Final answer:
The genotype of a wavy haired individual cannot be determined without specific genetic information, but assuming incomplete dominance, it would be 'SC'. For the pea plants, the phenotype ratio of offspring from a dwarf (homozygous recessive) and a tall (heterozygous) cross is 1 tall : 1 dwarf. Males cannot be carriers of red-green color blindness as it is an X-linked recessive trait.
Step-by-step explanation:
Genotype of Wavy Haired Individual and Punnett Square Analysis
The question seems to refer to a genetic trait with incomplete dominance, where wavy hair is the result of having one allele for straight hair and one for curly hair. However, the actual genotype of wavy hair cannot be determined without specific information about the genetics of hair texture. Assuming incomplete dominance, if 'S' represents straight hair and 'C' represents curly hair, then a wavy haired individual would have the genotype 'SC'. For simplicity, we will use this assumption to explain how a Punnett Square is used.
To determine the offspring's possible genotypes and phenotypes when crossing two wavy haired individuals ('SC' x 'SC'), we create a Punnett Square:
The resulting phenotypic ratio from this cross would be:
For the phenotype ratios question from the list (number 23), using a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous), the phenotypic ratio of the offspring would be:
This simplifies to a ratio of 1 tall : 1 dwarf.
In response to question 24, a human male cannot be a carrier of red-green color blindness; males are either affected or unaffected since the trait is X-linked recessive and males have only one X chromosome.