Unit 5 · Unit 5 end-of-unit test (also the test-out)
Suggested time: about 80 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.
A cell in an animal’s testis holds 2C of DNA before it copies its DNA. It then goes through meiosis I and meiosis II, with no copying between the two divisions.
Which of the following is the amount of DNA in each cell after meiosis II?
A plant’s body cells hold 12 chromosomes. In one dividing cell, six X-shaped chromosomes stand in a single row across the equator, each still two sister chromatids. To count chromosomes, count centromeres.
Which stage is the cell in?
A student looks at a homologous pair in a cell in prophase I: two X-shaped chromosomes of the same length lying side by side. The student says: “These two X’s are the two sister chromatids of one chromosome.” The student is wrong.
Which of the following is the reason the student is wrong?
A plant’s body cells hold 18 chromosomes. Suppose its eggs and its sperm were made by mitosis instead of meiosis, and an egg and a sperm then fused.
How many chromosomes would the zygote hold?
An animal’s body cells hold eight chromosomes. In one dividing cell, four X-shaped chromosomes move toward each pole, each still two sister chromatids joined at the centromere.
Which stage is the cell in?
An animal’s body cells hold 16 chromosomes. One cell has just finished meiosis I and cytokinesis. To count chromosomes, count centromeres.
Which of the following describes each of the two cells?
An animal’s body cells hold 14 chromosomes.
With no crossing over, how many different combinations of maternal and paternal chromosomes can this animal’s gametes carry?
The four chromatids of one paired homologous pair in a plant are numbered 1 to 4. Each chromatid carries alleles at two positions, D or d and E or e. The table gives each chromatid’s alleles before and after prophase I. One crossover happened.
Which two chromatids exchanged pieces in the crossover?
A plant’s body cells hold 14 chromosomes. One of its zygotes holds 15, and the plant that grows from that zygote develops differently from its parents.
Compared with its parents’ cells, which of the following describes what the plant’s cells make?
An animal’s body cells hold six chromosomes, in three homologous pairs.
What is the probability that one of its gametes carries the maternal chromosome of every pair?
An animal’s body cells hold 24 chromosomes. In one cell, one pair fails to separate during meiosis, and every other pair separates normally. The four cells the meiosis produces are counted.
Which set of counts shows that the failure happened in meiosis I?
In oregano, suppose purple flowers (F) are dominant to white flowers (f). The Punnett square drawn here shows the four offspring genotypes of one cross, with the two parents’ gametes along the edges hidden.
Which cross produced this square?
In loquats, suppose orange flesh (R) is dominant to pale flesh (r). An orange-fleshed tree grew from a seed of a pale-fleshed tree.
What is the orange-fleshed tree’s genotype?
In cyclamens, suppose dark leaves (D) are dominant to pale leaves (d).
Which cross gives offspring in a 1 : 1 genotypic ratio and a 1 : 1 phenotypic ratio?
In primulas, suppose tall stems (A) are dominant to short stems (a) and red flowers (B) to white flowers (b), and the two genes sit on different chromosomes. Two AaBb plants are crossed.
What is the probability that an offspring shows the recessive trait for one gene or for both genes?
A plant is heterozygous, Aa, for one gene. Researchers read the alleles in 500 of its pollen grains. Most grains carry A or a, but 4% of them carry both A and a.
Which event most likely produced the grains that carry both A and a?
In gladioli, suppose scarlet flowers (S) are dominant to white flowers (s). A breeder crosses two Ss plants and predicts 3 scarlet : 1 white. The 220 offspring are counted in the table.
Which of the following is the chi-square value for the breeder’s counts?
A breeder tests the two offspring classes of a test cross against 1 : 1 and gets a chi-square value of 4.5. The table drawn here is the formula sheet’s.
Which of the following gives the verdict on the null hypothesis at each p value?
In delphiniums, suppose a breeder crosses two plants heterozygous for two genes on different chromosomes, sorts the 272 offspring into four classes and tests the counts against 9 : 3 : 3 : 1. The chi-square value is 8.4. The table drawn here is the formula sheet’s.
Which of the following is the verdict on the null hypothesis at p = 0.05?
A rare autosomal recessive condition is recorded in the family drawn here; a filled shape shows the condition. Write R for the ordinary allele and r for the allele that causes the condition. II-4 married into the family.
Which of the following people could be either RR or Rr?
A student wants to decide from a family record whether a rare trait is dominant or recessive.
Which of the following observations, on its own, shows that the trait is dominant?
The ABO gene has three alleles: Iᴬ and Iᴮ are codominant, and i is recessive to both.
Which pair of parents can have children of all four blood types, A, B, AB and O?
In sesame, suppose two genes sit 20 map units apart on one chromosome, and 1 map unit is 1% recombination. A plant carrying A with B on one homolog and a with b on the other is test-crossed with an aabb plant, and 500 offspring are grown.
About how many of the 500 offspring are recombinant?
A man has a trait caused by a recessive allele on his X chromosome. He has a son and a daughter, and both are free of the trait. The son has a son, and the daughter has a son.
Which grandson could have received the allele from the man?
In azaleas, suppose one dominant allele, F, of a single gene gives frilled petals, red leaf veins and a hairy stem; plants without F have plain petals, green veins and a smooth stem. A breeder crosses an Ff plant with an ff plant and grows 400 offspring.
Which result does the breeder expect?
On a karyotype from a man, the chromosomes are sorted by size into 23 pairs. The X and the Y differ in length and carry mostly different genes, yet the karyotype puts them together as the last pair.
Why are the X and the Y counted as a pair?
In nilgai, a large antelope, suppose one gene sets a marker on the surface of red blood cells, and the marker comes in two forms, F and S. Every calf of a true-breeding F-marker line mated with a true-breeding S-marker line carries both markers, F and S, on its red cells. Two of these both-marker animals are mated, and the markers of their 96 offspring are recorded in the table.
Which deviation from Mendel’s 3 : 1 ratio does this cross show?
A health survey of one country records an inherited trait in about 1 in 25 men and about 1 in 600 women.
Which mode of inheritance fits the survey?
A gardener splits one chive plant into clumps, so every clump has the same DNA, and grows the clumps in three groups with the same light, temperature and soil. A protein in the root cells pumps salt out of the cell. The table gives the amount of that protein per root cell in each group.
Which of the following explains the pattern in the table?
A gardener roots 48 cuttings of one elderberry bush and grows 16 in full shade, 16 in partial shade and 16 in full sun, with the same water and soil. After eight weeks she measures every leaf and finds each group’s mean leaf length. The bars drawn here carry ±2SE error bars.
Which groups differ in mean leaf length?
(a)(i) Describe what happens at a chiasma in prophase I that gives a chromatid a new combination of alleles. (1 point)
A full-credit answer: A chromatid of the maternal homolog and a chromatid of the paternal homolog break at the same point.
The two pieces are exchanged.
Each of those chromatids now carries a combination of alleles that neither homolog had.
Check the box for each point your answer earns
Accept with or without: the name crossing over.
Common slip: Describing sister chromatids swapping pieces. Sister chromatids are identical copies, so a swap between them changes no combination.
(a)(ii) Crossing over aside, explain why the pollen grains of one jute plant differ in which chromosomes they carry. (1 point)
A full-credit answer: Each homologous pair lines up at metaphase I facing either way, regardless of the other pairs.
So which member of each pair a grain receives, the maternal or the paternal chromosome, is decided pair by pair.
With seven pairs, the grains carry many different mixtures of maternal and paternal chromosomes.
Check the box for each point your answer earns
Accept with or without: the count, 2⁷ = 128 kinds of grain.
Common slip: Answering with crossing over. The task rules a crossover out; the other shuffle inside meiosis is the orientation of the pairs.
(b)(i) Identify the independent variable in the researchers’ experiment. (1 point)
A full-credit answer: The growth temperature of the cuttings.
Check the box for each point your answer earns
(b)(ii) Identify the control group. (1 point)
A full-credit answer: The cuttings grown at 22 °C, the temperature of the nursery the parent plant grew in.
Check the box for each point your answer earns
Do not award: the 15 °C group chosen as the lowest temperature; the control is the condition the plant grew in before the change.
Common slip: Naming the 15 °C group because it is the coolest. The control is the parent plant’s own condition, 22 °C.
(b)(iii) Justify the researchers’ decision to examine the same number of cells at each temperature. (1 point)
A full-credit answer: The count is a total over the cells examined.
More cells would give more chiasmata whatever the temperature.
With 50 cells in every group, a difference between the totals comes from the temperature and from nothing else, so the totals can be compared directly.
Check the box for each point your answer earns
Do not award: ‘it is fair’ with no link to the totals.
Common slip: Saying only that it is fair. The point is that a total depends on how many cells were counted.
(c)(i) Describe the relationship between the growth temperature and the number of chiasmata counted. (1 point)
A full-credit answer: As the growth temperature rises, the number of chiasmata in 50 cells falls: 470 at 15 °C, 395 at 22 °C and 210 at 30 °C.
Check the box for each point your answer earns
Accept with or without: the counts 470, 395 and 210.
(c)(ii) Identify the smallest number of chiasmata one jute cell needs for every one of its homologous pairs to be held together at metaphase I. (1 point)
Write down the values in the question:
chromosomes in a body cell = 14
chiasmata needed per homologous pair = 1
Write down the equation:
Substitute the values into the equation:
A full-credit answer: Jute has 14 chromosomes, so 7 homologous pairs.
One chiasma holds one pair.
So the cell needs at least 7 chiasmata.
(d)(i) Predict how the proportion of pollen grains with an abnormal chromosome number in the plants grown at 30 °C compares with the proportion in the plants grown at 22 °C. (1 point)
A full-credit answer: A larger proportion of the 30 °C plants’ pollen grains has an abnormal chromosome number.
Check the box for each point your answer earns
(d)(ii) Justify your prediction in part (d)(i). (1 point)
A full-credit answer: Chiasmata hold the two homologs of a pair together until anaphase I.
At 30 °C the cells formed fewer chiasmata, so some pairs were held by none.
A pair held by no chiasma can move to one pole together at anaphase I, so both homologs enter one cell.
Meiosis II then gives grains with an extra chromosome and grains with one missing.
Check the box for each point your answer earns
Accept with or without: the means per cell — 210 ÷ 50 = 4.2 chiasmata at 30 °C, fewer than the 7 pairs, against 395 ÷ 50 = 7.9 at 22 °C, about one per pair.
Do not award: ‘fewer crossovers give less variation’; the task is about the chromosome number.
Common slip: Arguing from variation. Fewer crossovers do give fewer new combinations, but that changes no chromosome count.
(a) Explain why III-3 and III-4 are free of the disorder although their father, II-3, has it. (1 point)
A full-credit answer: II-3’s mitochondria carry the faulty gene.
At fertilization the sperm’s few mitochondria break down, so a zygote’s mitochondria all come from the egg.
II-4 is unaffected, so her eggs carried normal mitochondria.
So III-3 and III-4 received no faulty mitochondria.
Check the box for each point your answer earns
Common slip: Saying II-3 passed the gene on his Y or on his X. The gene is in mitochondrial DNA, outside the chromosomes.
(b) Identify every individual in generation III who can pass the disorder to their children. (1 point)
A full-credit answer: III-2 and III-6.
III-2 and III-6 have the disorder and are female, so their eggs will carry the faulty mitochondria.
III-1 and III-5 have the disorder, but a sperm’s mitochondria do not persist, so they will pass it to none of their children.
Check the box for each point your answer earns
Common slip: Listing every affected person in generation III. The affected men, III-1 and III-5, pass no mitochondria on.
(c) A student claims that the disorder is X-linked recessive. Based on the student’s claim, predict which individuals of generation III would be affected. (1 point)
A full-credit answer: III-1 and III-5.
Under the claim an affected mother carries the allele on both X chromosomes, so every son receives one and is affected.
Every daughter would receive an ordinary X from her unaffected father, so III-2 and III-6 would be unaffected carriers.
II-3 would pass his X to III-3, who would carry it unaffected, and his Y to III-4.
Check the box for each point your answer earns
Do not award: a list that includes a daughter.
Common slip: Predicting every child of an affected mother. Under the claim a daughter also receives her unaffected father’s X, which masks the allele.
(d) Justify why the pedigree contradicts the student’s claim. (1 point)
A full-credit answer: III-2 and III-6 have the disorder and are daughters.
Under X-linked recessive inheritance each would need the allele on both X chromosomes, one of them from her father.
Their fathers, II-2 and II-6, are unaffected, so each father’s only X carries the ordinary allele.
So the daughters could not have received a second copy from their fathers.
Check the box for each point your answer earns
Do not award: ‘the trait appears in every generation’ alone.
Common slip: Arguing only that daughters are affected. A daughter can show an X-linked recessive trait; the point is that her father would have to be affected too.
(a) Describe the difference between the results of the two crosses. (1 point)
A full-credit answer: In cross 1 every son is pale-finned and every daughter is dark-finned.
In cross 2 every son and every daughter is dark-finned.
So the two crosses differ in their sons.
Check the box for each point your answer earns
Accept with or without: the counts.
(b) Make a claim about where the gene for fin color sits. (1 point)
A full-credit answer: The fin-color gene sits on the X chromosome, and the pale allele is recessive.
Check the box for each point your answer earns
Accept with or without: the pale allele recessive.
(c) Support your claim in part (b) with evidence from cross 1. (1 point)
A full-credit answer: In cross 1 the mother is pale-finned and the father dark-finned.
A son’s only X comes from his mother, so every son received her X carrying the pale allele and is pale-finned.
Every daughter received an X from each parent, so each carries the father’s dark allele, which masks the pale allele, and is dark-finned.
A gene on an autosome would give sons and daughters the same fins.
Check the box for each point your answer earns
Accept with or without: a gene on an autosome would not sort the offspring by sex.
Common slip: Saying only that the trait is more common in males. The evidence is that every son took the mother’s trait and no daughter did.
(d) A daughter from cross 1 mates with a dark-finned male. Calculate the probability that their first offspring is a pale-finned son. (1 point)
Write down the values in the question:
Write down the equation:
Substitute the values into the equation:
A full-credit answer: The daughter is XᴰXᵈ, a carrier, and the male is XᴰY.
The probability that the offspring is a son is ½.
The probability that the egg carries Xᵈ is ½.
P(pale-finned son) = ½ × ½ = ¼ = 0.25.