Genetics: Inheritance, Sex Linkage, Linkage Mapping, Population Genetics and Disorders

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. In Labrador retrievers, allele B (black pigment) is dominant to b (brown, called chocolate). At a second, unlinked locus, genotype ee prevents pigment deposition in the coat, so ee dogs are yellow whatever their B genotype, while E_ dogs are black or chocolate as set by the B locus. A BbEe dog is mated with a Bbee dog. Among 48 puppies, the expected number of chocolate puppies is ____.

   Numerical answer — type the value.

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   Answer: 6

   Chocolate needs bb together with at least one E. From Bb × Bb, P(bb) = 1/4; from Ee × ee, P(Ee) = 1/2. P(chocolate) = 1/4 × 1/2 = 1/8, and 48 × 1/8 = 6. Yellow (ee) puppies make up half the litter (24) and black ones 3/8 (18). Applying the 9 : 3 : 4 ratio, which holds only for BbEe × BbEe, wrongly gives 9.

2. Two mice heterozygous at two unlinked coat-colour loci are crossed. The progeny are 92 agouti, 29 black and 39 albino. Which kind of gene interaction best explains this result?

   1. Dominant epistasis, which gives a 12 : 3 : 1 ratio
   2. Complementary gene action, which gives a 9 : 7 ratio
   3. Recessive epistasis, which gives a 9 : 3 : 4 ratio
   4. Dominant suppression, which gives a 13 : 3 ratio
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   Answer: C — Recessive epistasis, which gives a 9 : 3 : 4 ratio

   With 160 progeny one sixteenth is 10, so 92 : 29 : 39 is close to 90 : 30 : 40, i.e. 9 : 3 : 4. This is recessive epistasis: cc mice are albino whatever their agouti-locus genotype, so the A_cc (3/16) and aacc (1/16) classes merge into 4/16 albino. A 12 : 3 : 1 ratio needs a dominant masking allele, and 9 : 7 and 13 : 3 give only two phenotypes, not three.

3. In the cross AaBbCcDd × AaBbCcDd, with all four genes unlinked, what fraction of the progeny is heterozygous at exactly two of the four loci?

   1. 1/16
   2. 1/4
   3. 81/256
   4. 3/8
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   Answer: D — 3/8

   At each locus Aa × Aa gives a heterozygote with probability 1/2 and a homozygote with probability 1/2. There are C(4,2) = 6 ways to choose which two loci are heterozygous, each with probability (1/2)⁴ = 1/16, so the fraction is 6/16 = 3/8. The value 1/16 counts only one particular pair of loci, and 81/256 is the fraction showing all four dominant phenotypes.

4. Which of the following statements about the classic experiments on the genetic material are correct?

   1. In the Avery–MacLeod–McCarty experiment, treating the transforming extract with DNase abolished transformation
   2. In the Hershey–Chase experiment, most of the ³⁵S label entered the bacteria and appeared in progeny phage
   3. In the Meselson–Stahl experiment, after one generation in ¹⁴N medium all the DNA had hybrid density
   4. Griffith's experiment by itself identified DNA as the transforming principle
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   Answer: A — In the Avery–MacLeod–McCarty experiment, treating the transforming extract with DNase abolished transformation; C — In the Meselson–Stahl experiment, after one generation in ¹⁴N medium all the DNA had hybrid density

   DNase, but not protease or RNase, destroyed transforming activity, and one round of semiconservative replication makes every duplex half ¹⁵N and half ¹⁴N. In the Hershey–Chase experiment it was ³²P (DNA), not ³⁵S (protein), that entered the cells and reached the progeny. Griffith showed only that a transforming principle existed; its identity as DNA came from Avery and colleagues in 1944.

5. Which pair of observations shows that the Y chromosome determines maleness in humans but not in Drosophila?

   1. XXY humans are male, whereas XXY flies are female
   2. XO humans are female, and XO flies are also female
   3. XXY humans are female, whereas XXY flies are male
   4. XO humans are male, whereas XO flies are female
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   Answer: A — XXY humans are male, whereas XXY flies are female

   In humans a Y (carrying SRY) makes an XXY individual male despite the two X chromosomes. In Drosophila sex follows the X : A ratio, so an XXY fly (ratio 1.0) is female and an XO fly (ratio 0.5) is male, though sterile. The pairings that make XO flies female or XXY humans female are simply wrong.

6. Red–green colour blindness is X-linked recessive. A colour-blind man marries a woman with normal vision whose parents both have normal vision but whose brother is colour-blind. The probability that their first child is a colour-blind daughter is ____ (to three decimal places).

   Numerical answer — type the value.

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   Answer: 0.125

   The woman's brother got his X from their mother, who must be a carrier, so the woman is a carrier with probability 1/2. A daughter always receives the father's X, which carries the allele, and is colour-blind only if she also receives the mother's mutant X. P = P(woman is a carrier) × P(child is a daughter) × P(mother transmits the allele) = 1/2 × 1/2 × 1/2 = 1/8 = 0.125. Leaving out the 1/2 for the sex of the child gives 0.25.

7. In the snail Lymnaea peregra, shell coiling is controlled by a maternal-effect gene in which the dextral allele D is dominant to the sinistral allele d. A sinistral snail, used as the female parent, produces only dextral offspring. What is its genotype?

   1. dd, because sinistral coiling is recessive and the snail itself is sinistral
   2. DD, because every one of its offspring shows the dominant dextral coiling
   3. Dd, since its own shell reflects a dd mother and its broods reflect its own genotype
   4. It cannot be determined, because coiling direction is inherited through mitochondria
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   Answer: C — Dd, since its own shell reflects a dd mother and its broods reflect its own genotype

   With a maternal effect, a snail's own coiling is set by its mother's genotype and its broods' coiling by its own genotype. Being sinistral, it had a dd mother and so carries at least one d; producing dextral broods, it carries D. Hence it is Dd. Reading its phenotype as its genotype (dd) is the trap, and DD is impossible because it received d from its dd mother. The gene is nuclear, not mitochondrial.

8. A rare human disorder is caused by a mutation in mitochondrial DNA. Which of the following features are expected in pedigrees of this disorder?

   1. Affected women transmit the disorder to both sons and daughters
   2. Affected men do not transmit the disorder to any of their children
   3. Affected men transmit the disorder to all their daughters but to none of their sons
   4. Among the children of two unaffected carrier parents, about one quarter are affected
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   Answer: A — Affected women transmit the disorder to both sons and daughters; B — Affected men do not transmit the disorder to any of their children

   Mitochondria are inherited through the egg, so an affected mother passes mutant mtDNA to children of both sexes (severity varies with heteroplasmy), and an affected father passes it to none. Transmission from affected fathers to all daughters is the X-linked dominant pattern, and a one-quarter risk from two carrier parents is the autosomal recessive pattern.

9. A female heterozygous for three linked recessive mutations u, v and w is test-crossed to a male homozygous recessive for all three. The 2,000 progeny, classified by the alleles received from the female and listed in the order u, v, w, are: u + w 748; + v + 736; u v + 110; + + w 106; u + + 140; + v w 136; u v w 13; + + + 11. The coefficient of coincidence is ____ (to two decimal places).

   Numerical answer — type the value.

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   Answer: 0.67

   Parentals are the commonest pair (u + w, + v +) and the double crossovers the rarest (u v w, + + +). Comparing u + w with u v w shows that v has switched, so v is in the middle: u – v – w. RF(u–v) = (110 + 106 + 24)/2000 = 0.12; RF(v–w) = (140 + 136 + 24)/2000 = 0.15. Expected DCO = 0.12 × 0.15 × 2000 = 36; observed DCO = 24. CoC = 24/36 = 0.67 (interference 0.33). Leaving the DCOs out of the interval distances gives a wrong expected value.

10. A female heterozygous for three linked recessive mutations r, s and t is test-crossed, giving 1,000 progeny classified by the alleles received from the female (listed in the order r, s, t): r + + 375; + s t 371; r s + 88; + + t 86; r s t 38; + + + 36; r + t 4; + s + 2. Adding the two adjacent intervals, the map distance between the two genes at the ends of the map is ____ cM.

    Numerical answer — type the value.

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    Answer: 26

    Parentals: r + + and + s t; DCOs: r + t and + s +. Comparing r + + with r + t shows that t has switched, so t is in the middle: r – t – s. r–t = (38 + 36 + 6)/1000 = 8 cM; t–s = (88 + 86 + 6)/1000 = 18 cM. The end-to-end distance is 8 + 18 = 26 cM. Counting recombinants between r and s directly gives (88 + 86 + 38 + 36)/1000 = 24.8 cM, an underestimate because each double crossover restores the parental r–s combination.

11. In Drosophila, genes a and b are 20 map units apart as measured in females. A male heterozygous in coupling (AB/ab) is test-crossed to an ab/ab female. Which progeny are expected?

    1. 40% AB, 40% ab, 10% Ab and 10% aB, as from a female of the same genotype
    2. 50% AB and 50% ab, with no recombinant classes
    3. 25% each of AB, ab, Ab and aB, as for unlinked genes
    4. 45% AB, 45% ab, 5% Ab and 5% aB, since recombination is halved in males
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    Answer: B — 50% AB and 50% ab, with no recombinant classes

    Drosophila males do not undergo meiotic crossing over, so linkage in males is complete and only the parental gametes AB and ab are formed. The 40 : 40 : 10 : 10 split is what a female would give at 20 cM, which makes it the tempting answer. Halved recombination and independent assortment have no basis here.

12. An autosomal recessive disorder affects 1 in 3,600 newborns in a population at Hardy–Weinberg equilibrium. Using the exact expression 2pq, the expected number of heterozygous carriers among 3,600 people of this population is ____.

    Numerical answer — type the value.

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    Answer: 118

    q² = 1/3600, so q = 1/60 and p = 59/60. Carrier frequency 2pq = 2 × (59/60) × (1/60) = 118/3600, so 118 carriers are expected among 3,600 people, against 1 affected person. The shortcut 2q gives 120, which is only an approximation.

13. In a population at Hardy–Weinberg equilibrium, 10% of males show an X-linked recessive trait. What fraction of females are expected to be heterozygous carriers?

    1. 0.10
    2. 0.18
    3. 0.01
    4. 0.09
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    Answer: B — 0.18

    For an X-linked gene the frequency of affected males equals q, so q = 0.10 and p = 0.90. Carrier females are 2pq = 2 × 0.9 × 0.1 = 0.18, and affected females q² = 0.01. Taking the male frequency, 0.10, as the carrier frequency is the usual slip.

14. Which of the following processes, acting alone on an autosomal locus, change allele frequencies in a population?

    1. Inbreeding within a large population
    2. Genetic drift in a small population
    3. One generation of random mating in a population not at Hardy–Weinberg proportions
    4. Immigration from a population with a different allele frequency
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    Answer: B — Genetic drift in a small population; D — Immigration from a population with a different allele frequency

    Drift changes allele frequencies at random, and gene flow from a population with different frequencies shifts them towards the donor's values. Inbreeding moves alleles from heterozygotes into homozygotes without changing their frequencies, and one round of random mating restores p², 2pq and q² while leaving p and q unchanged.

15. A phenotypically normal man has a brother with cystic fibrosis (autosomal recessive); their parents are unaffected. He marries an unaffected woman with no family history from a population in which the carrier frequency is 1/25. The probability that their first child has cystic fibrosis is 1/N, where N = ____.

    Numerical answer — type the value.

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    Answer: 150

    The man's parents must both be carriers. Among their unaffected children AA : Aa = 1 : 2, so his carrier probability is 2/3. The woman's carrier probability is 1/25. If both are carriers the child is affected with probability 1/4. P = 2/3 × 1/25 × 1/4 = 2/300 = 1/150, so N = 150. Using 1/2 for the man gives 1/200, the common error.

16. A phenotypically normal woman has had two children with Down syndrome, and her own karyotype shows 45 chromosomes. What is the most likely explanation?

    1. She carries a balanced Robertsonian translocation, such as one joining chromosomes 14 and 21
    2. She has Turner syndrome, which raises the rate of nondisjunction of chromosome 21
    3. She is mosaic for trisomy 21, which appears as a missing chromosome in her karyotype
    4. Her eggs undergo nondisjunction because of advanced maternal age at each conception
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    Answer: A — She carries a balanced Robertsonian translocation, such as one joining chromosomes 14 and 21

    A balanced Robertsonian carrier has 45 chromosomes because two acrocentric chromosomes (often 14 and 21) are fused, yet has a normal amount of genetic material and is unaffected; her gametes can carry the fused chromosome plus a free 21, giving translocation Down syndrome with a high recurrence risk. Maternal age is the tempting answer, but it explains free trisomy and not a 45-chromosome mother. Turner syndrome (45,X) has its own phenotype, and trisomy mosaicism adds a chromosome rather than removing one.

17. A child carries a deletion of the region 15q11–q13 on the chromosome 15 inherited from the father. Which condition is expected?

    1. Angelman syndrome
    2. Prader–Willi syndrome
    3. Cri-du-chat syndrome
    4. Fragile X syndrome
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    Answer: B — Prader–Willi syndrome

    Because of genomic imprinting, several genes in 15q11–q13 are expressed from only one parental copy; loss of the paternal region gives Prader–Willi syndrome. Angelman syndrome is the tempting answer but results from loss of the maternal copy, which carries the maternally expressed UBE3A. Cri-du-chat is a 5p deletion, and fragile X is a CGG expansion in FMR1.

18. Which pairing of model organism and the genetic principle first demonstrated with it is correct?

    1. Neurospora crassa — one gene–one enzyme, from mutants unable to grow on minimal medium
    2. Caenorhabditis elegans — the first linkage map, from three-point crosses of eye-colour mutants
    3. Drosophila melanogaster — DNA as the genetic material, from transformation of its embryos
    4. Saccharomyces cerevisiae — semiconservative replication, from density labelling of its DNA with ¹⁵N
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    Answer: A — Neurospora crassa — one gene–one enzyme, from mutants unable to grow on minimal medium

    Beadle and Tatum used Neurospora mutants that needed a supplement to grow on minimal medium to tie single genes to single enzymes. The first linkage map came from Drosophila (Sturtevant), DNA as the genetic material from bacteria and phage, and semiconservative replication from E. coli (Meselson and Stahl).