Chapter 11 mastering biology

How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous?

Cross the green-pod plant with a yellow-pod plant.

Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel’s findings does her test cross illustrate?

law of segregation

During which part of meiosis (meiosis I or meiosis II) do the two alleles of a gene separate? During which phase does the separation occur?

meiosis I, Anaphase

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) and
the 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, gren wrinkled Plant 3: yellow round, RRYy, yellow round, green round Plant 4: green wrinkled, rryy, green wrinkled

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

From these results, determine the relationship between the mutant allele and its corresponding wild-type allele in each line.

Twist: The mutant allele is dominant to its corresponding wild type allele Forked: The mutant allele is dominant to its corresponding wild type allele Pale: The mutant allele is neither dominant nor completely recessive to its corresponding wild type allele

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).

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+

You continue your analysis by crossing the forked and twist lines. Your results are as follows:
Cross between twist and forked
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

You decide to designate the twist allele as FT to distinguish it from the forked allele F.

F^(T)F^(T) FF FF^(T)

If we assume that each square in the Punnett square represents one offspring of the heterozygous AaBbCc parents, then the squares below the Punnett square show the phenotypic frequencies of individuals with the same number of dark-skin alleles. These frequencies can then be plotted in a histogram. (Figure 1)
Which phenotype has the highest frequency?

the phenotype that results from having 3 dark-skin alleles

In order to discern patterns in the distribution of alleles, it is helpful to sketch a curve over the bars of the histogram, and a line through the highest bar. (Figure 2)
What pattern does the distribution of frequency values show in this histogram?

a normal distribution, characterized by frequencies that are symmetrically distributed around one central peak value

What is the best explanation for the curve’s shape?

Equal numbers of light-skin and dark-skin alleles result in a majority of the offspring having intermediate phenotypes.

You can use the Punnett square to see what would happen to the distribution of phenotype frequencies if one of the three genes were lethal when homozygous recessive. Select (Figure 3) from the drop-down menu. Notice that bb has been used as an example of a lethal genotype. Because bb individuals would not survive, those squares have been crossed out.
Now what are the phenotypic frequencies of the surviving offspring?

0=0 1=2 2=9 3=16 4=14 5=6 6=1

How does the shape of this curve compare with the curve in Figure 2? (Select Figure 2 from the drop-down menu.)

It is now skewed toward a higher number of dark-skin alleles.

What does this example indicate about the distribution of phenotype frequencies in a population?

A lethal combination of alleles in a population will result in phenotype frequencies that do not have a normal distribution.

Look over the pedigree you constructed in Part A.
Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia?

autosomal recessive

Jane and John are considering having another child. Given the pedigree you constructed and the mode of inheritance for galactosemia, what is the risk that their next child will have the disorder?

1/4 (because they are both heterozygotes)

If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactose- and galactose-free diet as soon as possible after birth. Even on this diet, affected individuals may still suffer from learning disabilities, ovarian failure (in young women), late-onset cataracts, and early death.
Which of the following tests would be most useful for Jane and John to have?

newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn’s blood)

Quantitative characters vary in a population along a continuum. How do such characters differ from the characters investigated by Mendel in his experiments on peas?

Quantitative characters are due to polygenic inheritance, the additive effects of two or more genes on a single phenotypic character. A single gene affected all but one of the pea characters studied by Mendel.

Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellow-round seeds (YYRR) and a plant with green-wrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation.

The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%.

Each chromosome in this homologous pair possesses a different allele for flower color. Which statement about this homologous pair of chromosomes is correct?

These homologous chromosomes represent a maternal and a paternal chromosome.

When a dominant allele coexists with a recessive allele in a heterozygote individual, how do they interact with each other?

They do not interact at all.