Final answer:
The phenotypic ratio for the offspring of two carriers for albinism is 3 non-albinistic:1 albinistic, meaning there's a 25% chance of albinism. In a dwarf-tall pea plant cross, the phenotypic ratio is 1 tall:1 dwarf. Males can be affected by red-green color blindness but are not typically carriers due to the X-linked nature of the trait.
Step-by-step explanation:
The inheritance pattern for albinism is an autosomal recessive trait. This means that to express the phenotype for albinism, an individual must inherit two copies of the recessive allele, one from each parent. In a scenario where both parents are carriers of the albinism trait, the Punnett square for predicting the genotypic ratio would consist of one square with two heterozygous parents (Aa x Aa).
The possible genotypes from such a cross would be AA, Aa, Aa, and aa. The phenotypic ratio of the offspring would be 3 non-albinistic: 1 albinistic. Specifically, this means there is a 75% chance that a child will not have albinism and a 25% chance that a child will have albinism if both parents are carriers.
For the cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous), the Punnett square would have the genotype of the dwarf plant as 'dd' and the tall plant as 'Dd'. The offspring genotypes would be Dd, dd, Dd, and dd. This results in a phenotypic ratio of 2 tall to 2 dwarf plants, or simplified, 1 tall:1 dwarf.
Concerning human males and red-green color blindness: yes, a human male can be a carrier of red-green color blindness. However, because this condition is X-linked, being a carrier commonly refers to females who have two X chromosomes, one of which carries the mutation. Males, having only one X chromosome, will express the trait if that single X chromosome carries the mutation.