Biology by Bradford
Unit 5: Heredity. Practice questions with scoring guidelines.
130 original AP-style questions on Unit 5, written to the CED learning objectives: 99 multiple-choice, 31 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 5.1 Meiosis 18 questions
- 5.2 Meiosis and Genetic Diversity 15 questions
- 5.3 Mendelian Genetics 36 questions
- 5.4 Non-Mendelian Genetics 62 questions
- 5.5 Environmental Effects on Phenotype 13 questions
- 5.6 Chromosomal Inheritance 12 questions
In the bank for Unit 5
- 99 multiple-choice
- 31 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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In Labrador retrievers, coat color is governed by two genes that assort independently. At the first gene, black (B) is dominant to chocolate/brown (b). A second gene controls whether any dark pigment is deposited in the hairs: dogs with at least one E allele deposit pigment, while ee dogs deposit no dark pigment and are yellow regardless of their genotype at the B gene. A breeder mates two dogs that are both BbEe.
A breeder crosses a yellow Labrador with a chocolate Labrador and, among the puppies, obtains several black puppies. Which pair of parental genotypes is consistent with producing black offspring from this cross?
- yellow Bbee × chocolate bbEe
- yellow bbee × chocolate bbEe
- yellow BBee × chocolate bbee
- yellow bbee × chocolate bbEE
Red-green color blindness is an X-linked recessive trait (Xᴺ = normal vision, Xⁿ = color blindness). A boy with Klinefelter syndrome (karyotype 47,XXY) is color-blind. His mother has normal vision and is a known carrier (XᴺXⁿ), and his father has normal vision (XᴺY). Ignoring crossing over, which event best explains the boy's chromosome makeup and phenotype?
| Individual | Sex-chromosome genotype | Phenotype |
|---|---|---|
| Mother | XᴺXⁿ | Normal vision (carrier) |
| Father | XᴺY | Normal vision |
| Son | XXY (47 chromosomes) | Color-blind |
- Nondisjunction during meiosis II in the mother, so an egg received two identical Xⁿ sister chromatids.
- Nondisjunction during meiosis I in the mother, so an egg received both of her X chromosomes (Xᴺ and Xⁿ).
- A new mutation in the boy's single X chromosome after fertilization, with no nondisjunction involved.
- Nondisjunction during meiosis I in the father, so an XᴺY sperm fertilized a normal Xⁿ egg.
A diploid animal has a chromosome number of 2n = 8. One primary spermatocyte from this animal completes meiosis I and then meiosis II, producing four sperm. Which statement correctly describes the chromosome content of a cell immediately after meiosis I is complete?
- Each cell contains 4 chromosomes, each made of a single chromatid, because the sister chromatids separated.
- Each cell contains 8 chromosomes, each made of a single chromatid, because only the chromatids separated.
- Each cell contains 4 chromosomes, each still made of two sister chromatids joined at the centromere.
- Each cell contains 8 chromosomes, each of two sister chromatids, as in the parent.
A student grew pea plants of three genotypes for a height gene, where T (tall) is completely dominant to t (short), under identical conditions and measured the height of mature plants. Five plants of each genotype were measured. The table shows the mean height and standard deviation (SD).
| Genotype | Mean height / cm (± SD) |
|---|---|
| TT | 180 (±8) |
| Tt | 175 (±9) |
| tt | 60 (±6) |
In humans and many other mammals, the frequency of offspring born with trisomy (three copies of one chromosome) increases with the age of the mother. Trisomy can result when nondisjunction during meiosis produces an egg that carries an extra chromosome. Researchers propose that eggs from older females undergo nondisjunction more often than eggs from younger females. The researchers work with a laboratory strain of mouse in which 2n = 40, so a normal egg contains 20 chromosomes. They can collect mature eggs from females of any age, stain the chromosomes, and count the number of chromosomes in each egg.
The model represents one pair of homologous chromosomes in a cell that is heterozygous at two linked genes (AaBb), with A and B on the maternal chromosome and a and b on the paternal chromosome. At left, the pair is shown in prophase I with a chiasma. At right are the four gametes produced from this cell after meiosis I and II. Shading shows the parental origin of the DNA.
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- A — Black puppies are BE; a chocolate dog is always bb, so the B allele must come from the yellow parent (making it Bbee), and the chocolate parent must supply an E allele (making it bbEe), producing some BbEe black offspring. The other pairings never provide both a B and an E allele.
- A — The color-blind boy must be XⁿXⁿY; his father could contribute only Xᴺ or Y, so both Xⁿ copies came from his carrier mother, and two identical Xⁿ alleles (sister chromatids) indicate failure of chromatid separation in her meiosis II (a maternal MI error would have delivered Xᴺ and Xⁿ, and a paternal XᴺY sperm would give a non-color-blind XᴺXⁿY boy).
- C — Meiosis I separates homologous chromosomes, so each daughter cell has the haploid number (n = 4), but each chromosome still consists of two sister chromatids until they separate in meiosis II.
- Bar graph with genotype on the x-axis and mean height on the y-axis, both labelled with units, a scale filling most of the grid, all means plotted correctly and ± SD error bars shown
- A correctly scaled and labelled bar graph with the points plotted and ± 1 SD bars drawn
- TT (or Tt) labelled as tallest and tt labelled as shortest
- Highest bar identified as TT (180 cm), lowest bar identified as tt (60 cm)
Total for part (A): 2 points
- TT and Tt plants have similar mean heights, both much taller than tt plants
- Plants with at least one T allele (TT and Tt) are tall, while tt plants are much shorter
- (180 − 60) ÷ 180 × 100 ≈ 66.7%
- tt plants are about 67% shorter than TT plants
- The ± SD ranges (172 to 188 cm, and 166 to 184 cm) overlap, so the means are not significantly different
- Not significantly different, because the error bars (± SD) overlap
Total for part (B): 3 points
- Approximately 60 cm, the same as the tt plants already measured
- About the same as the tt mean height shown, since the genotype tt is the same regardless of the parental cross
- Because T is completely dominant over t, phenotype (and therefore height) depends only on genotype (whether at least one T allele is present), not on which cross produced that genotype, so all tt individuals should have a similar height
- Phenotype is determined by genotype alone under complete dominance, so tt offspring from any cross should show the same short phenotype
Total for part (C): 2 points
- Because T shows complete dominance over t, a single copy of the dominant allele in a Tt plant is sufficient to produce the same tall phenotype as two copies in a TT plant; the recessive t allele's effect is masked, so height does not differ significantly between the heterozygous and homozygous dominant genotypes
- Complete dominance means one dominant allele is enough to produce the full tall phenotype, so Tt and TT plants look alike and should not differ significantly in height
- Increase the number of plants measured for each genotype
- Grow a larger sample of each genotype under identical conditions to reduce the standard deviation of the estimate
Total for part (D): 2 points
Total for question 4: 9 points
- If nondisjunction becomes more frequent as females age, then a higher percentage of the eggs collected from older females will contain an abnormal number of chromosomes (19 or 21 rather than 20) than the eggs collected from younger females.
- The percentage of eggs with a chromosome number other than 20 will increase with the age of the female from which the eggs were collected.
Total for part (A): 1 point
- Independent variable = the age of the female mouse; dependent variable = the percentage of her eggs that contain an abnormal number of chromosomes (not 20); constants (any two) = the mouse strain (all females genetically similar), diet and housing conditions, the number of eggs scored per female, the staining and counting method, and having the same person count chromosomes without knowing the age of the female.
- IV is female age, DV is the proportion of eggs with 19 or 21 chromosomes, and the females in every age group must come from the same strain and be kept under the same conditions, with eggs collected, stained and counted in the same way.
Total for part (B): 1 point
- Use at least ten females in each of several age groups (for example 3, 6, 12 and 18 months); collect a fixed number of mature eggs (for example 50) from each female; stain the chromosomes and count the number in every egg; classify each egg as normal (20) or abnormal (19, 21 or other); calculate the percentage of abnormal eggs for each female and the mean (±SE) for each age group; compare the means across age groups.
- Collect equal numbers of eggs from young, middle-aged and old females of the same strain kept under identical conditions, count the chromosomes in each stained egg, and record the percentage of eggs per female with an abnormal chromosome number; compare the mean percentage among the age groups.
Total for part (C): 1 point
- Support: the mean percentage of eggs with an abnormal chromosome number rises with age, for example a low percentage in 3-month-old females and a several-fold higher percentage in 18-month-old females, with both 19- and 21-chromosome eggs found (no increase with age would refute the hypothesis). During nondisjunction, a pair of homologous chromosomes fails to separate at anaphase I (or sister chromatids fail to separate at anaphase II), so one egg receives both copies of that chromosome (21) while another receives none (19); when a 21-chromosome egg is fertilized by a normal sperm carrying 20, the zygote has three copies of that chromosome (2n + 1 = 41), which is trisomy.
- If the hypothesis is correct, older females produce a greater proportion of eggs with 19 or 21 chromosomes than younger females; an egg with 21 chromosomes arises when homologs (meiosis I) or sister chromatids (meiosis II) both move to the pole that becomes the egg, and fertilization of that egg by a normal haploid sperm gives an embryo with an extra copy of one chromosome.
Total for part (D): 1 point
Total for question 5: 4 points
- Gametes 2 and 3; their chromosomes are part maternal and part paternal in shading, and each carries an allele combination (Ab or aB) not present on either parental chromosome.
- Gametes 2 and 3, because their chromatids switch shading at the crossover point and carry Ab and aB.
Total for part (A): 1 point
- Non-sister chromatids of the homologous pair break at the same point and exchange segments, so alleles from the maternal and paternal chromosomes end up on the same chromatid.
- Crossing over between non-sister chromatids swaps the segments below the chiasma, so A becomes linked to b and a to B.
Total for part (B): 1 point
- The proportion of recombinant gametes would decrease.
- Fewer recombinant gametes; most would be parental (AB or ab).
Total for part (C): 1 point
- A crossover only separates the alleles if it occurs between the two genes; with less distance between them there is less chance that a chiasma forms in that interval, so fewer chromatids are recombinant.
- Recombination frequency depends on the chance of a crossover between the loci, which falls as the distance between them shrinks.
Total for part (D): 1 point
Total for question 6: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 96
- Unit 2 Cell Structure and Function 105
- Unit 3 Cellular Energetics 112
- Unit 4 Cell Communication and Cell Cycle 104
- Unit 6 Gene Expression and Regulation 116
- Unit 7 Natural Selection 149
- Unit 8 Ecology 131
All eight units → · IB Biology subtopics → · Open the exam maker →