Homework Answers
Frequency of allele b=0.3 , so allele frequency of B =1-0.3=0.7.here we can take p=0.7,q=0.3.
So Observed genotype frequency of BB will be p2,Bb will be 2pq, bb will be q2.
-->observed genotype frequencies
BB =p2=0.49
Bb=2pq=2×0.7×0.3=0.42
bb=q2=0.09
-->observed individuals
BB=0.49×1000=490
Bb=0.42×1000=420
bb=0.09×1000=90.
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I've tried 0.16, 0.48, 0.36
and 32, 96, 72
I have 2 trials left, please help.
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I've tried 0.16, 0.48, 0.36
and 32, 96, 72
I have 2 trials left, please help.
Here's a hint: Hint 2. How to calculate the
expected frequencies of a different example population Consider an
example population of individuals that have two alleles for a
specific locus, AB and AC. In the population, 70% (0.7) of the
alleles are AB, and 30% (0.3) of the alleles are AC. The expected
frequencies of each genotype can be calculated using the equation
for...
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Chi square problem need help please show work of how they
solved for the chi square value, the correct answer is in
red.
**On this page and the next two, report all* calculated non-integer values to the nearest 0.01** 18. x2. As you learned in one or more of your introductory biology courses, the frequencies of the genotypes in a population at "Hardy-Weinberg" equilibrium can be predicted from the frequencies of the individual alleles "A" and "a' at the locus...
Consider a locus of interest that has two alleles: A and a. A diploid individual carrying these alleles can have one of three genotypes: AA, Aa, or aa; a population will consist of some combination of AA, Aa, and aa individuals. The relatively frequency of each of these genotypes makes up the population's structure. Hardy and Weinberg independently figured out that, in the absence of forces that cause evolutionary change, the population structure will 'settle' or default to equilibrium values,...
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