Asked by yo pap
Unattached earlobes are dominant to attached earlobes. Cleft chin is dominant to no cleft. Parents that are heterozygous for both traits are crossed.
Determine the genotype of the following offspring.
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All Answers 1
Answered by
GPT-5 mini
AI
I can do that — but I need to know the phenotype of each offspring (whether each has attached or unattached earlobes, and whether each has a cleft chin or not). Could you give those (or upload the picture)?
Meanwhile, here is a quick reference you can use once you supply the phenotypes.
Notation
- E = unattached earlobes (dominant), e = attached (recessive)
- C = cleft chin (dominant), c = no cleft (recessive)
- Parents: EeCc × EeCc
Genotype probabilities from EeCc × EeCc
- EECC 1/16
- EECc 2/16
- EEcc 1/16
- EeCC 2/16
- EeCc 4/16
- Eecc 2/16
- eeCC 1/16
- eeCc 2/16
- eecc 1/16
Phenotype classes and possible genotypes
- Unattached earlobes, cleft chin (E_ C_): EECC, EECc, EeCC, EeCc (total 9/16)
- Unattached earlobes, no cleft (E_ cc): EEcc, Eecc (3/16)
- Attached earlobes, cleft chin (ee C_): eeCC, eeCc (3/16)
- Attached earlobes, no cleft (ee cc): eecc (1/16)
If you tell me which phenotype each numbered offspring (1–5) shows, I’ll give the possible genotype(s) for each — and I can state probabilities if you want.
Meanwhile, here is a quick reference you can use once you supply the phenotypes.
Notation
- E = unattached earlobes (dominant), e = attached (recessive)
- C = cleft chin (dominant), c = no cleft (recessive)
- Parents: EeCc × EeCc
Genotype probabilities from EeCc × EeCc
- EECC 1/16
- EECc 2/16
- EEcc 1/16
- EeCC 2/16
- EeCc 4/16
- Eecc 2/16
- eeCC 1/16
- eeCc 2/16
- eecc 1/16
Phenotype classes and possible genotypes
- Unattached earlobes, cleft chin (E_ C_): EECC, EECc, EeCC, EeCc (total 9/16)
- Unattached earlobes, no cleft (E_ cc): EEcc, Eecc (3/16)
- Attached earlobes, cleft chin (ee C_): eeCC, eeCc (3/16)
- Attached earlobes, no cleft (ee cc): eecc (1/16)
If you tell me which phenotype each numbered offspring (1–5) shows, I’ll give the possible genotype(s) for each — and I can state probabilities if you want.
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