Using Punnett Squares Coursework Example
Free Using Punnett Squares Coursework Example
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Punnett Squares Exploration
First Cross
RR
RRRRR
rRrRr
Genotypic ratio: 50:50
Phenotypic ratio: 50:50
Second Cross
R r
RRRRr
r Rr rr
Genotypic ratio: 25:50:25
Phenotypic ratio: 25:75
Third Cross
r r
r rr rr
r rr rr
Genotypic ratio: 1:1
Phenotypic ratio: 1:1
Questions
What conclusions can you draw about the results of your first cross between a homozygous red bull (RR) and a heterozygous red cow (Rr)?
This case involves a pure Red homozygous dominant Bull and a mixed hybrid heterozygous Cow. The probability of offspring to result from the Punnett square is 50% for the heterozygous gene (RR) and 50% for the homozygous gene (Freidenreich et al. 2326). Moreover, R is a pure dominant trait hence there is a 100% chance that all offspring, in this case, will be Red. The genotype ratio is probability 50:50 while the phenotype ratio is 50:50 as well.
What conclusions can you draw about the results of your second cross between a heterozygous red bull (Rr) and a heterozygous red cow (Rr)?
This case involves a Mixed hybrid parent (Bull) and a mixed hybrid parent (Cow). The probability of offspring to result from the Punnett square is 25% for the homozygous RR (pure dominant), 25% homozygous rr (pure recessive) and 50% for the heterozygous gene Rr (hybrid) (Freidenreich et al. 2335). The genotype ratio is probability 25:50:25 whereas the phenotype ratio is 25:75.
What conclusions can you draw about the results of your third cross between a homozygous white bull (rr) and a homozygous white cow (rr)?
The case involves a pure red homozygous recessive Bull (rr) and a pure red recessive cow (rr). The probability of youngone to result from the Punnett square is 100% pure homozygous recessive since all parent is pure Red homozygous recessive (Freidenreich et al. 2340). The genotype ratio is probability 1:1 while the phenotype ratio is 1:1 as well.
Work Cited
Freidenreich, Hava Bresler, Ravit Golan ...
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