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A biology assignment, Cheat Sheet of Biology

A biology assignment about the harvy

Typology: Cheat Sheet

2022/2023

Uploaded on 11/21/2023

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Haely Flores 11/16/23
Assignment 9
Hardy Weinberg
At Hardy-Weinberg equilibrium, allele frequencies do not change from one generation to the
next. Hardy-Weinberg equilibrium occurs only if a population meets all of the following
assumptions:
natural selection does not occur
no mutations
the population is large enough to eliminate random changes in allele frequencies
individual mate at random
no migration
Assuming the assumptions of Hardy-Weinberg equilibrium are met, two equations represent
the relationship between allele frequencies and genotype frequencies.
p+q=1
p2+2pq+q2=1
p is the frequency of the dominant allele and q is the frequency of the recessive allele.
2pq fraction of heterozygous individuals
Let us take for example
In a class of 19 students, 12 can roll their tongue, and 7 can’t.
Possible individual genotypes are:
1. RR - can roll (Homozygous dominant)
2. Rr – can also roll (heterozygous dominant individual)
3. Rr – can’t roll (Homozygous recessive)
For both RR and Rr the fraction of the class that can roll is 12/19 = 0.63158
But we don’t what fraction is just RR or just Rr.
For rr, the fraction of the class that can’t roll is 7/19 = 0.36842
Here we are sure of the fraction of the class that is rr.
So, r x r = r2 = 0.36842
r = square root of 0.36842 = 0.60697
Remember p + q = 1
R + r = 1
pf3

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Haely Flores 11/16/

Assignment 9

Hardy Weinberg

At Hardy-Weinberg equilibrium, allele frequencies do not change from one generation to the next. Hardy-Weinberg equilibrium occurs only if a population meets all of the following assumptions:

  • natural selection does not occur
  • no mutations
  • the population is large enough to eliminate random changes in allele frequencies
  • individual mate at random
  • no migration Assuming the assumptions of Hardy-Weinberg equilibrium are met, two equations represent the relationship between allele frequencies and genotype frequencies.
  • p+ q = 1
  • (^) p^2 + 2 pq + q^2 = 1 p is the frequency of the dominant allele and q is the frequency of the recessive allele. 2pq fraction of heterozygous individuals Let us take for example In a class of 19 students, 12 can roll their tongue, and 7 can’t. Possible individual genotypes are:
  1. RR - can roll (Homozygous dominant)
  2. Rr – can also roll (heterozygous dominant individual)
  3. Rr – can’t roll (Homozygous recessive) For both RR and Rr the fraction of the class that can roll is 12/19 = 0. But we don’t what fraction is just RR or just Rr. For rr , the fraction of the class that can’t roll is 7/19 = 0. Here we are sure of the fraction of the class that is rr. So, r x r = r^2 = 0. r = square root of 0.36842 = 0. Remember p + q = 1 R + r = 1

Since we got r = 0. 1 – 0.60697 = R 0.39302 = R R x R = R^2 which is the fraction of people in the class with Homozygous dominant for tongue rolling. R^2 = R x R. Therefore 0.39302 x 0.39302 = 0. Heterozygous dominant will be 2Rr , just like 2pq above Since we already know values for R and r We plug in. 2 x R x r 2 x 0.39302 x 0.60697 = 0.477102. Let us verify our result. Remember p^2 + 2pq + q^2 = 1 Same way R^2 + 2Rr + r^2 = 1 0.154464 + 0.477102 + 0.36842 = 0.99998672 approximately 1