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anonymous

  • one year ago

PLEASE HELP! I just need someone to check my answer...it shouldn't take long. In a laboratory experiment of sexually reproducing insects, a certain trait is determined by a single gene. The dominant allele has complete dominance over the recessive allele, resulting in two different observable phenotypes. Scientists created a new population of this insect by crossing 100 homozygous dominant individuals with 100 homozygous recessive individuals. The following results are from the first five generations of this population: Generation Number of individuals with dominant trait Number of individuals with recessive trait Total number of individuals 1 100 100 200 2 325 0 325 3 375 125 500 4 300 250 550 5 325 425 750 Answer each part of this question completely in the space provided. (10 points) Using the data, explain the changes in phenotypic frequency from generation 1 to 3. Do you think this population is in Hardy-Weinberg equilibrium? Support your answer. What conditions must be met for a population to be in Hardy-Weinberg equilibrium?

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  1. anonymous
    • one year ago
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    My answer: Since the dominant allele has complete dominance over the recessive allele and in generation one, all individuals are homozygous for their specific alleles, generation two will have no homozygous recessive individuals. It will however, have some (or in this case all) heterozygous individuals displaying the dominant trait but who are also carrying the recessive allele. (Rr) When the second generation go to mate, due to the increase in genetic variation and the mechanism of sexual reproduction that shuffles existing alleles and deals them at random to produce individual genotypes, there will be several different combinations of the allele for the third generation, as seen in the graph. (RR, Rr, rR, and rr) The individuals with the recessive trait reappears because two heterozygous individuals were able to mate and gave both of their recessive alleles to their offspring. I don't think this population is in Hardy-Weinberg equilibrium though. The conditions which need to be met for a population to be in Hardy-Weinberg equilibrium are: No mutations Random mating No natural selection Extremely large population size No gene flow If a population remains in equilibrium, phenotypic frequencies do not change. However, if a population violates one of the 5 conditions above, trait frequencies will begin to alter. Obviously non-random mating is taking place here because if mating was random, some of the homozygous recessive individuals in the first generation would have mated with each other and produced homozgous recessive individuals in the second generation. But there aren't any. Also, the population is relatively small, especially in the beginning so genetic drift could have occurred as well.

  2. anonymous
    • one year ago
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    Yep that is pretty darn good!

  3. anonymous
    • one year ago
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    gimme a sec to verify

  4. anonymous
    • one year ago
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    This is what i put as the answer and let me know if you have questions. WARNING THIS IS NOT THE ANSWER JUST FOR COMPARING

  5. anonymous
    • one year ago
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    1. Since the dominant allele has complete dominance that explains why in generation 2 all the individuals were dominant. In generation 3 by then some individuals would have been homozygous recessive making the gene dominant which means that the homozygous recessive individuals will start rising in numbers. So overall the mixing of 2 populations explains the change that is occurring. 2. No it is not because it started with 2 different populations breeding, and it does not fulfil the condition p^2 + 2pq + q^2 = 1 so it does not conform to the Hardy-Weinberg equilibrium. 3. Mutation is not occurring, natural selection is not happening, all mating is random, and all are producing dame number of offspring.

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