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Unformatted text preview: ffspring Dihybrid cross shows parental and Dihybrid cross shows parental and recombinant types Dihybrid cross produces a Dihybrid cross produces a predictable ratio of phenotypes The law of independent assortment The law of independent assortment During gamete formation different pairs of alleles segregate independently of each other Summary of Mendel's work Summary of Mendel's work
Inheritance is particulate – not blending There are two copies of each trait in a germ cell Gametes contain one copy of each trait (haploid) Alleles (different forms of the trait) segregate randomly Alleles are dominant or recessive – thus the difference between genotype and phenotype Laws of probability for Laws of probability for multiple genes
P gametes F1 gametes
RYTS RyTS rYTS ryTs RRYYTTSS X rryyttss RYTS RrYyTtSs
RYTs RyTs rYTs rYtS RYtS RytS rYts rYts RYts Ryts rYTS ryts ryts X RrYyTtSs
RYTS RyTS rYTS ryTs RYTs RyTs rYTs rYtS RYtS RytS rYts rYts RYts Ryts rYTS ryts F2 What is the ratio of different genotypes and phenotypes? 0 •Punnet Square method – 2^4 = 16 possible gamete combinations for each parent •Thus, a 16 x 16 Punnet Square with 256 genotypes • That’s one big Punnet Square Loci Assort Independently – So we can look at each locus independently to get the answer 0 P1 F1 RRYYTTSS × rryyttss RrYyTtSs × RrYyTtSs What is the probability of obtaining the genotype RrYyTtss?
Rr x Rr 1RR:2Rr:1rr 2/4 Rr Yy x Yy 1YY:2Yy:1yy 2/4 Yu Probability of obtaining individual with Rr and Yy and Tt and ss. 2/4 x 2/4 x 2/4 x ¼ = 8/256 P1 F1 RRYYTTSS × rryyttss RrYyTtSs × RrYyTtSs What is the probability of obtaining a completely homozygous genotype? Genotype could be RRYYTTSS or rryyttss
Rr x Rr 1RR:2Rr:1rr ¼ RR ¼ rr (...
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- Spring '03