How would I create a bar graph using this hardy Weinberg application ? q^2=23/66= 0.59%q= √0.59 = 0.77p=1-qp=1-0.77= 0.23p^2=(0.23)^2 = 0.052pq=2(0.23) (0.77) = 0.35P^2+2pq=0.35 + 0.05 = 0.4Hardy-Weinberg Equations: p+q=1 p² +2pq+q²=1Homozygous recessive genotype frequency/ Recessive phenotype frequencyRecessive allele frequencyDominant allele frequencyHomozygous dominant genotype frequencyHeterozygote frequencyDominant trait phenotypic frequency Genotypic FrequencyAllelic FrequencyPhenotypic FrequencyFrequency found in Population (%)010Dominant2030Recessive4050Heterozygous607080Fig D. Application of Hardy-Weinberg. Genotypic, allelic and phenotypic frequency was calculated with the usesame

Hardy Weinberg's principle explains why a dominant allele doesn't eventually replace all other…

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Biochemistry

How would I create a bar graph using this hardy Weinberg application

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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