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1.01 Genotype and phenotype are two terms that are often confused by beginning biology students, but understand them is crucial to interpreting may genetics problems.
(a) Using your own words, explain the difference between genotype and phenotype. You can look up the definitions first in the textbook glossary, but be sure you can state what these terms mean in language that you understand.
Answer: An individual's phenotype is the expression of the trait that can be observed directly without knowing about the genetic mechanism that underlies the trait. Examples: red flowers, tall stems, hairy body, color-blind. The genotype refers to the actual alleles that are carried by the individual. Genotypes can be abbreviated like AA, aa, Aa, where each letter stands for one allele. A genotype can be homozygous or heterozygous.
(b) A heterozygous tall pea plant is crossed with a dwarf pea plant. What are the phenotypes and genotypes of each parent? Note: To answer this, you will need to look up which of these traits is dominant and which is recessive (if you do not already know).
Answer: Tall is dominant over dwarf. So the tall parent has genotype Tt (heterozygous). The dwarf plant has genotype tt(it must be homozygous because is has the recessive phenotype).
(c) Considering all seven of the pea traits that Mendel tested, is it possible for a heterozygous plant to have the dominant phenotype? Explain.
Answer: Yes. Heterozygous plants always have the dominant phenotype. When at least one of the two alleles is the dominant one, the phenotype is dominant. This is true for all of Mendel's traits because they demonstrated complete dominance.
(d) Is it possible for a heterozygous plant to have the recessive phenotype? Explain.
Answer: No. Heterozygous plants always have the dominant phenotype. Only homozygous recessive plants will exhibit the recessive phenotype.
1.02 When an individual expresses the dominant phenotype for a trait, two genotypes are possible, homozygous for the dominant allele or heterozygous. In a case where the genotype is not known for sure (because the dominant phenotype is present), it is indicated as A-, meaning that there is at least one dominant allele present, but the second allele may be A or a. Suppose you have a single pea plant with round seeds (R-). What cross would you make to determine if it has genotype RR or Rr? Explain how the results of your cross would enable you to determine the genotype?
Answer: You could cross it with a plant that has the recessive phenotpye (rr). This test parent can only produce r gametes, so if the unknown plant was RR, all the offspring would be Rr and have the dominant phenotype. If the unknown plant was Rr,about one-half the offspring would get a r from both parents (rr) and show the recessive phenotype.
1.03 In cucumbers, orange fruit is dominant to cream colored fruit. A true-breeding plant with orange fruit is crossed with a true-breeding plant with cream fruit. The F1 plant are then crossed with each other to produce the F2 generation.
(a) Using allele codes of your choice, show the genotypes and phenotypes of the parents (P generation). You must define what each allele code means.
Answer: (M = orange, m = cream) The orange plant is MM; the cream plant is mm. (The parents are both homozygous because we are told they are true-breeding.)
(b) What are the genotypes and phenotypes are possible in the F1 plants?
Answer: All of the F1 plants are heterozygous (Mm) and have orange fruit.
(c) Using a Punnett square, determine which genotypes and phenotypes are possible in the F2 generation and calculate the expected ratio of each.
Answer:
(d) What would be the expected genotype and phenotype ratios for a backcross between the F1 and the orange parent?
Answer:
(e) What would be the expected genotype and phenotype ratios for a backcross between the F1 and the cream parent?
Answer:
1.04 Maple syrup urine disease is a rare human genetic disease that runs in families. It is so called because a metabolic error causes the urine of affected babies to have an odor like maple syrup. The babies die at an early age if not treated. The parents are almost never affected with these symptoms. If we assume that the disease is caused by a single gene, is it a recessive or a dominant gene? Explain your answer.
Answer: It must be recessive because two unaffected parents can have a child with the disease. In this case the parents are “carriers,” meaning that they carry the recessive gene but do not express it. That is, the parents are heterozygous.
1.05 In one of Mendel's experiments, a true-breeding yellow-seeded plant was crossed with a true-breeding green-seeded plant. All F1 plants were yellow-seeded. Mendel then obtained an F2 generation by allowing F1 plants to self-pollinate.
(a) If one of the F1 plants were crossed with a green-seeded plant, what proportion of the progeny would you expect to be yellow?
Answer: This is a backcross. You would expect 1/2 the offspring to be yellow and 1/2 to be green.
(b) If a number of the yellow-seeded F2 plants were allowed to self-pollinate, approximately what proportion of these crosseswould you expect to produce only yellow progeny?
Answer: One-third. Construct a Punnett square for the F2 generation. Note that of the 3 boxes with the yellow phenotype, 1 box is YY and 2 are Yy. The YY plants will produce only yellow progeny when selfed, so 1/3 of the yellow F2s will produce only yellow offspring.
1.06 The first child of two normally pigmented parents has albinism, a recessive trait that results from a lack of the pigment melanin.
(a) Given that the normal allele is A and the albino allele is a, show the phenotypes and genotypes of the parents and child.
Answer: The pedigree chart below summarizes the family
(b) What is the probability that a second child from these parents will be an albino? (The probability is the fraction of her offspring that you would expect to have PKU. We will define probability and discuss probability calculations in detail later.)
Answer: Make a Punnett square with both parents heterozygous to get this answer. The probability is 1/4 because only one box of the four would have the disease.
(c) What is the probability that the second child will carry at least one copy of the albino allele?
Answer: Use the same Punnett square. Three boxes out of four have at least on copy of the albino allele. The probability is 3/4.
(d) The couple has a second child who has normal pigmentation. What is the probability that this child is a carrier?
Answer: Use the same Punnett square, but just examine the three boxes that would be normal (not aa). Two of those three boxes are Aa (carriers). The probability is 2/3.
1.07 A cross is made between a homozygous tall pea plant and a homozygous short pea plant.
Answer: This is a standard monohybrid cross and follows the same pattern as examples we have seen in class.
(a) What types of gametes could each plant produce?.
Answer: Tall plant produces T gametes; Short plant produces t gametes.
(b) What genotypes and phenotypes would occur in the F1 generation?.
Answer: All the F1s would be tall and heterozygous.
(c) An F1 plant is allowed to self-pollinate. Using a Punnett square, calculate the expected ratios of genotypes and phenotypes in the F2 generation.
Answer: Genotype ratio - 1/4 TT, 1/2 Tt, 1/4 tt. Phenotype ratio - 3/4 tall, 1/4 short
(d) 756 plants are produced in the F2 generation. Based on your expected ratios, how many of these would you expect to be tall and how many short?
Answer: Expected tall, 756 x 3/4 = 567; expected short 756 x 1/4 = 189
(e) The numbers actually obtained are 571 tall and 185 short. Calculate the actual phenotypic ratio. How well does this agree with your expected ratio?
Answer: Actual ratio is 571/185 = 3.09:1. This is good agreement with the expected ratio.
1.08 A woman with the rare recessive disease phenylketonuria (PKU), who had been treated with a diet having low levels of the amino acid phenylalanine, was told that it was unlikely her children would inherit PKU because her husband did not have it. However, her first child had PKU.
(a) What is the most likely explanation?
Answer: The husband must be a carrier (genotype Aa), otherwise all their children would be expected to be normal (but all would be carriers)
(b) Assuming this explanation is true, what would be the probability of her second child having PKU?
Answer: The cross now looks like this: aa x Aa. If you do a Punnett square for this cross, it will be one column by two rows. The two boxes will be Aa and aa. so their is a 1/2 chance that the next child will have PKU. Note that the fact that they have already had a PKU child does not change the probability of any future child will have it. That probability will always be 1/2. Of course we would not have know this if the first child would have been normal. The fact that the first child has PKU is what tells us that the father is Aa.
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