In humans, freckles are encoded by a dominant allele. A woman is heterozygous for freckles. According to the law of segregation, allele pairs separate during gamete formation and then randomly unite during fertilization. Which of the following would apply to this woman's child?

Answers

Answer 1

The child having an equal chance of inheriting the dominant allele or the

recessive allele for freckles from her mother is what applies.

The law of segregation by Mendel states that diploid organisms have a pair

of alleles and are passed to their offspring at random.

The woman being heterozygous means she has the dominant and recessive

allele present. Due to the law of segregation stating the alleles are passed

at random to the offspring, the child will have an equal chance of inheriting

the dominant allele or the recessive allele for freckles .

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----------------------------------------------------------------------------------------------

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For each offspring genotype in the Punnett square you just completed, determine the phenotype. In other words, what is the predicted fur color and eye color of the offspring? Using your Punnett square from the last step, fill in the predicted fraction for each phenotype in the data table below.

Black Fur and Black Eyes Black Fur and Red Eyes White Fur and Black Eyes White Fur and Red Eyes
Predicted Fraction
/16
/16
/16
/16

Answers

Assuming diallelic genes, expressing complete dominance and not interacting, the predicted fractions are 4/16 Black Fur, Black Eyes - 4/16 Black Fur, Red Eyes - 4/16 White Fur, Black Eyes - 4/16 White Fur, Red Eyes

Available data:

Two traits are involved in this cross  ⇒  fur color and eye color.

Black fur color is dominant over white fur color

Black eyes is dominant over red eyes

The cross is between a h0m0zyg0us recessive parent with a dihybrid one.

What can we deduct about this information?

Assuming diallelic genes coding for each trait, expressing complete dominance, and no interaction between genes, we can propose the following genotypes and phenotypes.

         Genotype                               Phenotype          

BBEE, BbEE, BBEe, BbEe        Black fur, Black eyes

bbEE, bbEe                              White fur, Black eyes

BBee, Bbee                              Black fur, Red eyes

bbee                                         White fur, Red eyes

How to perform the cross?

Cross: h0m0zyg0us recessive parent with a dihybrid parent

Parental phenotypes)    White fur, Red eyes   x   Black fur, Black eyes

Parentals genotype)    bbee     x     BbEe

Gametes)  be, be, be, be        

                 BE, Be, bE, be

Punnett square)       BE             Be            bE           be

                     be     BbEe        Bbee         bbEe       bbee

                     be     BbEe        Bbee         bbEe       bbee

                     be     BbEe        Bbee         bbEe       bbee

                     be     BbEe        Bbee         bbEe       bbee

F1)

4/16 = 1/4 = 25% of the progeny is expected to be dihybrid, BbEe, expressing black fur and black eyes

4/16 = 1/4 = 25% of the progeny is expected to be heter0zyg0us for fur color and h0m0zyg0us recessive for eye color, expressing black fur and red eyes.

 

4/16 = 1/4 = 25% of the progeny is expected to be heter0zyg0us for eye color and h0m0zyg0us recessive for fur color, expressing white fur and black eyes.

4/16 = 1/4 = 25% of the progeny is expected to be h0m0zyg0us recessive, bbee, expressing white fur and red eyes.

Predicted fraction per phenotype

Black Fur and Black Eyes  ⇒  4/16 BbEeBlack Fur and Red Eyes  ⇒  4/16 BbeeWhite Fur and Black Eyes  ⇒  4/16 bbEeWhite Fur and Red Eyes  ⇒  4/16 bbee

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