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In a family from Bohol, Tatay Ramon has dimples and is heterozygous (Dd) and Nanay Sisa has no dimples (dd). What are the parent genotypes and phenotypes, and what are the F1 genotypes, phenotypes, ratios, and percentages?

In a family from Bohol, Tatay Ramon has dimples and is heterozygous (Dd) and Nanay Sisa has no dimpl...
Answer

Parent generation: Tatay Ramon is genotype $Dd$ (has dimples) and Nanay Sisa is genotype $dd$ (no dimples). Crossing $Dd \times dd$ produces F1 offspring with genotypes 50% $Dd$ and 50% $dd$ (genotypic ratio $1:1$). The phenotypes are 50% with dimples and 50% without dimples (phenotypic ratio $1:1$).

Explanation

What the problem is asking

You are doing a simple monohybrid cross for one trait (dimples). The key is that $D$ is the dominant allele (dimples) and $d$ is the recessive allele (no dimples).

Parent generation (P): genotype and phenotype

  • Tatay Ramon: genotype $Dd$, phenotype = has dimples (because $D$ is dominant)
  • Nanay Sisa: genotype $dd$, phenotype = no dimples

Making the Punnett square for $Dd \times dd$

Gametes:

  • Tatay Ramon ($Dd$) can make $D$ or $d$
  • Nanay Sisa ($dd$) can make only $d$

Punnett outcomes:

  • $D \times d \rightarrow Dd$
  • $d \times d \rightarrow dd$

So the F1 offspring are split evenly between $Dd$ and $dd$.

F1 results (genotype, phenotype, ratios, percentages)

F1 Genotypes: $Dd$, $dd$

Genotypic ratio: $1 \, Dd: 1 \, dd$

Genotypic percentage:

  • $Dd = 50\%$
  • $dd = 50\%$

F1 Phenotypes: has dimples ($Dd$), no dimples ($dd$)

Phenotypic ratio: $1$ with dimples $: 1$ without dimples

Phenotypic percentage:

  • With dimples = $50\%$
  • Without dimples = $50\%$
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Skills You Achive
punnett squares mendelian genetics genotype to phenotype probability

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