Answers
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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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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/16
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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 colorBlack eyes is dominant over red eyesThe 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 eyes4/16 = 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 phenotypeBlack 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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