Chapter 5: Population Genetics PDF

CHAPTER 5 POPULATION GENETICS (Hours: 1L + 3T)  CHAPTER 5: POPULATION GENETICS 5.1 Gene Pool Concept 5.2 Hardy-Weinberg Law UPS 2 PSPM 1 7 MCQ 6 marks 2  LEARNING OUTCO...

CHAPTER 5 POPULATION GENETICS (Hours: 1L + 3T)  CHAPTER 5: POPULATION GENETICS 5.1 Gene Pool Concept 5.2 Hardy-Weinberg Law UPS 2 PSPM 1 7 MCQ 6 marks 2  LEARNING OUTCOMES 5.1 Gene pool concept a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium. (C3) 3 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Population Genetics The study of the genetic variation in a population of a species and the evolutionary forces that act on it.. Population A group of individuals of the same species living together in the same place at the same time. that can freely interbreed with each other and produce fertile offspring in nature. 4 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Genetic Variation Examples 5 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Gene Pool The total of all alleles for all loci present in a population. Consist of all copies of every type of allele at every locus in all members of the population. 6 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Calculations Involve Allele frequency Dominant allele frequency Recessive allele frequency Genotype frequency Homozygous dominant genotype frequency Homozygous recessive genotype frequency Heterozygous genotype frequency Phenotype frequency Dominant phenotype frequency Recessive phenotype frequency = homozygous recessive genotype frequency 7 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium ATTENTION Attention **Tips Please follow decimal places stated in the question for your calculation and answer. If the question doesn’t state decimal places, then follow this guide: ⮚ 1-99 1 decimal place ⮚ 100-999 2 decimal places ⮚ 1000 and above 3 decimal places 8 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Gene Pool Example: In a population of 80 rabbits, coat colour character is controlled by a gene. The 2 alleles involved are B & b. 40 black rabbits with the genotype BB 30 black rabbits with the genotype Bb 10 white rabbits with the genotype bb Calculate the gene pool Total number of alleles = 2 X total number of individuals = 2 X 80 = 160 alleles 9 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Allele Frequency The proportion of a specific allele relative to all alleles of one gene in a population. Example : Allele frequency of purple flower, R or allele frequency of white flower, r Formula: total number of allele R Frequency allele of purple flower, (R) = total number of alleles total number of allele r Frequency allele of white flower, (r) = total number of alleles 10 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 1 : Allele Frequency In a population of 80 rabbits, there are : 40 black rabbits (genotype BB) has 80 allele B 30 black rabbits (genotype Bb) has 30 allele B and 30 allele b 10 white rabbits (genotype bb) has 20 allele b 11 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 1 : Allele Frequency In a population of 80 rabbits, there are : 40 black rabbits (genotype BB) has 80 allele B 30 black rabbits (genotype Bb) has 30 allele B and 30 allele b 10 white rabbits (genotype bb) has 20 allele b Total number of alleles = 2 X 80 = 160 Alleles total number of black fur allele Frequency of allele for black fur (B), = total number of alleles (40 x 2) + 30 110 = = 160 160 = 0.7 12 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 1 : Allele Frequency In a population of 80 rabbits, there are : 40 black rabbits (genotype BB) has 80 allele B 30 black rabbits (genotype Bb) has 30 allele B and 30 allele b 10 white rabbits (genotype bb) has 20 allele b Total number of alleles = 2 X 80 = 160 alleles total number of white fur allele Frequency of allele for white fur (b) = total number of alleles (10 x 2) + 30 50 = = 160 160 = 0.3 13 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 2 : Allele Frequency In a population of 500 wildflowers, allele A is dominant allele for red flower and allele a is recessive allele for white flower. There are 320 homozygous dominant plant, 160 heterozygous and 20 white flowers. Calculate:- (320 x 2) + 160 800 a) Frequency of dominant allele (A) = = 500 x 2 1000 = 0.80 b) Frequency of recessive allele (a) (20 x 2) + 160 200 = = 500 x 2 1000 = 0.20 14 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Genotype Frequency The proportion of a particular genotype relative to all genotypes of the gene in a population. Example: Frequency of homozygous dominant genotype Frequency of homozygous recessive genotype Frequency of heterozygous genotype 15 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Genotype Frequency Formula: Frequency of Total number of homozygous dominant genotype homozygous dominant = genotype Total number of individuals Frequency of Total number of homozygous recessive genotype homozygous recessive = Total number of individuals genotype Frequency of Total number of heterozygous genotype heterozygous genotype = Total number of individuals This formula is applied when: - The number of organisms for all three genotypes are given. 16 - The population does not meet the Hardy-Weinberg condition. Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 1 : Genotype Frequency In a population of 80 rabbits, there are : 40 black rabbits (genotype BB) 30 black rabbits (genotype Bb) 10 white rabbits (genotype bb) Frequency of homozygous dominant genotype (BB), number of homozygous dominant individuals 40 = = = 0.5 total number of individuals 80 Frequency of heterozygous genotype (Bb), number of heterozygous individuals 30 = = = 0.4 total number of individuals 80 Frequency of homozygous recessive genotype (bb), number of homozygous recessive individuals 10 = = = 0.1 total number of individuals 80 17 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 2 : Genotype Frequency In a population of 500 wildflowers, allele A is dominant allele for red flower and allele a is recessive allele for white flower. There are 320 homozygous dominant plant, 160 heterozygous and 20 white flowers. Calculate:- a) Frequency of homozygous dominant genotype (AA) = 320 = 0.64 500 b) Frequency of heterozygous genotype (Aa) = 160 = 0.32 500 c) Frequency of homozygous recessive genotype (aa) = 20 = 0.04 500 18 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Phenotype Frequency The proportion of a particular phenotype relative to all phenotypes in the population. E.g: Dominant phenotype frequency (sum of homozygous dominant & heterozygous individuals) Formula: Dominant phenotype frequency: total number of dominant phenotype = total number of individuals total number of (homozygous dominant + heterozygous) = total number of individuals 19 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 1 : Phenotype Frequency The proportion of a particular phenotype relative to total number of individuals in the population a) Genotype BB = 40 rabbits (Black) b) Genotype Bb = 30 rabbits (Black) c) Genotype bb = 10 rabbits (White) E.g: Phenotype frequency of black rabbit = (40 + 30) / 80 = 0.9 Phenotype frequency of white rabbit = 10 / 80 = 0.1 20 Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies, genotype frequencies and genetic equilibrium Example 2 : Phenotype Frequency In a population of 500 wildflowers, allele A is dominant allele for red flower and allele a is recessive allele for white flower. There are 320 homozygous dominant plant, 160 heterozygous and 20 white flowers. Calculate:- 320 + 160 a) Dominant phenotype frequency = = 0.96 500 b) Recessive genotype frequency = 20 = 0.04 500 21  LEARNING OUTCOMES 5.2 Hardy-Weinberg Law a) State the Hardy-Weinberg Law (C1) b) Explain the five assumptions of Hardy-Weinberg Law for genetic equilibrium (C3) i. Large population size ii. Random mating iii. No mutation iv. No migration and, v. No natural selection c) Calculate allele and genotype frequencies (C3) 22 Learning Outcomes : 5.2 a) State the Hardy-Weinberg Law HARDY-WEINBERG LAW in 1908, G. H. Hardy (an English mathemathician) and W. Weinberg (a German physician) independently identified a mathematical relationship between alleles and genotypes in populations. This relationship has been called: The Hardy-Weinberg Equilibrium (The Hardy-Weinberg Law) and it concerns allele frequency 23 Learning Outcomes : 5.2 a) State the Hardy-Weinberg Law HARDY-WEINBERG LAW In a population that exists in genetic equilibrium, the alleles and genotypes frequencies will remain constant from one generation to next generation under certain conditions are met. 24 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW Assumptions of Hardy-Weinberg Law A population is said to be in genetic equilibrium if these five conditions are met. 5 conditions for Hardy-Weinberg Equilibrium. i. Large population size ii. Random mating iii. No mutation iv. No migration v. No natural selection 25 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW i. Large population size A large population produces a large sample of successful gametes. So that chance in fluctuation in the gene that can cause phenotype frequencies to change over time can be avoided (genetic drift can be avoided). 26 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW ii. Random mating Each individual in a population has an equal chance of mating with any individual of the opposite sex. Mate at random & does not select their mate. If individuals preferentially choose mates with certain genotypes, including close relatives (inbreeding), random mixing of gametes does not occur; causing changes in genotype frequencies. 27 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW iii. No mutation There must be no net mutations. Allele and genotype frequencies may change through the loss or addition of alleles through mutation. The gene pool is modified if mutations occur or if entire genes are deleted or duplicated. 28 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW iv. No migration There can be no migration between individuals into or out of the population. So, no exchange of alleles with other populations that might have different allele frequencies. 29 Learning Outcomes : 5.2 b) Explain the five assumtions of Hardy-Weinberg Law for genetic equilibrium. HARDY-WEINBERG LAW v. No Natural Selection No individuals will have a reproductive advantage over another individual. All individuals have the same potential to reproduce sexually. If natural selection occurs, certain phenotypes are favored over the others Certain genotypes has greater fitness / more advantage to survive; this will cause changes in allele frequencies 30 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION 1) Allele frequencies p+q=1 p = frequency of dominant allele q = frequency of recessive allele 2) Genotype frequencies p2 + 2pq + q2 = 1 p2 = frequency of homozygous dominant genotype 2pq = frequency of heterozygous genotype q2 = frequency of homozygous recessive genotype 31 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. SUMMARY These frequencies can be calculated by using these formulae : Allele frequency p+q=1 Genotype frequency p2 + 2pq + q2 = 1 Dominant phenotype frequency p2 + 2pq Recessive phenotype frequency q2 32 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips Determine the type of information. They may be in the form of: ✔ number of individuals ✔ percentage ✔ ratio ✔ probability ✔ fractional ✔ total number of allele 33 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips Identify any information of recessive traits, we always begin calculations by calculating the frequency of homozygous recessive genotype (q2). NEVER START calculation with information of dominant trait. If allele frequency is required, answer should be either in the form of: p (dominant) q (recessive) 34 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips If genotype frequency is required, answer should be either in the form of: p2 q2 2pq If number of heterozygous individual or the carrier is required, answer should be: 2pq x number of individual **Rounding number rules applied: If the next digit is less than 5, keep the rounding digit the same. 35 If the next digit is 5 or more, increase the rounding digit by 1. Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips If number of individual with dominant trait is required, answer should be: (p2 + 2pq) x number of individual If heterozygous or carrier frequencies is required, answer should be : 2pq If percentage of heterozygous or carrier is required, answer should be : 2pq x 100 36 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips Must label p, q, p2, 2pq, and q2 Please follow decimal places stated in the question for your calculation and answer. If the question doesn’t state decimal places, then follow this guide: ⮚ Tens – 1 decimal place ⮚ Hundreds – 2 decimal places ⮚ Thousands and above 3 decimal places 37 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. HARDY-WEINBERG EQUATION **Tips Must label p, q, p2, 2pq, and q2 Please follow decimal places stated in the question for your calculation and answer. If the question doesn’t state decimal places, then follow this guide: ⮚ Tens – 1 decimal place ⮚ Hundreds – 2 decimal places ⮚ Thousands and above 3 decimal places 38 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 1 In a population of 300 plants, the total number of recessive allele is 60. Calculate the recessive and dominant allele frequency. Frequency of recessive allele, q = total number of recessive allele total number of alleles Frequency of recessive allele, q = 60 = 0.10 600 p+q=1 Frequency of dominant allele, p = 1 - 0.1 = 0.90 39 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 2 A wildflower population with 2 varieties contrasting in flower colour. An allele for pink flowers (A) are completely dominant to an allele for white flowers, (a). Frequency of recessive allele, a is 20%. Find the frequency of dominant allele, A. Frequency of recessive allele, q = 20% = 0.20 p + q = 1; Frequency of dominant allele, p = 0.80 40 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 3 In a gene pool, there is 60% allele H. Calculate:- Frequency of dominant allele, p = 60 100 = 0.60 Frequency of recessive allele, q = 1 - 0.6 = 0.40 41 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 3 In a gene pool, there is 60% allele H. Calculate:- The percentage of heterozygote in that population? ⮚ Frequency of dominant allele, p = 0.60 ⇒p+q=1 ⮚ Frequency of recessive allele, q = 1 – 0.6 = 0.40 ⮚ Frequency of heterozygous genotype, 2pq = 2(0.6) (0.4) = 0.48 ⮚ Percentage of heterozygous genotype = 0.48 X 100 = 48 % 42 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. Learning Outcomes : 5.1 (a) Explain population genetics, gene pool, allele frequencies and genetic equilibrium Example 4 In a population, 84% have the ability to roll their tongue. Ability to roll the tongue is determined by a dominant allele, L. The inability to roll the tongue is due to recessive allele, l. Find the allele frequency for L and l. Find the genotype frequency for homozygous dominant, heterozygous & homozygous recessive. The percentage of individuals that cannot roll their tongue : = 100 % - 84 % = 16 % If there are 100 person, 16 individuals cannot roll their tongue = 16 = 0.16 100 43 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. Example 4 In a population, 84% have the ability to roll their tongue. Ability to roll the tongue is determined by a dominant allele, L. The inability to roll the tongue is due to recessive allele, l. Frequency of homozygous recessive genotype, q2 = 16 = 0.16 100 Frequency of recessive allele, q = √ 0.16 = 0.40 p+q=1 Frequency of dominant allele, p = 1 - 0.4 = 0.60 Frequency of homozygous dominant genotype, p2 = (0.6)2 = 0.36 Frequency of heterozygous genotype, 2pq = 2 (0.6) (0.4) = 0.48 44 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. Example 5 In Kuala Lumpur, one out of 10,000 babies were born with albino disease. What is the frequency of albino and normal individual in KL? How many people are actually a carrier to the albino recessive allele? Show your calculation and answer in four decimal places. Frequency of homozygous recessive genotype, q2 1 = = 0.0001 10,000 Frequency of recessive allele, q. = √ 0.0001 = 0.0100 // 0.01 p+q=1, Frequency of dominant allele, p = 1 – q = 1 – 0.0100 = 0.9900 // 0.99 Frequency of normal individual, p2 + 2pq = (0.9900)2 + 0.0198 = 0.9801 + 0.0198 = 0.9999 45 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. Example 5 In Kuala Lumpur, one out of 10,000 babies were born with albino disease. What is the frequency of albino and normal individual in KL? How many people are actually a carrier to the albino recessive allele? Show your calculation and answer in four decimal places. Frequency of heterozygous genotype, 2pq = 2(0.9900) (0.0100) = 0.0198 Number of carrier individuals = 0.0198 X 10,000 = 198 46 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 6 The allele y occurs with a frequency of 0.8 in a population of clams. Give the frequencies of the genotypes YY, Yy and yy. Show your calculation. Frequency of recessive allele, q = 0.8 p + q = 1, Frequency of dominant allele, p = 1 - q = 1 – 0.8 = 0.2 47 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXAMPLE 6 Now, use the other equation to get the genotype frequencies Frequency of homozygous dominant genotype, p2 = (0.2)2 = 0.04 Frequency heterozygous genotype, 2pq = 2(0.2)(0.8) = 0.32 Frequency of homozygous recessive genotype, q2 = (0.8)2 = 0.64 48 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. QUESTION 7 In a population of 500 wildflowers, 20 are white flowers or having recessive phenotype and the rest have dominant phenotype. What are the genotype frequency of Aa individuals, frequency of dominant phenotype and frequency of recessive phenotype individuals ? 49 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. QUESTION 7 Solution Frequency of homozygous recessive genotype, q2 = 20/500 =√0.04 Frequency of allele, q = 0.20 p+q=1 Frequency of dominant allele, p = 1 – q = 1 – 0.20 = 0.80 50 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. QUESTION 7 Solution Frequency of homozygous dominant genotype, p2 = (0.80)2 = 0.64 Frequency of heterozygous genotype, 2pq = 2 (0.80)(0.20) = 0.32 Frequency of dominant phenotype = p2 + 2pq = 0.64 + 0.32 = 0.96 51 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. QUESTION 8 Albinism in mice is a recessive allele due to a mutant gene. In a population of mice, 16% are albino. Assume that the population in the Hardy-Weinberg equilibrium, what is the frequency of carrier? Frequency of homozygous recessive genotype, q2 = 0.16 = √0.16 Frequency of recessive allele, q = 0.40 p+q=1 Frequency of dominant allele, p = 1- p = 1- 0.40 = 0.60 So, frequency of the carrier, 2pq = 2 (0.60)(0.40) 52 = 0.48 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. QUESTION 9 In a population of 400 people who dwell on a Pacific island, there are 16 people with homozygous recessive trait. What is the number of heterozygous individual? Frequency of homozygous recessive genotype, q2 = 16/400 = √0.04 Frequency of recessive allele, q = 0.20 p+q=1 Frequency of dominant allele, p = 1 - q = 0.80 Number of heterozygous people in the population = 2pq x 400 = 2(0.800)(0.200) x 400 = 128 people 53 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 1 Thalassemia is an inherited autosomal recessive blood disease in human. Thalassemia major is a severe form of anaemia due to homozygous recessive condition, while thalassemia minor is a mild form of anaemia shown in individuals with heterozygous genotype. In a population of 12 750, two individuals are suffering from thalassemia major. a) Determine the frequencies of the dominant and recessive alleles in the population. (Calculate up to five decimal places). b) How many individuals would be suffering from thalassemia minor in the population? c) If 1000 normal individuals migrated out of the population, what would be the new frequencies of the dominant and recessive alleles? 54 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 1 a) Determine the frequencies of the dominant and recessive alleles in the population. (Calculate up to five decimal places). Frequency of homozygous recessive genotype, q2 = 2/12750 = √0.00016 Frequency of recessive allele, q = 0.01265 p+q=1 Frequency of dominant allele, p = 1 – q = 1 – 0.01265 = 0.98735 55 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 1 b) How many individuals would be suffering from thalassemia minor in the population? Frequency of individuals with thalassemia minor, 2pq = 2 (0.98735)(0.01265) = 0.02498 Number of individuals with thalassemia minor; = 0.02498 X 12 750 = 318 individuals 56 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 1 c) If 1000 normal individuals migrated out of the population, what would be the new frequencies of the dominant and recessive alleles? New population =12 750 – 1000 = 11 750 Frequency of recessive allele = Total recessive allele Total of all alleles = (2x2) + 318 11 750 x 2 = 322 23 500 = 0.014 57 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 1 c) If 1000 normal individuals migrated out of the population, what would be the new frequencies of the dominant and recessive alleles? New population =12 750 – 1000 = 11 750 Frequency of dominant allele = Total dominant allele Total of all alleles = (11 430 x2) + 318 11 750 x 2 = 23 178 23 500 = 0.986 58 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 2 In a population, 1101 samples were MM, 1496 were MN and 503 were NN. (Calculate up to four decimal places). a) Calculate the allele frequencies of M & N b) Calculate the genotype frequencies of MM, MN and NN a) Calculate the allele frequencies of M and N Frequency of allele M = (1101 x 2) + 1496 3698 =0.5965 = 3100 x 2 6200 Frequency of allele N = (503 x 2) + 1496 2502 =0.4035 = 3100 x 2 6200 59 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 2 b) Calculate the genotype frequencies of MM, MN and NN Frequency of genotype MM = 1101/3100 = 0.3552 Frequency of genotype MN = 1496/3100 = 0.4826 Frequency of genotype NN = 503/3100 = 0.1623 60 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 3 In a population of 1800 birds, a sampling shows that 1200 birds have the dominant phenotype. a) What is the genotype frequency for heterozygous individuals. b) Individuals of the population randomly mate over generations. If in a new generation the number of this population increases to 2300, how many would be expected to have homozygous dominant genotype? 61 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 3 a) What is the genotype frequency for heterozygous individuals. Number of recessive phenotype = 1800 -1200 = 600 Frequency of homozygous recessive genotype, q2 = 600 / 1800 = √0.333 Frequency of recessive allele, q = 0.577 p+q=1 Frequency of dominant allele, p = 1 - q = 1 - 0.577 = 0.423 Frequency of heterozygous genotype, 2pq = 2(0.577)(0.423) = 0.488 62 Learning Outcomes : 5.2 c) Calculate allele and genotype frequencies. EXTRA QUESTION 3 b) Individuals of the population randomly mate over generations. If in a new generation the number of this population increases to 2300, how many would be expected to have homozygous dominant genotype? Frequency of homozygous dominant genotype, p2 = (0.423)2 = 0.179 Number of birds with homozygous dominant genotype = 0.179 X 2300 = 411.7 = 412 birds 63  REFERENCES Campbell N.A , Urry L.A,Cain M.L, Wasserman S.A, Minorsky P.V, Reece J.B Biology, 12th ed. (2021), Pearson Education, Inc. Solomon E.P & Martin C.E., Martin C.W., Berg, L.R, Biology, 11th ed. (2019) Thomson Learning, Inc. ** Notes: All reference is referred to hardcopy books (including pages) A slight difference between paging in e-book and hardcopies 64  REFERENCES (FIGURE) Figure 1 - https://www.quia.com/jg/2543185list.html Figure 2 - https://www.gktoday.in/major-gene-pool-centres/ Figure 3 - https://www.palomar.edu/anthro/synthetic/synth_2.htm Figure 4 - Adapted from Campbell,11th ed., page 488 65