Assuming a single diallelic gene coding for the trait and expressing complete dominance, the phenotypes, genotypes, and probabilities of getting each of them vary according to the parentals' genotypes. See the options below.
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Since I do not have the genotypes from #2, I will provide <em>different potential scenarios </em>for this question.
I advise you to <em>check on them</em> and see <em>which one matches the genotype from #2.</em>
Let us assume a single diallelic gene is coding for the trait and expresses complete dominance.
<h3 /><h3><u>SCENARIO 1</u> ⇒ long nose is the dominant trait</h3>
Let us say that
- L is the dominant allele and codes for long nose
- l is the recessive allele and codes for stubby nose
Since long nose is dominant over stubby nose
If Dobby comes from a purebred family and has long nose, his genotype must be LL.
And if Viola has stubbi nose, her genotype must be ll.
<u>Cross 1</u>
Parentals) LL x ll
Gametes) L L l l
Punnett square) L L
l Ll Ll
l Ll Ll
F1) Genotype ⇒ 100% heter0zyg0us Ll
Phenotype ⇒ 100% long-nosed
- <em>There is 100% chances for a child to have a long nose</em>
- <em>There is 0% chances for a child to have a stubby nose</em>
- <em>These children are not purebred</em>
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<h3><u>SCENARIO 2</u> ⇒ Stubby nose is the dominant trait</h3>
Let us say that
- S is the dominant allele and codes for stubby nose
- s is the recessive allele and codes for long nose
Since stubby nose is dominant over long nose
If Dobby comes from a purebred family and has long nose, his genotype must be ss.
And if Viola has stubbi nose, her genotype must be either SS or Ss.
There are two possible crosses.
<u>Cross 1</u> : Violet is h0m0zyg0us dominant SS
Parentals) SS x ss
Gametes) S S s s
Punnett square) S S
s Ss Ss
s Ss Ss
F1) Genotype ⇒ 100% heter0zyg0us Ss
Phenotype ⇒ 100% stubby-nosed
- <em>There is 100% chances for a child to have a stubby nose</em>
- <em>There is 0% chances for a child to have a long nose</em>
- <em>These children are not purebred</em>
<u>Cross 2</u>: Violet is heter0zyg0us, Ss
Parentals) Ss x ss
Gametes) S s s s
Punnett square) S s
s Ss ss
s Ss ss
F1) Genotype ⇒ 50% heter0zyg0us Ss and 50% h0m0zyg0us recessive ss
Phenotype ⇒ 50% stubby-nosed and 50% long-nosed
- <em>There is 50% chances for a child to have a stubby nose</em>
- <em>There is 50% chances for a child to have a long nose</em>
- <em>These children are not purebred</em>
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You can learn more about single gene crosses at
brainly.com/question/12653314?referrer=searchResults