Practice questions · Topic 5.4

Unit 5 · Practice for the Topic 5.4 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Non-Mendelian genetics, summed up

Video coming soon

Heterozygotes that show, 1 : 2 : 1 in phenotypes; linked genes, recombinants and map units; the X and the Y, and why recessive X-linked traits are mostly male; one gene with many effects; organelle genes from the mother; deciding the mode from a pedigree or from counts.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation. The first free-response question walks you through a linked test cross one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Question 1

Suppose that in one breed of dog, true-breeding curly-coated dogs mated with true-breeding straight-coated dogs give only wavy-coated puppies, and two wavy-coated dogs mated give curly, wavy and straight puppies.

Which kind of dominance does coat texture show?

A. Complete dominance B. Codominance C. Incomplete dominance D. A merging of the two alleles into one wavy-coat allele

Question 2

Suppose that in a species of lizard one gene sets scale color. The allele Sᴮ gives black scales and the allele Sʸ gives yellow scales, and the two alleles are codominant.

What do the scales of an SᴮSʸ lizard look like?

A. Every scale is black B. Every scale is yellow C. Every scale is one shade in between black and yellow D. Black scales and yellow scales, each scale one color

Question 3

A type B child has a type A mother and a type B father.

Which two alleles of the ABO gene does the mother carry?

A. Iᴬ and Iᴬ B. Iᴬ and i C. Iᴮ and i D. Iᴬ and Iᴮ

Question 4

In a species of gecko, a dark body (D) is dominant to a pale body (d) and a long tail (L) to a short tail (l). A DdLl gecko is crossed with a ddll gecko. One offspring has a pale body and a long tail.

Which gamete did the DdLl parent give that offspring?

A. DL B. Dl C. dL D. dl

Question 5

Four Punnett squares drawn for the cross, numbered 1 to 4.

In humans, a recessive allele on the X chromosome, written Xᵃ, causes a trait, and Xᴬ is the ordinary allele. A woman who carries the allele, XᴬXᵃ, and a man with the trait, XᵃY, have children. Four squares drawn for this cross are numbered 1 to 4 below, with the woman’s gametes meant to sit along the top edge and the man’s down the left edge.

Which square is drawn correctly?

A. Square 1 B. Square 2 C. Square 3 D. Square 4

Question 6

In a species of mammal, a gene sits on the Y chromosome, and one of its alleles gives a patterned coat. A patterned male has two sons and two daughters. One of his daughters later has a son.

Which of these five descendants carry the patterned-coat allele?

A. The two sons only B. The two daughters only C. The two sons and the daughter’s son D. All four children and the daughter’s son

Question 7

Red-green color blindness comes from a recessive allele on the X chromosome. In a large town the allele is uncommon.

Which of the following describes the people in the town who are color-blind?

A. Men only B. Many men and a few women C. About as many women as men D. Many women and a few men

Question 8

A family's records follow a rare disorder across three generations. Every child of an affected woman is affected, sons and daughters alike. Affected men also have sons and daughters, and all of their children are unaffected.

Which explanation fits the records?

A. Both parents pass the disorder on equally through the mitochondria in their gametes B. The disorder is autosomal dominant, because every affected parent passes it on C. The disorder is written in mitochondrial DNA, which the egg supplies and the sperm does not D. The disorder is X-linked recessive, because affected mothers pass an X to every child

Question 9

A Punnett square for XᴰXᵈ × XᴰY: the woman’s gametes along the top edge, the man’s down the left edge.

In humans, a recessive allele on the X chromosome, written Xᵈ, causes a trait, and Xᴰ is the ordinary allele. The Punnett square below is for a woman who carries the allele, XᴰXᵈ, and a man whose X carries the ordinary allele, XᴰY.

Which of the couple’s children show the trait?

A. Half of the sons and none of the daughters B. Half of the daughters and none of the sons C. Half of the sons and half of the daughters D. All of the sons and none of the daughters

Question 10

In honeybees, a worker develops from an egg that a sperm fertilized, and a drone, a male bee, develops from an egg that no sperm fertilized. A worker’s body cells each hold 32 chromosomes.

How many chromosomes does each of a drone’s body cells hold?

A. 8 B. 16 C. 32 D. 64

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Conceptual Analysis · 5 points

Suppose a breeder works with a plant in which tall (T) is dominant to dwarf (t) and a hairy stem (H) to a smooth stem (h). The breeder crosses a TtHh plant, bred from a true-breeding tall hairy line and a true-breeding dwarf smooth line, with a tthh plant and counts 600 offspring. The counts are in the table below, with the chi-square table from the formula sheet beneath it. Against a 1 : 1 : 1 : 1 ratio the breeder has already calculated χ² = 311.

Top: the four classes among 600 offspring of a TtHh plant, bred from true-breeding tall hairy and dwarf smooth lines, crossed with a tthh plant. Bottom: the chi-square table from the AP Biology formula sheet.

(a) Identify the two parental classes and the two recombinant classes, and justify the choice. (1 point)

Hint: Which two combinations of alleles did the TtHh plant inherit, one from each of its true-breeding parents?

A full-credit answer: The parental classes are tall hairy (261) and dwarf smooth (255), because the TtHh plant inherited T with H from one true-breeding parent and t with h from the other, so those are the combinations its chromosomes carried; the recombinant classes are tall smooth (44) and dwarf hairy (40), because those combinations were on neither of its chromosomes and are the two small classes.

Check the box for each point your answer earns

tall hairy and dwarf smooth as parental (the combinations the parent's chromosomes carried, the two large classes); tall smooth and dwarf hairy as recombinant (new combinations, the two small classes).

Common slip: Calling the two large classes the recombinants.

(b) Calculate the number of recombinant offspring. (1 point)

Hint: Add the counts of the two recombinant classes you named in part (a).

Add the two recombinant classes:

recombinant offspring = 44 + 40 = 84

A full-credit answer: Recombinant offspring = 44 + 40 = 84.

(c) Calculate the recombination frequency between the two genes. (1 point)

Hint: Which count goes on top of the fraction, and which total goes underneath, before you multiply by 100?

Write down the values in the question:

recombinant offspring = 84

total offspring = 600

Write down the equation:

recombination frequency = \frac{recombinant offspring}{total offspring} \times 100 %

Substitute the values into the equation:

recombination frequency = \frac{84}{600} \times 100 % = 14.0 %

A full-credit answer: Recombination frequency = 84 ÷ 600 × 100 % = 14.0 %.

(d) State the map distance between the two genes, with its unit. (1 point)

Hint: How many map units is one percent of recombination?

map units

Convert the recombination frequency to a map distance:

1 % recombination = 1 map unit

map distance = 14.0 map units

A full-credit answer: The two genes are 14.0 map units apart, because one percent of recombination is one map unit.

(e) Identify the degrees of freedom and the critical value at p = 0.05, and determine whether the null hypothesis of independent assortment is rejected and what that says about the two genes. (1 point)

Hint: How many classes are there, which row and column of the table do you read, and which is larger, the calculated value or the critical value?

A full-credit answer: There are four classes, so 3 degrees of freedom.
The critical value at p = 0.05 is 7.81.
χ² = 311 is far larger than 7.81, so the null hypothesis is rejected.
So the counts do not fit 1 : 1 : 1 : 1, and the two genes are not assorting independently.
Therefore the two genes are linked.

Check the box for each point your answer earns

3 degrees of freedom and a critical value of 7.81, AND the decision (reject the null hypothesis) with its ground (311 is larger than 7.81), so the genes do not assort independently (not ‘accept’ or ‘prove’).

Common slip: Reading the column for four degrees of freedom. Degrees of freedom are the number of classes minus one.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

A condition caused by one gene is recorded in one family in the pedigree below; a filled shape shows the condition. II-4 married into the family.

A family pedigree for a condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-4 married into the family.

(a) Make a claim about the mode of inheritance of the condition. (1 point)

A full-credit answer: Claim: the condition is autosomal dominant.

Check the box for each point your answer earns

the claim, autosomal dominant; no reasoning is required for this point.

Accept: dominant, on an ordinary pair of chromosomes.

Accept as extra reasoning, if given: II-2, an unaffected daughter of two affected parents, rules out X-linked recessive and non-nuclear inheritance; as the daughter of the affected I-1, who gave her his one X, she rules out X-linked dominant too.

Common slip: Writing recessive because the condition appears in every generation. Appearing in every generation is a hint, not a proof; the decisive family decides. Writing X-linked or maternal: II-2 is an unaffected daughter of two affected parents, which neither an X-linked recessive allele nor mitochondrial DNA can produce, and her affected father I-1 gave her his one X, so an X-linked dominant allele would have made her affected as well; the allele is on an ordinary pair.

(b) Support the dominant-or-recessive part of your claim with evidence from one family in the pedigree. (1 point)

A full-credit answer: Evidence: I-1 and I-2 both have the condition, yet their daughter II-2 does not.
If the allele were recessive, both affected parents would be homozygous for it.
Then every child would receive the allele from both parents and be affected.
An unaffected child of two affected parents is possible only if the allele is dominant and both parents are heterozygous.
So this family supports the claim that the allele is dominant.

Check the box for each point your answer earns

the evidence (two affected parents, I-1 and I-2, have an unaffected child, II-2) AND the reasoning that links it to the claim (a recessive allele in two affected parents would reach every child, so the allele must be dominant).

Common slip: Giving the evidence without the link, or the reasoning without naming the family. Support a claim needs both.

(c) Using A for the dominant allele and a for the recessive allele, identify the genotypes of I-1, I-2 and II-2, and explain how the pedigree fixes them. (1 point)

A full-credit answer: II-2 does not have the condition, so she carries no A: she is aa, and she received an a from each parent.
I-1 and I-2 have the condition, so each carries an A, and each also passed an a to II-2, so both are Aa.

Check the box for each point your answer earns

II-2 aa; I-1 Aa and I-2 Aa, because each shows the condition (carries A) and each passed an a to their unaffected daughter.

Common slip: Writing I-1 and I-2 as AA. An AA parent has no a to pass on, so an aa child rules AA out.

(d) Calculate the probability that the next child of II-3 and II-4 has the condition. (1 point)

Write down the values in the question:

II-3 is Aa, II-4 is aa

P (A from II-3) = \frac{1}{2}

P (a from II-4) = 1

Write down the equation:

P (A and B) = P (A) \times P (B)

Substitute the values into the equation:

P (A a) = \frac{1}{2} \times 1 = \frac{1}{2} = 0.5

A full-credit answer: The probability is $\frac{1}{2}$ , which is 0.5.
II-3 has the condition and has an unaffected daughter, III-2, so she is Aa; II-4 is unaffected, so he is aa.
Half of II-3's gametes carry A, and every child who receives it has the condition.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

Take the end-of-topic test