat each gene as its own monohybrid cross and multiply: P(Aa) × P(Bb) × P(cc) = 1/2 × 1/2 × 1/4 = 1/16. The value 9/64 is the fraction with the dominant A and B phenotypes and the recessive c phenotype (A_B_cc), which covers several genotypes, not just this one.
2. The F2 of a dihybrid pea cross is expected to show a 9:3:3:1 ratio. The counts are 98 round yellow, 30 round green, 25 wrinkled yellow and 7 wrinkled green. Calculate the chi-square value and decide whether the data fit the 9:3:3:1 hypothesis. The critical value for 3 degrees of freedom at p = 0.05 is 7.81.
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Answer: χ² ≈ 2.44, below 7.81, so the data fit a 9:3:3:1 ratio (the hypothesis is not rejected)
With 160 seeds the expected counts are 90, 30, 30 and 10, so χ² = 64/90 + 0/30 + 25/30 + 9/10 ≈ 0.71 + 0 + 0.83 + 0.90 = 2.44. Four classes give 3 degrees of freedom, and because 2.44 is less than 7.81, the deviation is small enough to be explained by chance.
3. In a flowering plant, gene C is needed to make any pigment, so cc plants have white flowers. A second gene sets the pigment color: in plants with at least one C allele, P_ gives purple flowers and pp gives red flowers. What ratio of purple to red to white is expected among the offspring of CcPp × CcPp?
- A 9 : 4 : 3
- B 9 : 3 : 4
- C 12 : 3 : 1
- D 9 : 6 : 1
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Answer: B. 9 : 3 : 4
Nine-sixteenths of the offspring are C_P_ (purple) and 3/16 are C_pp (red), while every cc plant is white whatever its P genotype, which adds 3/16 + 1/16 = 4/16. This is recessive epistasis: cc masks the P gene and merges the last two classes of the 9:3:3:1 ratio.
4. Two true-breeding lines of a normally purple-flowered plant each have white flowers because of a recessive mutation. When the two white lines are crossed with each other, all of the F1 plants have purple flowers. What does this show?
- A The two mutations are in different genes
- B The two mutations are in the same gene
- C The F1 plants show incomplete dominance
- D The two mutations are codominant alleles of one gene
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Answer: A. The two mutations are in different genes
This is a complementation test. If both mutations were in the same gene, the F1 would inherit two defective copies of it and stay white; purple F1 plants mean each line supplies a working copy of the gene the other line lacks (aaBB × AAbb gives AaBb).
5. A rare trait shows this pattern across many families: affected fathers with unaffected mothers pass the trait to all of their daughters and to none of their sons, while affected mothers pass it to about half of their children of either sex. What is the most likely mode of inheritance?
- A Autosomal dominant
- B Autosomal recessive
- C X-linked recessive
- D X-linked dominant
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Answer: D. X-linked dominant
A father gives his X chromosome to every daughter and his Y to every son, so a dominant allele on his X shows up in all of his daughters and none of his sons. Under autosomal dominance about half of his children of each sex would be affected, and an X-linked recessive allele from him would make his daughters carriers rather than affected.
6. Two unaffected parents have a child with an autosomal recessive disorder. The affected child's unaffected sister later has children with a man who is a known heterozygous carrier. What is the probability that their first child will be affected?
Show answer
Answer: 1/6
Both parents must be carriers (Aa), and because the sister is unaffected she is not aa, so she is AA or Aa in a 1:2 ratio, a 2/3 chance of being a carrier. Two carriers have an affected child with probability 1/4, so the answer is 2/3 × 1/4 = 1/6.
7. Hemophilia A is X-linked recessive. A woman with normal clotting whose father had hemophilia A has children with a man who has normal clotting. If they have a son, what is the probability that he has hemophilia A?
- A 0
- B 1/4
- C 1/2
- D 1
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Answer: C. 1/2
The woman must have inherited her father's X chromosome with the hemophilia allele, and because she clots normally she is a heterozygous carrier. A son receives his only X from his mother, so each son has a 1/2 chance of being affected; 1/4 is instead the chance that a child of unknown sex is an affected son.
8. A plant heterozygous for two linked genes was produced by crossing an AABB parent with an aabb parent. When it is testcrossed to an aabb plant, the offspring are 412 AaBb, 398 aabb, 92 Aabb and 98 aaBb. What is the map distance between genes A and B?
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Answer: 19 map units (19 cM)
The heterozygote received AB from one parent and ab from the other, so the Aabb and aaBb offspring, which came from Ab and aB gametes, are the recombinants. The recombination frequency is (92 + 98)/1000 = 0.19, and 1% recombination equals 1 map unit.
9. A plant heterozygous for three linked genes, from a cross of true-breeding ABC and abc parents, is testcrossed to an aabbcc plant. The offspring, listed by the gamete each received from the heterozygote, are: A B C 354, a b c 346, A b c 76, a B C 84, A B c 58, a b C 62, A b C 9, a B c 11. Which gene lies in the middle?
- A Gene A
- B Gene B
- C Gene C
- D It cannot be determined without the map distances
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Answer: B. Gene B
The two rarest classes, A b C and a B c, are the double crossovers. Compared with the parental combinations A B C and a b c, they differ only at B, and a double crossover switches only the middle gene, so the order is A–B–C (about 18 cM from A to B and 14 cM from B to C).
10. One pair of homologous chromosomes fails to separate during meiosis I, and meiosis II then proceeds normally. What chromosome content do the four resulting gametes have?
- A Two n + 1 and two n − 1
- B Two normal (n), one n + 1 and one n − 1
- C Two normal (n) and two n + 1
- D All four n + 1
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Answer: A. Two n + 1 and two n − 1
Both homologs go to the same daughter cell in meiosis I, so that cell's two products each carry an extra chromosome (n + 1) and the other cell's two products each lack one (n − 1). Two normal gametes plus one n + 1 and one n − 1 is what nondisjunction in meiosis II produces instead.
11. How many Barr bodies would you expect in the somatic cells of a person with the karyotype 47,XXY (Klinefelter syndrome)?
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Answer: One
X inactivation leaves only one X chromosome active in each somatic cell, so the number of Barr bodies is the number of X chromosomes minus one. A 47,XXY person has two X chromosomes and therefore one Barr body, the same as a typical 46,XX female.
12. An autosomal recessive disorder affects 1 in 2,500 newborns in a population that is in Hardy-Weinberg equilibrium for this gene. What fraction of the population are heterozygous carriers?
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Answer: About 0.039 (3.9%, roughly 1 in 26 people)
The frequency of affected people is q² = 1/2,500 = 0.0004, so q = 0.02 and p = 0.98. Carriers make up 2pq = 2 × 0.98 × 0.02 = 0.0392, about 1 in 26 people, far more than the 1 in 2,500 who are affected.
13. A large population starts mating mostly with close relatives, with no selection, mutation or migration. What is the main effect of this inbreeding over the following generations?
- A It increases the frequency of the recessive allele
- B It increases the frequency of heterozygotes
- C It keeps genotype frequencies at Hardy-Weinberg proportions
- D It increases the frequency of homozygotes without changing allele frequencies
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Answer: D. It increases the frequency of homozygotes without changing allele frequencies
Relatives often carry copies of the same ancestral allele, so their offspring are more likely to be homozygous, which moves genotype frequencies away from Hardy-Weinberg proportions. Inbreeding only rearranges alleles into genotypes, so allele frequencies stay the same unless another force, such as selection against the homozygotes it exposes, acts on them.
14. An E. coli strain has a lacI mutation that makes a repressor unable to bind the operator. The cells are grown with no glucose. How is the lac operon expressed?
- A It is not transcribed, with or without lactose
- B It is transcribed only when lactose is present
- C It is transcribed whether or not lactose is present
- D It is transcribed only after glucose is added
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Answer: C. It is transcribed whether or not lactose is present
Normally the repressor sits on the operator and blocks transcription until lactose, as allolactose, pulls it off. A repressor that cannot bind the operator never blocks RNA polymerase, so the operon is expressed constitutively, and with no glucose present the cAMP-CAP activator keeps transcription high.
15. A single-base substitution in a gene's coding region changes an mRNA codon from UGG (tryptophan) to UGA. What kind of mutation is this?
- A Missense mutation
- B Nonsense mutation
- C Silent mutation
- D Frameshift mutation
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Answer: B. Nonsense mutation
UGA is a stop codon, so translation ends early and the protein is cut short, which defines a nonsense mutation. A missense mutation would swap one amino acid for another, a silent mutation would keep the same amino acid, and a frameshift requires an insertion or deletion whose length is not a multiple of three.