To determine the phenotype ratio of hitchhiker’s thumb (H) and free earlobes (F) from a cross of both parents that are heterozygous for these traits (HhFf), we can use a Punnett square.
When we set up the cross, the gametes from each parent will be HF, Hf, hF, and hf. By filling out the Punnett square, we can see the possible combinations of alleles for the offspring:
- HHFF – Hitchhiker’s thumb & Free earlobes
- HHFf – Hitchhiker’s thumb & Free earlobes
- HhFF – Hitchhiker’s thumb & Free earlobes
- HhFf – Hitchhiker’s thumb & Free earlobes
- Hhff – Hitchhiker’s thumb & Attached earlobes
- hhFF – Attached thumb & Free earlobes
- hhFf – Attached thumb & Free earlobes
- hhff – Attached thumb & Attached earlobes
Counting the phenotypes:
- Hitchhiker’s thumb & Free earlobes: 4 (HHFF, HHFf, HhFF, HhFf)
- Hitchhiker’s thumb & Attached earlobes: 1 (Hhff)
- Attached thumb & Free earlobes: 2 (hhFF, hhFf)
- Attached thumb & Attached earlobes: 1 (hhff)
So, the overall phenotype ratio can be summarized as:
- 4 Hitchhiker’s thumb & Free earlobes
- 1 Hitchhiker’s thumb & Attached earlobes
- 2 Attached thumb & Free earlobes
- 1 Attached thumb & Attached earlobes
This gives us a phenotype ratio of 4:1:2:1 for the traits. Therefore, the final answer for the phenotype ratio of these characteristics when crossing two heterozygous parents is 4:1:2:1.