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2.01 Mendel's analysis of dihybrid crosses led to the principle of independent assortment, a very important idea related to the idea that genes act as particles.
2.01 Mendel's analysis of dihybrid crosses led to the principle of independent assortment, a very important idea related to the idea that genes act as particles.
(a) Using your own words, explain what is meant by independent assortment. You can look up the definition first in the textbook glossary, but be sure you can state what these terms mean in language that you understand.
(a) Using your own words, explain what is meant by independent assortment. You can look up the definition first in the textbook glossary, but be sure you can state what these terms mean in language that you understand.
Answer: When two genes (two pairs of alleles) are being considered, the segregation of the first pair is independent of the segregation of the second pair. This applies to more than two pairs as well. Because of independent assortment, all allele combinations that are possible in the gametes from one parent are equally likely. This is why a Punnett square for a dihybrid cross correctly predicts the outcome of the cross.
Answer: When two genes (two pairs of alleles) are being considered, the segregation of the first pair is independent of the segregation of the second pair. This applies to more than two pairs as well. Because of independent assortment, all allele combinations that are possible in the gametes from one parent are equally likely. This is why a Punnett square for a dihybrid cross correctly predicts the outcome of the cross.
(b) Why is it necessary to use a dihybrid cross to demonstrate independent assortment?
(b) Why is it necessary to use a dihybrid cross to demonstrate independent assortment?
Answer: With a monohybrid cross, only one allele pair is considered. There is no other allele pair for the first one to be independent of. So one must have consider two (or more) allele pairs in the same cross to demonstrate that one is independent of the other.
Answer: With a monohybrid cross, only one allele pair is considered. There is no other allele pair for the first one to be independent of. So one must have consider two (or more) allele pairs in the same cross to demonstrate that one is independent of the other.
2.02 Two of the traits Mendel studied in peas were flower position and pod color. From monohybrid crosses, he determined that axial flowers (A) is dominant to terminal (a), and green pods (P) is dominant to yellow (p). Suppose that you pollinate the flowers of a plant with axial flowers and green pods with pollen taken from a plant with terminal flowers and yellow pods. Both of these parent plants are known to be true-breeding. F1 plants are then allowed to self pollinate to produce the F2 generation.
2.02 Two of the traits Mendel studied in peas were flower position and pod color. From monohybrid crosses, he determined that axial flowers (A) is dominant to terminal (a), and green pods (P) is dominant to yellow (p). Suppose that you pollinate the flowers of a plant with axial flowers and green pods with pollen taken from a plant with terminal flowers and yellow pods. Both of these parent plants are known to be true-breeding. F1 plants are then allowed to self pollinate to produce the F2 generation.
(a) What are the genotypes and phenotypes of the parents (P generation)?
(a) What are the genotypes and phenotypes of the parents (P generation)?
Answer: The axial/green plant is AAPP; the terminal/yellow plant is aapp. (The parents are both homozygous for both genes because we are told they are true-breeding.)
Answer: The axial/green plant is AAPP; the terminal/yellow plant is aapp. (The parents are both homozygous for both genes because we are told they are true-breeding.)
(b) What genotypes and phenotypes are possible in the F1 plants?
(b) What genotypes and phenotypes are possible in the F1 plants?
Answer: All of the F1 plants are heterozygous for both genes (AaPp) and would have axial flowers and green pods (both dominant phenotypes).
Answer: All of the F1 plants are heterozygous for both genes (AaPp) and would have axial flowers and green pods (both dominant phenotypes).
(c) Using a Punnett square, determine which genotypes and phenotypes are possible in the F2 generation and calculate the expected ratio of each.
(c) Using a Punnett square, determine which genotypes and phenotypes are possible in the F2 generation and calculate the expected ratio of each.
Answer:
Answer:
(d) Would you expect the results of this cross to be different if, in the P generation, the pollen had come from the axial/green plant and the terminal/yellow plant was used as the female? Explain.
(d) Would you expect the results of this cross to be different if, in the P generation, the pollen had come from the axial/green plant and the terminal/yellow plant was used as the female? Explain.
Answer: No. Mendel found that reciprocal crosses always produced similar results.
Answer: No. Mendel found that reciprocal crosses always produced similar results.
2.03 A cross was made between fruit flies of genotypes AAbb and aaBB.
2.03 A cross was made between fruit flies of genotypes AAbb and aaBB.
(a) Show the Punnett square for the expected F2 progeny types.
(a) Show the Punnett square for the expected F2 progeny types.
Answer: F1 generation is all AaBb because the parents are homozygous. For the F2 generation, AaBb x AaBb:
Answer: F1 generation is all AaBb because the parents are homozygous. For the F2 generation, AaBb x AaBb:
(b) What proportions of A-B-, A-bb, aaB-, and aabb progeny do you expect in the F2?
(b) What proportions of A-B-, A-bb, aaB-, and aabb progeny do you expect in the F2?
Answer: A-B- 9/16, A-bb 3/16, aaB- 3/16, aabb 1/16
Answer: A-B- 9/16, A-bb 3/16, aaB- 3/16, aabb 1/16
2.04 In a particular cross, one parent has genotype RrYy.
2.04 In a particular cross, one parent has genotype RrYy.
(a) What is the probability of this parent producing each of the following types of gametes: RY, Ry, rY, and ry?
(a) What is the probability of this parent producing each of the following types of gametes: RY, Ry, rY, and ry?
Answer: These are the four possible gametes, which would be produced in equal numbers. The probability of any one of them is 1/4.
Answer: These are the four possible gametes, which would be produced in equal numbers. The probability of any one of them is 1/4.
(b) If the other parent has genotype rryy, what is the probability that an offspring will be RrYy? Rryy? RRYy?
(b) If the other parent has genotype rryy, what is the probability that an offspring will be RrYy? Rryy? RRYy?
Answer: The Punnett square for this cross looks like:
Answer: The Punnett square for this cross looks like:
The probabilities are: RrYy - 1/4, Rryy - 1/4, RRYy - 0
The probabilities are: RrYy - 1/4, Rryy - 1/4, RRYy - 0
2.05 Consider the following crosses in Drosophila. The two traits being investigated involve eye color and the presence or absence of wing crossveins. The outcomes of four crosses are shown below.
2.05 Consider the following crosses in Drosophila. The two traits being investigated involve eye color and the presence or absence of wing crossveins. The outcomes of four crosses are shown below.
(a) Which eye color is dominant and which is recessive?
(a) Which eye color is dominant and which is recessive?
Answer: We can consider the two traits separately. First looking at the eye color trait, in cross 2, both parents have red eyes, but offspring can have either red or orange. This means that both parents must be heterozygous, so red is dominant.
Answer: We can consider the two traits separately. First looking at the eye color trait, in cross 2, both parents have red eyes, but offspring can have either red or orange. This means that both parents must be heterozygous, so red is dominant.
(b) Which wing vein type is dominant and which is recessive?
(b) Which wing vein type is dominant and which is recessive?
Answer: Now looking at just the wing trait-in cross 4, both parents have crossveins, but some offspring are crossveinless. This means that both parents must be heterozygous, so crossveins is dominant.
Answer: Now looking at just the wing trait-in cross 4, both parents have crossveins, but some offspring are crossveinless. This means that both parents must be heterozygous, so crossveins is dominant.
(c) What is the most likely genotype of each parent for each of the four crosses?
(c) What is the most likely genotype of each parent for each of the four crosses?
Answer: Cross 1: rrVv x rrVv, cross 2: RrVv x Rrvv, cross 3: R-vv x R-Vv (at least one parent must be RR, cross 4: RrVv x RrVv
Answer: Cross 1: rrVv x rrVv, cross 2: RrVv x Rrvv, cross 3: R-vv x R-Vv (at least one parent must be RR, cross 4: RrVv x RrVv
2.06 Consider the following cross: Aa Cc FF Yy Rr NN x AA CC Ff Yy Rr nn.
2.06 Consider the following cross: Aa Cc FF Yy Rr NN x AA CC Ff Yy Rr nn.
(a) List all possible different types of gametes the each parent can produce.
(a) List all possible different types of gametes the each parent can produce.
Answer: Female: ACFYRN ACFYrN ACFyRN ACFyrN
Answer: Female: ACFYRN ACFYrN ACFyRN ACFyrN
AcFYRN AcFYrN AcFyRN AcFyrN
AcFYRN AcFYrN AcFyRN AcFyrN
aCFYRN aCFYrN aCFyRN aCFyrN
aCFYRN aCFYrN aCFyRN aCFyrN
acFYRN acFYrN acFyRN acFyrN
acFYRN acFYrN acFyRN acFyrN
Male: ACFYRn ACFYrn ACFyRn ACFyrn
Male: ACFYRn ACFYrn ACFyRn ACFyrn
ACfYRn ACfYrn ACfyRn ACfyrn
ACfYRn ACfYrn ACfyRn ACfyrn
(b) What size (how many boxes) would the Punnett square have to be in order to analyze the progeny of this cross?
(b) What size (how many boxes) would the Punnett square have to be in order to analyze the progeny of this cross?
Answer: 16 x 8 = 128 boxes
Answer: 16 x 8 = 128 boxes
2.07 Some very complicated crosses are being done involving 32 different genes in mice. Calculate the total number of different gametes that could be produced by each of the following:
2.07 Some very complicated crosses are being done involving 32 different genes in mice. Calculate the total number of different gametes that could be produced by each of the following:
(a) How many different sperm could a male produce that is homozygous for all 32 genes
(a) How many different sperm could a male produce that is homozygous for all 32 genes
Answer: 1 gamete (no heterozygous genes)
Answer: 1 gamete (no heterozygous genes)
(b) How many different eggs could a female produce that is heterozygous for 3 genes and homozygous for the other 29 genes?
(b) How many different eggs could a female produce that is heterozygous for 3 genes and homozygous for the other 29 genes?
Answer: 23 = 8 gametes
Answer: 23 = 8 gametes
(c) How many different eggs could a female produce that is heterozygous for 10 genes and homozygous for the other 22 genes?
(c) How many different eggs could a female produce that is heterozygous for 10 genes and homozygous for the other 22 genes?
Answer: 210 = 1024 gametes
Answer: 210 = 1024 gametes
(d) How many different sperm could a male produce that is heterozygous for all 32 genes?
(d) How many different sperm could a male produce that is heterozygous for all 32 genes?
Answer: 232 = 4, 294, 967, 296 gametes - WOW!
Answer: 232 = 4, 294, 967, 296 gametes - WOW!
2.08 There are two genes in fruit flies that determine eye color and wing shape. Red eyes is dominant to sepia (brown) eyes, and normal wings is dominant to vestigial (shriveled) wings. Describe a procedure you could use to determine the genotype of a fly that has red eyes and normal wings. This fly may be homozygous or heterozygous for either gene.
2.08 There are two genes in fruit flies that determine eye color and wing shape. Red eyes is dominant to sepia (brown) eyes, and normal wings is dominant to vestigial (shriveled) wings. Describe a procedure you could use to determine the genotype of a fly that has red eyes and normal wings. This fly may be homozygous or heterozygous for either gene.
Answer: The unknown fly could be mated with a fly that has both recessive traits (sepia/vestigial), a typical test cross. Then look for any recessive traits to show up in the offspring. If sepia offspring are produced, then the parent was heterozygous for the eye color gene. If not it was homozygous. If vestigial offspring are produced, then the parent was heterozygous for the wing gene. If not it was homozygous.
Answer: The unknown fly could be mated with a fly that has both recessive traits (sepia/vestigial), a typical test cross. Then look for any recessive traits to show up in the offspring. If sepia offspring are produced, then the parent was heterozygous for the eye color gene. If not it was homozygous. If vestigial offspring are produced, then the parent was heterozygous for the wing gene. If not it was homozygous.
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