Heredity: Unit Diagnostic (Biology Unit 5)

What this diagnostic is for

Use this page before you revise Unit 5, or right after you finish it. There is one question per topic, with extra ones for the larger topics. Each answer names the study guide to read if you got it wrong.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, so your score does not predict an exam score. Treat it as a map of what to revisit.

Calculator and data. A calculator is allowed on both sections of the Biology exam; a basic one is enough here. Chi-square: χ² = Σ (o − e)² ÷ e, df = categories − 1, critical values at p = 0.05: 3.84 (1 df), 5.99 (2 df), 7.81 (3 df). All organisms with letter-named genes, and all data, are fictional.

How to work. Answer every question before opening any answer, and write a short reason for each choice.

Question 1 (multiple choice · 5.1)

Which statement is true of anaphase in mitosis and of anaphase II in meiosis, but not of anaphase I?

(A) Sister chromatids separate and move to opposite poles.
(B) Homologous chromosomes move to opposite poles.
(C) Each pole receives a full diploid set of chromosomes.
(D) Each chromosome moving towards a pole is made of two chromatids.

Answer and explanation

Answer: (A). In mitosis and meiosis II, the proteins holding sister chromatids together break down and the sisters are pulled apart. In anaphase I they stay joined.

(B) describes anaphase I only; the shared stage names make it tempting.
(C) fits mitosis in a diploid cell, but meiosis II happens in a haploid cell.
(D) is the mark of anaphase I.

If you missed this: read “Mitosis and meiosis compared” in the meiosis study guide.

Question 2 (multiple choice · 5.2)

A fictional moth has 2n = 6. Ignoring crossing over, what fraction of its gametes contain at least one chromosome that the moth inherited from its father?

(A) 1/8
(B) 3/8
(C) 1/2
(D) 7/8

Answer and explanation

Answer: (D). With n = 3 pairs there are 2³ = 8 equally likely combinations. Only the all-maternal one, probability (1/2)³ = 1/8, has no paternal chromosome. So 1 − 1/8 = 7/8.

(A) is the all-maternal (or all-paternal) gamete.
(B) is the chance of exactly one paternal chromosome (3 × 1/8).
(C) is the chance for one pair only; all three pairs must be combined.

If you missed this: read “Independent assortment: shuffling whole chromosomes” in the meiosis and genetic diversity study guide.

Question 3 (multiple choice · 5.3)

Which event in meiosis is the physical basis of Mendel’s law of independent assortment?

(A) The random orientation of each homologous pair at metaphase I, which does not depend on how other pairs line up
(B) The separation of sister chromatids at anaphase II
(C) Crossing over between sister chromatids in prophase I
(D) The random fusion of gametes at fertilization

Answer and explanation

Answer: (A). Genes on different chromosomes travel with their own pair. Each pair faces the poles at random, independently of the others, so alleles of one gene are sorted independently of alleles of the other.

(B) is also a separation, but sister chromatids carry the same alleles (unless crossing over has happened).
(C) has the wrong chromatids: crossing over is between non-sister chromatids, and it mixes alleles within one chromosome.
(D) adds variation, but only after gametes are made.

If you missed this: read “The law of independent assortment” in the Mendelian genetics study guide.

Question 4 (multiple choice · 5.3)

Two parents are both carriers (Aa) of a recessive condition. They have two children. What is the probability that exactly one of the two children has the condition?

(A) 1/16
(B) 3/16
(C) 3/8
(D) 1/2

Answer and explanation

Answer: (C). Each child: P(affected) = 1/4, P(unaffected) = 3/4. Exactly one affected can happen two ways: first only (1/4 × 3/4 = 3/16) or second only (3/16). These are mutually exclusive, so add: 3/8.

(A) is the chance that both children are affected.
(B) counts only one of the two orders.
(D) simply adds 1/4 + 1/4.

If you missed this: read “The rules of probability” in the Mendelian genetics study guide.

Question 5 (multiple choice · 5.4)

Which single observation in a family would rule out X-linked recessive inheritance of a rare trait?

(A) An affected son whose mother is unaffected
(B) An affected daughter whose father is unaffected
(C) Many more affected males than affected females
(D) An affected man whose sons are all unaffected

Answer and explanation

Answer: (B). An affected daughter needs the recessive allele on both X chromosomes, so her father’s only X must carry it, and he would be affected too.

(A) fits: a carrier mother passes the allele to her son.
(C) is the typical X-linked recessive pattern, so it supports it.
(D) also fits: fathers give sons a Y, not an X.

If you missed this: read “Sex-linked traits” in the non-Mendelian genetics study guide.

Question 6 (multiple choice · 5.4)

In a test cross AaBb × aabb, the four phenotype classes are almost equal, and χ² against 1 : 1 : 1 : 1 is well below the critical value. A student concludes: “Genes A and B must be on different chromosomes.” Which is the best evaluation?

(A) The conclusion is certain, because linked genes always give two large and two small classes.
(B) The conclusion is not certain. Genes far apart on the same chromosome recombine so often that they also give close to 50% recombinants.
(C) The conclusion is wrong, because equal classes show that one gene affects both traits.
(D) The conclusion is wrong, because equal classes show that both genes are on the X chromosome.

Answer and explanation

Answer: (B). The farther apart two genes are, the more often a crossover falls between them. Recombination frequency cannot rise above about 50%, which looks exactly like independent assortment. The data fit separate chromosomes but do not prove it.

(A) holds only for genes close enough to give well under 50% recombinants.
(C) describes pleiotropy, which keeps two traits together.
(D) has no support: the data do not compare the sexes.

If you missed this: read “Linked genes and gene mapping” in the non-Mendelian genetics study guide.

Question 7 (multiple choice · 5.5)

A baby has two recessive alleles for phenylketonuria (PKU), so it cannot break down the amino acid phenylalanine. Fed a low-phenylalanine diet from birth, it develops without the brain damage PKU can cause. Which statement is correct?

(A) The diet changed the baby’s genotype from homozygous recessive to heterozygous.
(B) The baby’s genotype is unchanged; the diet changed the environment, so the same genotype produced a different phenotype.
(C) The diet made the recessive allele behave as a dominant allele.
(D) Because the condition was prevented, the child cannot pass a PKU allele to its own children.

Answer and explanation

Answer: (B). The child still has two PKU alleles. The diet stops phenylalanine building up. Phenotype depends on genotype and environment.

(A) is the common slip: the environment does not change the DNA sequence.
(C) misuses “dominant”, which describes how alleles act in a heterozygote.
(D) treats a phenotype as inherited. Every gamete the child makes carries a PKU allele.

If you missed this: read “Genotype is not the whole story” and “Examples you should be able to explain” in the environmental effects on phenotype study guide.

Question 8 (short answer · 5.2)

A fictional worm has 2n = 4. Gene A is on the long pair of chromosomes and gene B is on the short pair. The worm inherited A and B from its mother, and a and b from its father.

(a) Ignoring crossing over, list every kind of gamete the worm can make. (b) Without crossing over, one cell gives four gametes of only two kinds (for example two AB, two ab). Explain why. (c) In a different cell, the four gametes are AB, aB, Ab and ab. Name the event that must have happened, and say when and between which chromatids.

Answer and explanation

(a) 2² = 4 kinds: AB, Ab, aB, ab, each 1/4.

(b) In one cell the pairs line up in only one way at metaphase I. If both maternal chromosomes face the same pole, one cell gets A and B and the other a and b. Sister chromatids are identical, so meiosis II gives two copies of each. The other orientation happens in other cells.

(c) Crossing over in prophase I, between non-sister chromatids of one pair, between the gene and the centromere. The sister chromatids then carry different alleles (for example A and a), and meiosis II separates them. A crossover in either pair works.

If you missed this: read “Crossing over” and “Independent assortment” in the meiosis and genetic diversity study guide.

Question 9 (short answer · 5.3)

In a hypothetical snail, striped shell (S) is dominant to plain (s), and tall spire (T) is dominant to short (t). The genes are on different chromosomes. A cross SsTt × ssTt gives 240 offspring: 98 striped tall, 24 striped short, 85 plain tall and 33 plain short.

(a) Use the product rule to give the expected ratio and counts. (b) Calculate χ² and state your conclusion.

Answer and explanation

(a) Ss × ss: 1/2 striped. Tt × Tt: 3/4 tall. Multiply: striped tall 3/8, striped short 1/8, plain tall 3/8, plain short 1/8, a 3 : 1 : 3 : 1 ratio (not 9 : 3 : 3 : 1, as only one parent is Ss). Expected: 90, 30, 90, 30.

(b)

Class	o	e	(o − e)² ÷ e
striped tall	98	90	0.711
striped short	24	30	1.200
plain tall	85	90	0.278
plain short	33	30	0.300
Total	240	240	χ² = 2.489

df = 3; critical value 7.81. Since 2.489 < 7.81, fail to reject the null hypothesis: the data fit independent assortment with complete dominance.

If you missed this: read “The chi-square test” and Worked example 1 in the Mendelian genetics study guide.

Question 10 (short answer · 5.4)

In a hypothetical mammal (XX females, XY males), a recessive X-linked allele Xʳ gives a rough coat; the dominant allele Xᴿ gives a smooth coat. A smooth-coated female whose father had a rough coat mates with a rough-coated male.

(a) Give the genotypes of both parents. (b) Predict the fraction of daughters and of sons with rough coats. (c) In the reciprocal cross, a rough-coated female mates with a smooth-coated male. Predict the offspring, and explain why the two crosses give different results.

Answer and explanation

(a) Her father gave her his Xʳ; she is smooth, so she is XᴿXʳ. The male is XʳY.

(b) Daughters get Xʳ from the father and Xᴿ or Xʳ from the mother: 1/2 rough. Sons get Xᴿ or Xʳ from the mother: 1/2 rough.

(c) XʳXʳ × XᴿY: all daughters smooth (XᴿXʳ carriers) and all sons rough (XʳY). Fathers give their X only to daughters; sons get their X from the mother. Autosomal genes give the same result both ways, so a difference points to sex linkage.

If you missed this: read “Sex-linked traits” and Worked example 3 in the non-Mendelian genetics study guide.

Your next step

Mark each question right or wrong. A short answer missed in one part counts as missed.

Topic	Question(s)	If you missed it, read
5.1 Meiosis	1	Study guide 5.1
5.2 Meiosis and genetic diversity	2, 8	Study guide 5.2
5.3 Mendelian genetics	3, 4, 9	Study guide 5.3
5.4 Non-Mendelian genetics	5, 6, 10	Study guide 5.4
5.5 Environmental effects on phenotype	7	Study guide 5.5

How to use your result

Wrong answers first. Read the named section, then do that topic’s practice questions.
Lucky guesses count as misses if your reason did not match the explanation.
Missed Questions 1, 3 and 8? All three depend on what chromosomes do in meiosis I. Start with Topic 5.1 before the genetics topics.
Missed Questions 2 and 4? Both use the product and sum rules; rework them in Topic 5.3.
No misses? Go straight to the mixed unit review for longer, exam-style questions.
Come back in a week and check that your reasons are now complete.