Using the information on the box provided, which of the following statements is true? First Generation Second Generation Third Generation p=0.2 p=0.2 p=0.4 q=0.8 q=0.8 |q=0.6 O The allele frequency in the gene pool of 2nd and 3rd is constant. O The 2nd and 3rd generation is in Hardy-Weinberg equilibrium O The population in 1st and 2nd generation is evolving. O The 2nd and 3rd generation is an evolving population.

Using the information on the box provided, which of the following statements is true? First Generation Second Generation Third Generation p=0.2 p=0.2 p=0.4 q=0.8 q=0.8 |q=0.6 O The allele frequency in the gene pool of 2nd and 3rd is constant. O The 2nd and 3rd generation is in Hardy-Weinberg equilibrium O The population in 1st and 2nd generation is evolving. O The 2nd and 3rd generation is an evolving population.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 11QP: Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four...

See similar textbooks

Similar questions

What do you expect to happen to the frequnecy of the F allele after 1000's of generations It will remain the same as in the initial population it will increase until it becomes fixed (a frequency of 1.0) it will be lost from the population totally (frequency of 0.0) it will approach a frequency of 1.0 but never quite reach that level. There is not enough information to anwer this question. Type here to search PrtSc InsertDelete F10 F11 F12 F9 F6 F7 F5
Consider a sample of 100 individuals sampled from a population in Hardy-Weinberg equilibrium that are genotyped at a single locus. Of these 100 individuals, 25 individuals are A¡A1, 45 are A1A2, and 30 are A2A2. What is the actual (observed) allele frequency of the A1 allele in this sample? (you may use a calculator if needed) O A. 0.325 O B. 0.275 O C. 0.475 O D. 0.525
There is an outbred population of mice living in a farmer’s field. Mutations occur at a genecalled nurseryrhyme that cause a totally recessive form of blindness (h=0). These blind micedo not survive to reproduce because the farmer’s wife cuts off their tail (and other bits) witha carving knife. Surveying the field for baby mice you find that 3/10,000 mice are blind.A. Assuming that the population mates at random, what is the mutation rate of blindnesscausing alleles?B. Following more careful study you now find that there is actually a 20% reduction in theviability of heterozygotes for these mutations. What would you now estimate as themutation rate for this gene?
Populations in Hardy-Weinberg equilibrium show no change in allele frequencies from one generation to the next, assuming that certain conditions are met. Which condition is being violated in this image? *** O No gene flow O No genetic drift O No mutation O Random mating 4
This Punnett square shows allele combinations for all possible genetic crosses in the wildflower population described in Part A. CR E p= 0.7 CW q= 0.3 p= 0.7 ? CW CR CR CR CW q= 0.3 ? ? CR CW CW CW ? What are the expected genotype frequencies in the offspring generation? O 0.49 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers) 0.49 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.30 CWCW (white flowers) O 0.09 CRCR (red flowers), 0.21 CRCW (pink flowers), 0.42 CWCW (white flowers) 0.09 CRCR (red flowers), 0.42 CRCW (pink flowers), 0.09 CWCW (white flowers)
Small population size causes genetic drift because ofchance sampling of different alleles from one generation to the next. We can predict how much geneticdrift occurs for a given population size using binomialsampling statistics. With a population of size N, wecan estimate that 95% of the time the allele frequency(p) in the next generation will be withinthe confidence interval of p ± 1.96 (√p(1 − p)2N ),where p(1 − p)2Nis an estimate of the statistical variancein allele frequencies from one generation to the nextwith random sampling of 2N alleles each generation. a. What is the confidence interval for p = 0.5 whenN = 100,000?b. What is the confidence interval for p = 0.5 whenN = 10?c. How are the results in parts (a) and (b) related tothe consequences of a population bottleneck?
Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four phenotypic classes of the ABO blood groups are found to be A = 0.33, B = 0.33, AB = 0.18, and i = 0.16. What is the frequency of the i allele?
How Can We Measure Allele Frequencies in Populations? In a population where the females have the allelic frequencies A = 0.35 and a = 0.65 and the frequencies for males are A = 0.1 and a = 0.9, how many generations will it take to reach HardyWeinberg equilibrium for both the allelic and the genotypic frequencies? Assume random mating and show the allelic and genotypic frequencies for each generation.
How Can We Measure Allele Frequencies in Populations? Drawing on your newly acquired understanding of the HardyWeinberg equilibrium law, point out why the following statement is erroneous: Because most of the people in Sweden have blond hair and blue eyes, the genes for blond hair and blue eyes must be dominant in that population.
You sample a population of blue-bellied lizards in the Mexican desert. You calculate allele frequencies to be 0.42 (A1) and 0.58 (A2). You genotype 1000 individuals, and discover that 130 are homozygous for the A, allele, 290 are homozygous for the A2 allele, and 580 are heterozygous. O No, this population is not in genetic equilibrium because there are more homozygotes and fewer heterozygotes than predicted under Hardy Weinberg equilibrium. O No, this population is not in genetic equilibrium because there are more heterozygotes and fewer homozygotes than predicted under Hardy Weinberg equilibrium. Yes, this population is in genetic equilibrium because allele frequencies are not changing. Yes, this population is in genetic equilibrium because the observed genotype frequencies match those that are predicted under Hardy Weinberg equilibrium.
Below are individuals in a population broken down by genotype. (A) Complete the table AA Aa aa Total # individuals 305 650 55 # A alleles # a alleles (B) Determine the frequency of alleles A and a. Show your work. (C) Assuming the population is in Hardy-Weinberg equilibrium, determine the number of heterozygous individuals in the next generation with a population of 1200.
For a locus with two alleles (B and b) in a population at riskfrom an infectious neurodegenerative disease, 16 individualshave the genotype BB, 92 individuals have Bb and 12individuals have bb. Use the Hardy-Weinberg equation todetermine whether this population appears to be evolving.