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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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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
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