Cross Review Answers:
1. B - This problem requires the product rule to be applied in order to determine the results from two separate events. Therefore, the chances for getting heads on both coins is 1/2 * 1/2, for getting tails on both coins is 1/2 * 1/2, and for getting one head and one tail is the remaining 50%.
2. C - The only way for the recessive phenotype to be expressed is if both recessive alleles are present in the genotype.
3. D - The offspring of a cross between two hybrids should yield a 3:1 phenotypic ratio; therefore, approximately 90 offspring should be yellow.
4. A - This question is referring to a test cross; therefore, the homozygous recessive individual must be mated to the unknown.
5. C - Since the wife of the man had a father with blue eyes, she must be carrying the blue-eyed allele; therefore, their children have a 50/50 chance of having blue eyes.
6. A - The product rule must be applied to solve this problem. The probability of the gametes containing an A allele is 1/2; B allele - 1/1; C allele - 1/1; D allele - 1/1; E allele - 1/1; therefore, (1/2*1*1*1*1) yields only 2 distinct kinds of gametes.
7. B - The higher the frequency of crossing over between two genes, the farther apart they are on a chromosome; therefore, the correct sequence is represented by letter B.
8. E - By definition, a dominant allele is expressed whether it is homozygous dominant or heterozygous.
9. C - Codominance exists when the heterozygote expresses both phenotypes of the homozygotes as in Sickle Cell Anemia.
10. D - Incomplete dominance exists when the heterozygote expresses a phenotype intermediate of the homozygotes as in the pink heterozygotes of the Snapdragon Flowers.
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