Mastering bio ch.11

Part D - Assigning genotypes for codominant alleles You decide to designate the twist allele as FT to distinguish it from the forked allele F. Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels may be used once, more than once, or not at all.

(Twisted flower)F^T F^T (forked flower)FF (forked and twisted flower)FF^T

Part A - Determining relationships between alleles You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure wild-type line. The numbers in the F2 indicate the number of progeny in each phenotypic class.From these results, determine the relationship between the mutant allele and its corresponding wild-type allele in each line.Label each mutant line with the best statement from the list below. Labels may be used once, more than once, or not at all.

(Twisted leaves)The mutant allele is dominant to its corresponding wild-type allele (Forked leaves)The mutant allele is dominant to its corresponding wild-type allele (pale leaves)The mutant allele is neither dominant nor completely recessive to its corresponding wild-type allele

One character in peas that Mendel studied was yellow versus green seeds. A cross between a homozygous yellow line (YY) and a homozygous green line (yy) will result in F1 plants that are heterozygous (Yy) for this trait and produce yellow seeds. When an F1 plant undergoes meiosis, what gamete types will it produce, and in what proportions?

1/2 Y 1/2 y

In a mating of mice with the genotypes AYaBb x AYaBb , what is the probability that a live-born offspring will have yellow fur?

2/3

What would be the expected frequency of agouti brown offspring in the litter?

3/16

Use the completed Punnett square in Part B to answer the questions below about the F2 generation. Answer each question in the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

3/4 1/3 63/64 27/64

In a cross between two mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur?

Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BB or Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.

Part C - The effect of a third gene on fur color In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse.

Because the C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally. AaBbcc-Albino, AaBBCC-Agouti Black Aabbcc-Albino AAbbCc Agouti Brown aaBbCc-Solid color, Black AABBcc-Albino

Mendel studied pea plants dihybrid for seed shape (round versus wrinkled) and seed color (yellow versus green). Recall that the round allele (R) is dominant to the wrinkled allele (r) andthe yellow allele (Y) is dominant to the green allele (y). The table below shows the F1 progeny that result from selfing four different parent pea plants.Use the phenotypes of the F1 progeny to deduce the genotype and phenotype of each parent plant.

Plant 1: green round, Rryy, green round, green wrinkled Plant 2: yellow round, RrYy, yellow round, yellow wrinkled, green round, green wrinkled Plant 3: yellow round, RRYy, yellow round, green round Plant 4: green wrinkled, rryy, green wrinkled

Part C - Crossing the forked and twist mutants You continue your analysis by crossing the forked and twist lines. Your results are as follows:Which of the following statements best explains the outcome of this cross?

The forked mutation and the twist mutation are codominant alleles of the same locus.

Part B - Crossing the forked and pale mutants You continue your genetic analysis by crossing the forked and pale mutant lines with each other. The leaves of the F1 are light green (intermediate between pale and wild-type leaves) and forked. The F2 has six phenotypic classes, as shown below.You designate the forked mutant allele as F (wild type = f+ ) and the pale mutant allele as p (wild type = P). Consider the alleles for leaf color first. Drag the white labels to the white targets to identify the genotype of each F2 class. Remember that p (the pale mutant allele) and P (the wild-type allele) are incompletely dominant to each other. Consider the alleles for leaf shape next. Drag the blue labels to the blue targets to identify the genotype of each F2 class. Remember that F (the forked mutant allele) is dominant to f + (the wild-type allele).Labels may be used once, more than once, or not at all.

a) PP b)F_ c) Pp d) F_ e)pp f) F_ g) PP h) f+ f+ i) Pp j) f+ f+ k) pp l) f+ f+

Punnett squares are convenient ways to represent the types and frequencies of gametes and progeny in experimental crosses.This Punnett square shows the results of a Yy x Yy cross to form F2 progeny. Use your understanding of Mendel's law of segregation and the rules of probability to complete the Punnett square for this cross.-First identify the gametes. Use pink labels to identify the male and female gamete types and white labels to identify the gamete frequencies. -Then identify the F2 progeny. Use pink labels to identify the progeny genotypes and white labels to identify the progeny frequencies.

parent 1: 1/2 Y ; 1/2 y parent 2: 1/2 Y ; 1/2y 1/4 YY ; 1/4 Yy 1/4 Yy ; 1/4 yy

A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F1: [round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green]. Use this information to deduce the genotypes of the parent plants.

yellow round = RrYy yellow wrinkled= rrYy

For the cross in Part B, predict the frequencies of each of the phenotypes in the F1 progeny, and determine the genotype(s) present in each phenotypic class.

yellow round: 3/8, RrYY, RrYy (x2) yellow wrinkled: 3/8, rrYY, rrYy(x2) green round: 1/8, Rryy green wrinkled: 1/8, rryy