POGIL_The_Hardy_Weinberg_Equation_S__1_.docx

The Hardy-Weinberg Equation How can we make predictions about the characteristics of a population? Why? Punnett squares provide an easy way to predict the possible genotypes for an offspring, but it is not practi-cal to perform a Punnett square analysis on all possible combinations of all members of a population to predict what the population might look like in the future. For that we must turn to statistics. The Hardy-Weinberg equation is a tool biologists use to make predictions about a population and to show whether or not evolution is occurring in that population. Model 1 – Controlled (Selective) Mating Males Females Males Females Bb bb Bb bb bb Bb bb B b Bb bb Bb bb bb Bb bb B b Bb bb Bb bb bb Bb bb B b 1. This study source was downloaded by from CourseH on :06:06 GMT -06:00 1.How many mating pairs are illustrated in Model 1? In model 1 there’s 12 mating pairs. 2. Describe the parents in each mating pair in Model 1. Use terms such as homozygous, heterozygous, dominant, and recessive. The first group is heterozygous and homozygous, one parent is heterozygous and the other is homozygous. 3. Use two Punnett squares to determine the possible genotypes for offspring from the pairs. First couple b b B Bb Bb b bb bb The Hardy-Weinberg Equation 1 4. If each mating pair has one offspring, predict how many of the first generation offspring will have the following genotypes. BB 0 Bb 6 bb 6 5. Imagine the 24 beetles in Model 1 as a population in an aquarium tank. a. How likely is the pairing scenario in Model 1 to take place during the natural course of things within that tank? The pairing scenario in model 1 takes is very unlikely. b. Why is Model 1 labeled “Selective Mating”? Model one is labeled selective mating because they only mate with other members of the population who have a certain trait. Bb and Bb bb and bbb This study source was downloaded by from CourseH on :06:06 GMT -06:00 6. List two other pairings that might occur in the population in Model 1 if the beetles were allowed to mate naturally. If the population was allowed to mate naturally a pairing that might arise would be bb/bb and Bb/Bb. 7. If the population of beetles in Model 1 mated naturally would your prediction for the offspring in Question 4 still be valid? Explain. No I believe that my prediction would not be valid because it would not include the offspring from the other two types of pairings. 8. Discuss in your group the limitations of Punnett square predictions when it comes to large popu-lations. Summarize the key points of your discussion here. A Punnett square can only give predictions for one type of pairing.There’s no absolute way to know the each type of pairing that will naturally occur in a large population. Model 2 – Population Genetics Males Females Bb Bb bb bb Bb Bb bb bb Bb Bb bb bb Bb Bb bb bb Bb Bb bb bb Bb Bb bb bb 9. Compare the organisms in the population in Model 1 with the organisms in the population in Model 2. The organism have the same number of males and females, heterozygous and homozygous. 10. Individually match up twelve mating pairs from the population in Model 2 that might occur in a natural, random mating situation. This study source was downloaded by from CourseH on :06:06 GMT -06:00 Bb-bb-4 pairs bb-bb-4 pairs Bb-Bb- 4 pairs 11.Compare your set of mating pairs with other members of your group. Did your mating scheme match anyone else’s in the group? Yes my pairing set did match other members of my group. The Hardy-Weinberg Equation 3 Read This! When it comes to mating in natural populations with hundreds or even millions of individuals, it is dif-ficult, maybe even impossible, to think of all the mating scenarios. After several generations of leaving things up to nature, the alleles that are present in the population will become more and more randomized. Statistics can help biologists predict the outcome of the population when this randomization has occurred. If the population is particularly nonrandom to start, this randomization may take several generations. 12.How many total alleles are in the population in Model 2? The total amount of alleles in population 2 are 2. The total amount would be 2*24-48 13.What is the probability of an offspring from the Model 2 population getting a dominant allele? The probability of an offspring in model 2 getting a dominant allele is 12/48=0.25. 14.What is the probability of an offspring from the Model 2 population getting a recessive allele? The probability is 36/48=0.75 15. If p is used to represent the frequency of the dominant allele and q is used to represent the frequency of the recessive allele, then what will p + q equal? 0.25+0.75=1 16.Use your knowledge of statistics to calculate the probability of an offspring from the Model 2 population having each of these genotypes. Support your answers with mathematical equations. (Don’t forget there are two ways to get a heterozygous offspring—Bb or bB.) BB Bb bb 0.25*0.25 0.75*0.25 =0.1875 0.75*0.75 =0.0625 =0.5625 0.25*0.75=0.1875 0.1875+0.1875=0.375 Final 0.0625+0.375+0.5625=1 This study source was downloaded by from CourseH on :06:06 GMT -06:00 17.Check your answers in Question 16 by adding the three values together. Your sum should be equal to one. Explain why the sum of the three answers in Question 16 should be equal to one. The sum should be equal to one because there's only 3 possible outcomes and it represents 100% or 1.0. 18.Using p and q as variables, write formulas for calculating the probability of an offspring from a population having each of the following genotypes. BB p*p=p^2 Bb pq+qp=2pq bb q*q=q^2 Read This! The equations you have just developed, p + q = 1 and p2 + 2pq +q2 = 1, were first developed by G. H. Hardy and Wilhelm Weinberg. They represent the distribution of alleles in a population when • The population is large. • Mating is random. • All genotypes are equally likely to reproduce (there is no natural selection). • No organisms enter or leave the population (there is no immigration or emigration). • No mutations occur. In other words, the group of alleles available in the population must be very stable from generation to generation. If the distribution of genotypes in a population matches that predicted by the HardyWeinberg equation, then the population is said to be in Hardy-Weinberg equilibrium. If the distribution of genotypes in a population does not match that predicted by the Hardy-Weinberg equation, then the population is said to be evolving. 20.Consider the requirements for a population to be in Hardy-Weinberg equilibrium. In the natural world, are populations likely to be in Hardy-Weinberg equilibrium? Justify your reasoning. The conditions listed above about the Hardy-weinberg equilibrium would be very difficult or nearly impossible to find in the real world because it’s unlikely that natural selection won’t occur. 21. Sickle-cell anemia is a genetic disease. The Sickle-cell allele is recessive, but individuals with the homozygous recessive genotype (ss) often die prematurely due to the disease. This affects approximately 9% of the population in Africa. Use the Hardy-Weinberg equations to calculate the following: a. The frequency of the recessive allele in the population (q). q= √0.09 =0.30 This study source was downloaded by from CourseH on :06:06 GMT -06:00 b. The frequency of the dominant allele in the population (p). 1=0.30+p p=0.70 c. The frequency of homozygous dominant individuals in the African population. 0.7*0.7=0.49 d. The frequency of heterozygous individuals in the African population. 2*0.70*0.30+0.42 e. Based on this analysis, is the African population in Hardy-Weinberg equilibrium? Justify your answer. The members of the African population are not in Hardy-Weinberg equilibrium because the people who get sickle cell are homozygous recessive will die before they can reproduce and this is an example of natural selection. 22.Individuals with the heterozygous genotype (Ss) for Sickle-cell exhibit resistance to Malaria, a serious disease spread by mosquitoes in Africa and other tropical regions. a. Discuss with your group how this might affect the frequency of the recessive allele in the African population. Summarize your group’s conclusions here. This may affect the frequency because individuals with recessive allele have better fitness than homozygous individuals who are more likely to die from diseases like Malaria. a. How might this trait affect the values calculated in Question 21 and the population’s tendency toward Hardy-Weinberg equilibrium? The population's tendency toward Hardy-Weinberg equilibrium because the of the restanis in the recessive group will increase the q value and the p value will decrease. 23.Consider the beetle population in Model 2. Imagine a change occurred in the beetle’s ecosystem that made it easier for predators to spot the white beetles and six of the white beetles were lost. Predict the genotype frequency in the population after this event. p=0.33 q=o.66 BB=(P)^2+0.109 Bb=(2pq)=0.4356 bb=(q)^2=0.4356 24. Compare your answers to Question 22 with those of Question 16. How do your answers support the conclusion that the population is not in Hardy-Weinberg equilibrium? #16 Before Natural Selection BB=0.0625 Bb=0.376 bb=0.563 #22 After Natural Selection BB=0.109 Bb=0.4356 bb=0.4356 This study source was downloaded by from CourseH on :06:06 GMT -06:00 Extension Questions 25.The ability to taste PTC is due to a single dominant allele “T.” You sampled 215 individuals and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate the following frequencies. a. The frequency of the recessive allele. q 2 = 0.65 =0.6 The frequency would be 60% b. The frequency of the dominant allele. Since q2 = 0.6, then p + q = 1. p+q then p = 0.4 which 40% c. The frequency of the heterozygous individuals. The heterozygous is 48% 2pq=2*0.4*0.6=0.48 26. Sixty flowering plants are planted in a flowerbed. Forty of the plants are red-flowering homo-zygous dominant. Twenty of the plants are white-flowering homozygous recessive. The plants naturally pollinate and reseed themselves for several years. In a subsequent year, 178 red-flowered plants, 190 pink-flowered plants, and 52 white-flowered plants are found in the flowerbed. Use a chi-square analysis to determine if the population is in Hardy-Weinberg equilibrium. Phenotype 0 expected (o-e)2 / e white Pink 52 205 1 Red sum=455 The plants are not significantly different from what was expected. This study source was downloaded by from CourseH on :06:06 GMT -06:00 The Hardy-Weinberg Equation 7 This study source was downloaded by from CourseH on :06:06 GMT -06:00 Powered by TCPDF ()