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All the F1 will have wire hair.

The reason:

A homozygous wire-haired dog must be SS. It must have at least one S allele if it shows the dominant character, and "homozygous" means pure-breeding, so there is no other allele present.

A smooth-haired dog must be ss. Any organism displaying a recessive trait must be homozygous, as a dominant allele would mask the recessive one.

So the cross is:

SS x ss

and one parent must pass on the S gene, one the s. Therefore all the F1 must be Ss. Their phenotype is wire-haired, because they have a dominant allele, but they are all heterozygous, because they inherited a recessive gene from the smooth-haired parent.

This cross demonstrates Mendel's First Law.

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Q: In dogs wire hair S is dominant to smooth s In a cross of a homozygous wire-haired dog with a smooth-haired dog what will be the phenotype of the F1 generation?
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