Bio3TC17 Population Genetics QA

BioF63TC17-Population genetics-QA

Population Genetics

S1 A farmer has 2000 goats. 1500 of the goats are brown coated and the rest is white coated. Allele ‘A’ for the brown coat is dominant, while allele ‘a’ for the white coat is recessive. Assume that the population of goats are at genetic equilibrium and breeds randomly.

(a) Calculate the dominant allele and recessive allele frequency of the population. [2m]

(b) Calculate the genotype frequency in the F1 generation if the cows are left to breed randomly and sexually. [3m]

(c) If the farmer adds another 1000 homozygous dominant brown goats into the population, calculate the new dominant and recessive allele frequency. [5m]

S2 Resistance against a poison in a population of bats is controlled by the dominant allele D. 84% of the population in a test area is found to be resistant against the poison.

(a) What is the allele frequency for the dominant and recessive allele in the population if the population of bats are at genetic equilibrium? [3m]

(b) Calculate the expected number of bats with genotype DD, Dd and dd in a population of 500 bats in the test area. [4m]

(c) If all the bats in the population of 500 bats that does not have resistance against the poison are killed, what is the frequency for bats without resistance that are born in the next generation if the remaining bats in this generation breed randomly and sexually? [3m]

S3 In a population of rats, 36% have a brown coat. The yellow coat is controlled by the dominant allele K and the brown coat is controlled by the recessive allele k.

(a) Calculate the frequency of the dominant and recessive alleles. [5m]

(b) The gene for the yellow coat in rats is a lethal gene. What is meant by lethal gene? [2m]

(c) What happens to allele frequency for the brown and yellow coat in rats in the generation if the parental generation is at genetic equilibrium, breeds sexually and randomly? Explain your

answer. [3m] (STPM 2001

P2)

S4 (a) Explain what is meant by gene pool, allele frequency and genotype frequency. [6m]

(b) By using an example, show how genotype frequency in the F1 generation that is produced from a population at genetic equilibrium can be determined. [9m] (Clone STPM 1991 P1 07)

S5 (a) State the Hardy-Weinberg Law. [1m]

In a type of plant, the red coloured flower is controlled by the dominant allele (R), while the recessive allele (r) controls the white flower. A farmer bought 1000 plants from a randomly reproducing population. Of the 1000 plants, only 40 plants produced white flowers, while the rest produced red flowers.

(i) What is meant by randomly reproducing population? State the importance of a randomly reproducing population. [3m]

(ii)What are the requirements for a population to be at genetic equilibrium? [4m]

(iii) Calculate the dominant and recessive allele frequency in population of 1000 plants. [5m]

(iv) Calculate how many of the red flowered plants are heterozygote. [2m] Clone STPM 1997

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Genotype FrequencyRR 980Rr 840rr 180


S6 A population of diploid organisms consist of 2000 individuals. The table below shows the genotypes in the population.

(a) By using the Hardy-Weinberg equation, show that the allele frequency and genotype frequency in the F1 generation is constant. [9 m]

(b) State and explain three factors that can change the constant values of allele and genotype frequency in the F1 generation. [6 m]

Clone STPM 1990

7 (a) What does the Hardy–Weinberg principle predict? [3m]......................................................................................................................................................................................................

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The table shows the frequencies of some alleles in the population of cats in three cities.

(b) White cats are deaf. Would the Hardy–Weinberg principle hold true for white cats?Explain your answer. [2m]

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(c) What is the evidence from the table that non-agouti and blotched are alleles of different genes? [1m]

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(d) Hair length in cats is determined by a single gene with two alleles. The allele for long hair (h) is recessive. The allele for short hair (H) is dominant. Use the information in the table and the Hardy–Weinberg equation to estimate the percentage of cats in London that are heterozygous for hair length. Show your working. [2m]

8 Huntington’s disease is a human inherited condition resulting in gradual degeneration of nerve cells in the brain. It is caused by a dominant allele but usually no symptoms are evident until the person is at least 30 years old. It is very rare in most populations. However, in one isolated area in Venezuela, 48% of the population possess a genotype which gives rise to Huntington’s disease. Many of the inhabitants of this area can trace their origins back to a common ancestor 200 years ago.

(a) Use the Hardy-Weinberg equation to estimate the percentage of this Venezuelan population which is heterozygous for Huntington’s disease. Show your working.

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(b) Suggest why (i) there is such a high incidence of Huntington’s disease in this population;

[3m]

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(ii) Huntington’s disease has not been eliminated from this population by natural selection.

[3m]

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AQA June 2002

6 (a) (q2 = 0.52 / q = 0.72) (p = 1 – 0.72 = 0.28) p + q = 1 / p2 + 2pq + q2 = 1 ; Answer = 2pq / use of appropriate numbers; Answer = 40%; 3

(b) (i) Any three from: (MARK AS A WHOLE) Small founder population / common ancestor; Genetic isolation / small gene pool / no immigration / no migration / in-breeding; High probability of mating with person having H-allele;

(ii)Reproduction occurs before symptoms of disease are apparent; Genetic argument – Hh x hh _ 50% / Hh x Hh _ 75% affected offspring; No survival / selective disadvantage; 3 max Ignore ‘survival of the fittest’

AnswersS1 (a)According to the Hardy-Weinberg equation, q2 is the frequency of homozygous recessive genotype and q is recessive allele frequency.

(2000—1500) So, q2 = --------------------- , q=0.5

2000 According to the Hardy-Weinberg equation, p-q =1, so, p = 1-q = 1- 0.5 = 0.5 (b) According to the Hardy-Weinberg equation,

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p2 + 2pq + q2 = 1, in which, p2 = frequency of homozygote dominant genotype = (0.5)2 = 0.25 2pq = heterozygous genotype frequency = 2(0.5)(0.5) = 0.5 q2 = frequency of homozygote recessive genotype = (0.5)2 = 0.25 (c)

The number of goats with the following genotypes in the population: Homozygous dominant = 0.25 x 2000 = 500 Heterozygous = 0. 5 x 2000 1000 Homozygous recessive 0.25 x 2000 = 500 With the addition of 1000 homozygous dominant goats, the total homozygous dominant goats is 1500.

Genotype AA Aa aa Total alleleNumber of organisms 1500 1000 500 —Number of A allele 2 x 1500 =3000 x 2 x 1000 = 1000 — 4000Number of a allele — x 2 x 1000 = 1000 2 x 500 = 1000 2000

Total allele 3000 2000 1000 6000 4000

Dominant allele frequency, p = ---------- = 0.67 6000

2000 Recessive allele frequency, q = ------------- = 0.33 6000

S2 (a) According to the Hardy-Weinberg equation, q2 is the frequency of homozygous recessive genotype and q is recessive allele frequency. (84) Number of bats without resistance to poison = 1 - —---- = 0.16 100 So, q2 = 0.16, q = 0.4

According to the Hardy-Weinberg equation, P + q =1, so, p= 1—q= 1-0.4=0.6

(b) According to the Hardy-Weinberg equation, p2 + 2pq + q2 = 1, where, p2 = homozygote dominant genotype (DD) frequency = (0.6)2 = 0.36 2pq = heterozygous genotype (Dd) frequency = 2(0.6)(0.4) = 0.48 q2 = homozygote recessive genotype (dd) frequency = (0.4)2 = 0.16 Number of homozygote dominant genotype (DD) bats = 0.36 x 500 = 180 bats Number of heterozygote genotype (Dd) bats = 0.48 x 500 = 240 bats

(c)

Number of homozygote recessive genotype (dd) bats = 0.16 x 500 = 80 bats 600 Dominant allele (D) frequency, p = ----- = 0.7 14 840 240 Recessive allele (d) frequency, q = ----- = 0.286 840 Frequency of bats without resistance in next generation = q2 = (0.286)2 = 0.082

S3 (a) According to the Hardy-Weinberg equation, p2+2pq+q2=1 and p + q=1,where, p = dominant allele (K) frequency, q = recessive allele (k) frequency,

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Genotype DD Dd dd Total alleleNumber of organism 180 240 All dead —Number of allele 2x180=360 ½ x 2 x 240 =240 — 600Number of allele ½ x 2 x 240 =240 — 240Total allele 360 480 -- 840


p2 = homozygote dominant (KK) genotype frequency, 2pq = heterozygous genotype (Kk) frequency, q2 homozygote recessive (kk) genotype frequency. So, q2 = 0.36, q = 0.6 According to the Hardy-Weinberg equation, p + q=1, so, p =1-q=1 -0.6=0.4

(b) The lethal gene is a genotype that results in the death of the organism. In rats, the homozygous dominant genotype ‘KK’ for yellow coat is a lethal gene as it causes the death of the rat.

(c) The allele frequency for the brown and yellow coat alleles in the F1 generation will change.

The death of rats with the lethal gene (KK) causes gene to flow out of the population. The F, generation formed is not at genetic equilibrium, so the Hardy-Weinberg equation cannot be used for the population produced.

S4 (a) • Gene pool consists of all the alleles at all the gene loci in all the individuals of a population that breeds sexually. • Allele frequency is the ratio between the total number of an allele in a population to the total number of alleles in all the gene loci in a population. • Allele frequency = total number of an allele in population total number of alleles in population • Genotype frequency is the ratio between the total number of organisms with a certain genotype to the total number of organisms in a population. • Genotype frequency = total number of organisms with certain genotype total number of organisms • A natural population consists of 10000 individuals. A selected trait is controlled by two alleles, dominant allele (A) and recessive allele (a). Out of the 10 000 individuals, 6000 have genotype AA, 2000 have genotype Aa and 2000 have genotype aa.

Gene pool = 2 (6000) + 2 (2000) + 2 (2000) = 20 000 alleles

2(6000) + ½ (2 x 2000) Frequency of dominant allele (A) = --------------------------------

2(6000) + 2(2000) + 2(2000)

= 14000 20000

= 0.7

½ (2 x 2000) + 2(2000)Frequency of recessive allele (a) = ----------------------------------

2(6000) + 2(2000) + 2(2000) = 6000

20000 = 0.3

— Genotype frequency of AA = 6000 = 0.6 10 000

— Genotype frequency of Aa = 2000 = 0.2 10 000

— Genotype frequency of aa = 2000 = 0.2 10 000

(b) • According to the Hardy-Weinberg equation, p + q = 1, p2 + 2pq + q2 = 1 p = dominant allele frequency q = recessive allele frequency p2 = homozygous dominant genotype frequency q2 = homozygous recessive genotype frequency 2pq = heterozygous genotype frequency

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• In a population at genetic equilibrium, a trait is controlled by dominant allele (B) and recessive allele(b). In a population of 10000 organisms, 1600 organisms have genotype ‘bb’.

1600 So, q2 = --------- = 0.16

10000 q = 0.4 p+q = 1 p = 1-0.4=0.6

• If this population is left to reproduce randomly and sexually, the genotype frequency in the F1

generation produced are: Homozygous dominant (BB) genotype frequency =p2 = (0.6)2 = 0.36 Homozygous recessive (bb) genotype frequency = q2 = (0.4)2 = 0.16 Heterozygous (Bb) genotype frequency = 2pq = 2(0.4)(0.6) = 0.48 • For the population to be at genetic equilibrium, — population size is large — selection does not occur and mating between the organisms in the population is random — migration, emigration and immigration does not occur — genetic drift does not occur — no mutation occurs

S5 (a) The Hardy-Weinberg law states that allele frequency and genotype frequency of a population at genetic equilibrium remains constant from generation to generation.

(i) Randomly breeding population means that each organism in the population has an equal chance to mate and reproduce with another organism in the same population. Selection does not occur; each genotype is equally likely to pass genes to the offspring. If breeding is not random (selection occurs), the frequency of selected alleles will increase in subsequent generations.

(ii)• Population size is large. • Mating between organisms in the population is random. • Genetic drift does not occur. • Mutation does not occur. • Migration (emigration and immigration) does not occur. Gene flow into or out of the population does not occur.

(iii) According to the Hardy-Weinberg equation, p2 + 2pq + q2 = 1 q2 is the frequency of homozygous recessive genotype and q is the recessive allele (r) frequency. 40 Frequency of white flower plants in the population = —— = 0.04 1000

So, q2 = 0.04, q = 0.2 According to the Hardy-Weinberg equation, p + q =1, p = dominant allele (R) frequency so, p = 1—q = 1—0.2 = 0.8

(iv) According to the Hardy-Weinberg equation, 2pq = heterozygote genotype frequency, Number of heterozygote red flower plants = 2(0.8)(0.2) x 1000 = 320 plants.

S6 (a) According to the Hardy-Weinberg equation, p2 + 2pq + q2 = 1 and p + q = 1, where, p = dominant allele frequency, q = recessive allele frequency, p2 = homozygote dominant genotype frequency, 2pq = heterozygous genotype frequency, q2 = homozygote recessive genotype frequency. For the given population, 2800

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Dominant allele (R) frequency, p = -----— = 0.7 4000

1200 Recessive allele (r) frequency = q = ---------- = 0.3

4000 9 80Homozygote dominant genotype frequency = —---- = 0.49 2000 840Heterozygous genotype frequency = —----- = 0.42 2000 180 Homozygote recessive genotype frequency = --------- = 0.09 2000 If the population breeds sexually and randomly and produces 1000 organisms in the F1 generation, the number of organisms with specific genotype are as shown follows:

Homozygous dominant (RR) = p2 x 1000 = (0.7)2 x 1000 = 490 Heterozygous (Rr) = 2pq x 1000 = 2(0.7)(0.3) x 1000 = 420 Homozygous recessive (rr) = q2 x 1000 = (0.3)2 x 1000 = 90 • 1400

Dominant allele (R) frequency, p = ---------- = 0.7 2000 Recessive allele (r) frequency, q = = 0.3 2000 490Homozygote dominant genotype frequency = = 0.49

420Heterozygous genotype frequency = —---- = 0.42 1000

Homozygote recessive genotype frequency = —~— = 0.09Thus, the allele frequency and genotype frequency in the parental and F1 generations are constant.

(b) (i) Population is small. • In a small population, the death of a small number of organisms caii

cause large changes to the allele frequency and genotype frequency of the population.• In a large population, the same amount of deaths cause little effect to the allele frequency and

genotype frequency of the population.

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Genotype RR Rr rr Total alleleNumber of organisms 980 840 180 —Number of R allele 2 x 980 =1960 0.5x 2 x 840 =840 — 2800

Number of r allele 0 0.5x 2 x 840 =840 2 x 180

=360

1200

Total allele 1960 1680 360 4000


• In a small population, the effect of genetic drift (bottleneck effect or founder’s effect) is greater compared to a bigger population.(ii) Mating is non-random. • Selection occurs in the population. • Allele frequency of the allele that produces phenotype that increases survival and reproductive chances increases in subsequent generations.(iii) Migration into or out of the population, emigration or immigration occurs. Gene flow occurs resulting in changes in the allele frequency and genotype frequency.(iv) Genetic drift occurs. Population size is drastically reduced by random events such as natural disaster. The remaining population has different allele frequency and genotype frequency from the original population. (v) Mutation occurs. Mutation causes production of new alleles or new genotype in a population. This causes changes in the allele frequency or genotype frequency. (Any three)

Genotype RR Rr rr Total allele

Number of organisms 490 420 90 —Number of R alleles 2x490=980 1x2x420 — 1400

Number of r allele =420 x 2 x 420 =420

2 x 90=980

600

Total allele 980 840 180 2000

7 (a) The frequency/proportion of alleles (of a particular gene); Will stay constant from one generation to the next/over generations / no genetic change over time; Providing no mutation/no selection/population large/population genetically isolated/mating at random/no migration;

The three principles for marking are: What feature What happens to it Providing . . . Accept: genotype/explanation of genotype Accept: alternative wording, e.g. there is no gene flow/genetic drift for genetically isolated. (b) White/deaf cats unlikely to survive/selected against;

Will not pass on allele (for deafness/white fur) (to next generation)/will reduce frequency of allele; Accept: alternative wording, e.g. have a disadvantageous phenotype Neutral: will not breed (c) In Paris/London frequencies (of these alleles) add up to more than 1; Can be shown by correct figures to be more than 1 e.g. 0.71 + 0.78 = 1.49 Accept: more than100% (d) Two marks for correct answer of 44(.22);;

One mark for incorrect answer in which p/frequency of H determined as 0.67 and q/frequency of h as 0.33

OR Answer given as 0.44(22);

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Bio3TC17 Population Genetics QA