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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and 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

In petunias, a true-breeding red-flowered line crossed with a true-breeding white-flowered line gives only pink-flowered offspring, and two pink plants crossed give red, pink and white offspring.

Which kind of dominance does petunia flower color show?

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

Question 2

In petunias, red (Cᴿ) and white (Cᵂ) show incomplete dominance and a CᴿCᵂ plant is pink. Two pink petunias are crossed and 160 offspring flower.

How many of the 160 offspring are expected to be white?

A. 40 B. 80 C. 120 D. 160

Question 3

A woman is type AB and a man is type B. The man's mother was type O.

Which blood types are possible for their children?

A. AB only B. A, B or AB C. A, B, AB or O D. B or AB only

Question 4

The four classes of kits from a BbTt mouse (B with T on one chromosome, b with t on the other) crossed with a bbtt mouse.

In mice, black coat (B) is dominant to brown (b) and a normal tail (T) to a short tail (t). A BbTt mouse whose chromosomes carried B with T and b with t is crossed with a bbtt mouse. The kits are counted in the table below.

Which two classes are the parental combinations?

A. Black normal and black short B. Brown short and brown normal C. Black short and brown normal D. Black normal and brown short

Question 5

The four classes of kits from a BbTt mouse (B with T on one chromosome, b with t on the other) crossed with a bbtt mouse.

For the mouse cross in the table above, 180 black normal, 174 brown short, 24 black short and 22 brown normal kits were counted, 400 in all.

What is the map distance between the coat-color gene and the tail gene?

A. 0.115 map units B. 11.5 map units C. 13.0 map units D. 88.5 map units

Question 6

Three genes sit on one chromosome of a fly, written A, B and C. Their map distances are A to B 21 map units, B to C 15 map units, and A to C 6 map units.

In what order do the three genes sit along the chromosome?

A. A, C, B B. A, B, C C. B, A, C D. C, A, B

Question 7

Red-green color blindness comes from a recessive allele on the X chromosome. A color-blind woman has three sons with a man who sees colors as most people do.

What do the sons see, and why?

A. All three see colors as most people do, because their father's ordinary allele hides their mother's B. About half of them are color-blind, because each son receives one of his mother's two alleles C. None of them is color-blind, because each son received his father's X, which carries the ordinary allele D. All three are color-blind, because a son's only X comes from his mother and both of hers carry the allele

Question 8

A family's records follow a rare disorder across three generations. Every affected woman has affected sons and affected daughters. 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

The chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.

A breeder test-crosses a plant heterozygous for two genes and sorts the offspring into four classes. Against 1 : 1 : 1 : 1, chi-square comes out at 3.10. The table below is the formula sheet's.

What is the verdict, and what does it say about the two genes?

A. Fail to reject the null hypothesis, because 3.10 is smaller than 7.81; the counts are consistent with independent assortment B. Reject the null hypothesis, because 3.10 is smaller than 3.84; the genes are linked and sit on one chromosome C. Accept the null hypothesis, because 3.10 is so small that the counts prove that the genes sit on different chromosomes D. Fail to reject the null hypothesis, because 3.10 is smaller than 3.84; the two genes sit 3.10 map units apart

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

In tomato plants, tall (T) is dominant to dwarf (t) and a hairy stem (H) to a smooth stem (h). A breeder crosses a TtHh plant, whose chromosomes carry T with H and t with h, 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 tomato plant (T with H on one chromosome, t with h on the other) 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 heterozygous parent's own chromosomes carry?

A full-credit answer: The parental classes are tall hairy (261) and dwarf smooth (255), because the parent's chromosomes carried T with H and t with h; 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: 84 recombinant offspring.

(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: The recombination frequency is 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 state the verdict of the chi-square test and what it 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: Four classes give 3 degrees of freedom, and the critical value at p = 0.05 is 7.81. χ² = 311 is far larger than 7.81, so the breeder rejects the null hypothesis: the counts do not fit 1 : 1 : 1 : 1, and the two genes are not assorting independently. They are linked.

Check the box for each point your answer earns

3 degrees of freedom, critical value 7.81, reject the null hypothesis because 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) Identify the mode of inheritance of the condition. (1 point)

A full-credit answer: The condition is autosomal dominant.

Check the box for each point your answer earns

autosomal dominant.

Accept: dominant, on an ordinary pair of chromosomes.

Accept as the justification, if one is given: II-2, an unaffected daughter of the affected father I-1, rules out an X-linked dominant allele (a father gives his one X to every daughter), and as the unaffected child of two affected parents she rules out X-linked recessive and maternal inheritance 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) Justify the claim that the allele is dominant, using one family in the pedigree. (1 point)

A full-credit answer: I-1 and I-2 both have the condition, yet their daughter II-2 does not. If the allele were recessive, two affected parents would both be homozygous for it and could give a child nothing else, so every child would be affected. An unaffected child of two affected parents is possible only if the allele is dominant and both parents are heterozygous.

Check the box for each point your answer earns

two affected parents (I-1 and I-2) have an unaffected child (II-2), which a recessive allele cannot produce, so the allele is dominant.

Common slip: Arguing from the condition appearing in every generation. That is consistent with dominant but does not prove it.

(c) Write A for the allele that causes the condition and a for the ordinary 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 9 correct.

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