1. Freckles are due to dominant alleles (F), and no freckles are due to recessive alleles (f). Ted and Mary both have freckles, but their son Billy has no freckles. Give the genotypes of Ted, Mary and Billy. If Ted and Mary have additional children, what is the chance for freckles?
a. Ted X Mary is Ff X Ff. Each child has a 1/4 for FF, 1/4 for Ff, 1/4 for Ff, and 1/4 for ff genotype, resulting in 3/4 freckles and 1/4 no freckles phenotype. In percentages, 75% freckles and 25% no freckles children.
2. Pigmented eyes (generally brown) are dominant to blue eyes. If Ted has blue eyes and Mary is heterozygous, what is the chance that their offspring will have brown eyes?
a. If Ted has bb (recessive blue eyes) and Mary has Bb (heterozygous bown eyes), then the chances are a cross of Bb x bb, resulting in the usual 1/4 Bb, 1/4 bb, 1/4 Bb, and 1/4 bb, totalling to 1/2 Bb and 1/2 bb: thus 50 % heterozygous brown and 50 % recessive blue.
3. Widow’s peak is due to a dominant allele. If Ted is homozygous dominant and Mary has straight hairline, what is the chance that their offspring will have widow’s peak?
a. Ted is WW and Mary is ww (straight hairline assumes recessive alleles), then it is a cross of WW X ww. This results in all offspring 1/4 Ww, 1/4 Ww, 1/4 Ww, 1/4 Ww, or all Ww. 100% dominant heterozygous widow\'s peak.
4. If Fred and Wilma are both heterozygous for Widow’s peak, what is the chance that they will have a c chance their child will have a widow\'s peak.hild with widow’s peak?
a. Ww (Fred: paternal alleles) X Ww (Wilma: maternal alleles) cross wil produce: 1/4 WW, 1/4 Ww, 1/4 Ww, and 1/4 ww: phenotype resulting in 3/4 widow\'s peak and 1/4 straight hair line: thus 75%?
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