can somebody help me with this worksheet please!?

Can Somebody Help Me With This Worksheet Please!?
Can Somebody Help Me With This Worksheet Please!?
Can Somebody Help Me With This Worksheet Please!?

Answers

Answer 1

A test cross performed between the individual with unkown genotype and the h0m0zyg0us recessive individual. 9) hh. 10) Hh or HH. 11) HH x hh. 12) 100% Hh. 13) Hh x  hh. 14) 50%Hh and 50% hh. 15) -16) in the text.

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Available data:Single diallelic gene codes for hair in Guinea pigsAllele H is dominant and codes for having hairAllele h is recessive and codes for not having hair

For this gene we will assume complete dominance, meaning that the dominant alele completely hides the expression of the recessive allele.

According to this information, we can say that

Genotype      Phenotype

HH                  Hairy pig

Hh                   Hairy pig

hh                   Hairless pig

9) Geneviene does not have hair  ⇒  Its genotype is hh

10) Fred has hair ⇒ Its genotype is HH or Hh

Possible crosses

11 and 12)

Cross1: If Fred was genotype HH and bred with Genevieve

Parentals) HH    x    hh

Gametes)   H   H    h    h

Punnett square)    H    H

                        h   Hh   Hh

                        h   Hh   Hh

F1) 100% of the progeny is expected to have hair and be heter0zyg0us, Hh.

13 and 14)

Cross2: If Fred was genotype Hh and bred with Genevieve

Parentals) Hh    x    hh

Gametes)   H   hh    h    h

Punnett square)    H    h

                        h   Hh   hh

                        h   Hh   hh

F1) 50% of the progeny is expected to have hair and be heter0zyg0us, Hh.

     50% of the progeny is expected to be hairless and be h0m0zyg0us recessive, hh.

15)  

• If the individual with the unknown genotype (Fred) is heter0zyg0us, the phenotypic rate of the descendants is 50% heter0zyg0us -Hh- and 50% h0m0zyg0us recessive -hh-.

• But if the individual with the unknown genotype is h0m0zyg0us dominant, the phenotypic rate of the descendants is 100% heter0zyg0us for the trait, Hh.  

16)

The weakness is that even if Fred is heter0zyg0us for the trait, when making the cross it might occur that all individuals in the progeny are born h0m0zyg0us recessive, and there is a risk of missinterpreting the results.

So in that situation, many crosses must be done between fred and genevieve to se sure about Freds genotype.

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You can learn more about test cross at

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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.

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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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