I need help with question 5.

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dcbiostat eNotes educator| Certified Educator

In order to make such a Punnett Square, the genotypes of each parent need to be determined. A genotype is the allele pairs that an individual has for a trait. Alleles are variations of genes. These are represented by letters in your question. Alleles can be either dominant or recessive. In your question, the dominant allele is the capital letter and the recessive allele is the lowercase letter. Dominant alleles are considered to be “stronger” than recessive alleles. Phenotypes are the physical representation of a genotype.  Because dominant alleles are “stronger” than recessive alleles, only one dominant allele is needed to produce a dominant phenotype.

Genotypes can be homozygous dominant, heterozygous, or homozygous recessive. Homozygous dominant genotypes have two dominant alleles and will show the dominant phenotype. A heterozygous genotype has one dominant and one recessive allele. Because a heterozygous genotype contains a dominant allele, it will also show the dominant phenotype. Homozygous recessive genotypes contain two recessive alleles. Having a homozygous recessive genotype is the only way an individual can show the recessive phenotype.

The fact that the running black mouse had a mother who was a waltzing, brown mouse suggests that this mouse received dominant traits from his father and a recessive traits from his mother. Thus, this mouse is heterozygous for each trait. We will use RrBb to represent this mouse’s genotype. Since the second mouse is homozygous for running and waltzing, its genotype must be RRbb.

Thus, the cross that is being completed is RrBb x RRbb. The resulting Punnett Square is as follows:

 

                RB                          Rb                          rB                       rb

 

Rb          RRBb                    RRbb                     RrBb                     Rrbb

 

Rb          RRBb                    RRbb                     RbBb                     Rrbb

 

Rb          RRBb                    RRbb                     RrBb                     Rrbb

 

Rb          RRBb                    RRbb                     RrBb                     Rrbb

 

The genotype probabilities of offspring from this cross are as follows:

25% RRBb

25% RRbb

25% RrBb

25% Rrbb

 

The phenotype probabilities of offspring from this cross are as follows:

50% Running, black

50% Running, brown

 

 

 

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