BIO181 Chapter 15 MasteringBiology Homework

What can we observe in order to visualize Mendel’s Law of Segregation?
– sister chromatids separating during mitosis
– the behavior of sex-linked genes
– the replication of DNA
– homologous chromosomes separating during meiosis II
– homologous chromosomes separating during meiosis I

homologous chromosomes separating during meiosis I

Experimental technique: Reciprocal crosses
– w+ w+
– w+ w
– w w
– w+ Y
– w Y
– w
– w+
– Y

First half of reciprocal cross: a) w Y b) Y c) w+ d) w+ Y Second half of reciprocal cross: e) w+ Y f) Y g) w h) w Y

Experimental results: The F2 generation
– w+ w+
– w+ w
– w w
– w+ Y
– w Y
– w
– w+
– Y

a) w+ b) w c) w+ w+ d) w+ Y e) w+ w f) w Y

Experimental prediction: Comparing autosomal and sex-linked inheritance
– 0
– 25
– 50
– 75
– 100

1. 100, 0 2. 0, 100 3. 100, 0 4. 100, 0

Independent assortment of three genes
Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.
– 27:9:9:9:3:3:3:1
– 1:1:1:1:1:1:1:1
– 1:0:0:0:0:0:0:0
– 1:0:1:0:1:0:1:0

1:1:1:1:1:1:1:1

Now, suppose that the three tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how?
Which statement best predicts the results of the cross MmDdPp x mmddpp assuming that all three genes are linked?
– Lack of independent assortment means that you cannot predict the frequencies of phenotypes in the offspring.
– All eight possible phenotypes would occur in equal numbers in the offspring (1:1:1:1:1:1:1:1).
– All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.
– Only the parental phenotypes could possibly occur in the offspring.

All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.

Suppose that you perform the cross discussed in Part B: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results:

a) 12cM b) 5cM c) d d) p

In cosmos plants, purple stem (A) is dominant to green stem (a), and short petals (B) is dominant to long petals (b). In a simulated cross, AABB plants were crossed with aabb plants to generate F1 dihybrids (AaBb), which were then test crossed (AaBb X aabb). 900 offspring plants were scored for stem color and flower petal length. The hypothesis that the two genes are unlinked predicts the offspring phenotypic ratio will be 1:1:1:1.
Using the ratio of 1:1:1:1, calculate the expected number of each phenotype out of the 900 total offspring. Drag the correct values onto the data table. Labels may be used once, more than once, or not at all.

225, 225, 225, 225

The goodness of fit is measured by χ2. This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match.
The formula for calculating this value is
χ^2=∑(o−e)^2/e
where o = observed and e = expected.
The expected and observed data have been entered into the table below. Carry out the operations indicated in the top row. In the last column, enter your answers to two decimal places. Then add up the entries in the last column to find the χ^2 value.

Deviation (o-e): -5 -15 6 14 (o-e)^2 25 225 36 196 (o-e)^2/e .11 1 .16 .87 Sum 2.14

The χ2 value means nothing on its own–it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests the observed data is not consistent with the hypothesis, and thus the hypothesis should be rejected.
What is the hypothesis that you are testing?
– The two genes are unlinked and are assorting independently, leading to a 1:1:1:1 ratio of phenotypes in the offspring.
– The two genes are unlinked and are assorting independently, leading to a 1:1:0:0 ratio of phenotypes in the offspring.
– The two genes are linked and are assorting together, leading to a 1:1:1:1 ratio of phenotypes in the offspring.
– The two genes are linked and are assorting together, leading to a ratio of phenotypes in the offspring that deviates significantly from 1:1:1:1.

The two genes are unlinked and are assorting independently, leading to a 1:1:1:1 ratio of phenotypes in the offspring.

Between which values on the df = 3 line does your calculated χ^2 value lie?
– between 1.42 and 2.19
– between 1.42 and 2.37
– between 1.65 and 2.19
– between 2.37 and 3.66

between 1.42 and 2.37

What is the probability range that your data fit the expected 1:1:1:1 ratio of phenotypes?
– p < 0.70
– 1.42 < p < 2.37
– p = 0.50
– 0.50 < p < 0.70

0.50 < p < 0.70

Based on whether there are non-significant differences (p &gt; 0.05) or significant differences (p ≤ 0.05) between the observed and expected values, you can determine if the data are consistent with the hypothesis that the two genes are unlinked and assorting independently.
Do your results support the hypothesis that the stem color and petal length genes are unlinked and assorting independently, or do the observed values differ from the expected values enough to reject this hypothesis?

– The hypothesis is supported. Because p ≤ 0.05, the differences between the observed and expected values are not statistically significant.
– The hypothesis is not supported. Because p ≤ 0.05, the differences between the observed and expected values are statistically significant.
– The hypothesis is supported. Because 0.50 &lt; p &lt; 0.70, the differences between the observed and expected values are not statistically significant.
– The hypothesis is not supported. Because 0.50 &lt; p &lt; 0.70, the differences between the observed and expected values are statistically significant.

The hypothesis is supported. Because 0.50 < p < 0.70, the differences between the observed and expected values are not statistically significant.

The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes.

The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown.
Drag one label into the space to the right of each chromosome or pair of chromosomes. You can use a label once, more than once, or not at all.

deletion duplication translocation inversion duplication translocation inversion

Suppose a diploid cell with three pairs of homologous chromosomes (2n = 6) enters meiosis.
How many chromosomes will the resulting gametes have in each of the following cases?
Drag one label into each space at the right of the table. Labels can be used once, more than once, or not at all.

3 only 2 or 4 0 or 6 2, 3, or 4 0, 3, or 6

Suppose that a carrier of familial Down syndrome mated with a person with a normal karyotype. Which gamete from the carrier parent could fuse with a gamete from the normal parent to produce a trisomy-21 zygote?
Drag one of the white cells (representing gametes) to the white target in the diagram. Drag one of the pink cells (representing zygotes) to the pink target.

White: 14(green) 21(yellow) 21 (yellow) Pink: 14(green) 14(green) 21(yellow) 21(yellow) 21(yellow)

How are human mitochondria inherited?

– from the father only
– as linear DNA
– from the mother only
– as an X-linked trait
– without DNA

from the mother only

Complete the table by dragging the white labels to the gray targets, the blue labels to the blue targets, and the pink labels to the pink targets. Labels can be used once, more than once, or not at all.

a) AB b) ab c) aB d) Ab e) AaBb f) aabb g) aaBb h) Aabb i) non REC j) non REC k) REC l) REC

You have also identified a third recessive mutation that you call crimson. Wasps homozygous for crimson have a red abdomen. (Wild-type abdomens are yellow.)

You make pure-breeding crimson (cc) and triple-mutant (apricot, blunt, crimson) lines. To map the three loci, you perform a trihybrid (three-point) testcross: You first cross crimson wasps with (apricot, blunt) wasps to make an F1, and then you testcross the F1 with the triple-mutant line.

The F2 has eight phenotypes. For each F2 phenotype, determine the genotype of the gamete it received from the F1 trihybrid.

Complete the table by dragging the labels to the correct targets.

a) ABC b) abc c) abC d) ABc e) AbC f) aBc g) Abc h) aBC

The eight gamete classes for the testcross in Part B are shown in the table below. For each gamete, determine if it represents a recombinant (REC) or nonrecombinant (nonREC) gamete for the three pairs of loci: apricot and blunt (the A-B column), blunt and crimson (the B-C column), and apricot and crimson (the A-C column).
Complete the table by dragging the labels to the correct targets. Labels can be used once, more than once, or not at all.

non REC, REC, REC non REC, non REC, non REC REC, non REC, REC REC, REC, non REC

Use these data to complete the linkage map shown below.
Drag the blue labels to the blue targets to identify the allele loci indicated by vertical dashes. Note that the apricot allele is already labeled.
Drag two of the white labels to the gray targets to identify the distances between loci. Each white label gives the distance in map units. One map unit, or one centiMorgan (cM), is equivalent to a 1% recombination frequency.

a) 20mu b) 10mu c) blunt d) crimson

Your 28 year old friend is pregnant. Which of the following screenings is she most likely to initially have to test for Down syndrome?
– blood test
– amniocentesis
– chorionic villus sampling
– fetal DNA

Blood test

How is the fetal DNA used in this new screening tool unique?

– It is not from a cell, but is floating freely in the mother’s blood.
– It is taken out of the umbilical cord.
– It contains fragments of RNA.
– It is all mitochondrial DNA.

It is not from a cell, but is floating freely in the mother’s blood.

Which of the following is true?

– The chance of having a child with a chromosomal abnormality increases with the age of the mother.
– Age of the mother has no effect on the chance of having a child with a chromosomal abnormality.
– The chance of having a child with a chromosomal abnormality decreases with the age of the father.
– The chance of having a child with a chromosomal abnormality decreases with the age of the mother.

The chance of having a child with a chromosomal abnormality increases with the age of the mother.

Which of the following is true regarding this new fetal DNA test?

– A decreased number of false positive tests for Down syndrome and an increased number for Trisomy 18.
– An increased number of false positive tests for both Down syndrome and Trisomy 18.
– A decreased number of false positive tests for both Down syndrome and Trisomy 18.
– An increased number of false positive tests for Down syndrome and an decreased number for Trisomy 18.

A decreased number of false positive tests for both Down syndrome and Trisomy 18.

Which of the following best summarizes current research on this new fetal DNA test?

– It works well for finding all genetic abnormalities.
– It works well for finding errors in chromosome number.
– It is too risky to the mother to continue the research.
– It works well, but only for finding sex-linked disorders.

It works well for finding errors in chromosome number.

At the time of Mendel’s pea plant experiments, no one knew how organisms formed gametes. As Mendel studied the inheritance of two different characters, he wondered how the alleles for the two characters segregated into gametes. Mendel had two hypotheses for how this might work.
The figure below shows the experiment that Mendel used to distinguish between these two hypotheses. The results of the experiment confirmed that the alleles for these characters undergo independent assortment.

Drag the terms to the appropriate blanks to complete the sentences below. Not all terms will be used.

1. yyrr 2. YyRr 3. all possible combinations 4. YR, Yr, yR, yr 5. four, 9 to 3 to 3 to 1

In the figure below, you can see Mendel’s experiment again, this time superimposed on the events of meiosis and fertilization. How does chromosomal inheritance during meiosis explain Mendel’s law of independent assortment?
Drag the labels to their appropriate locations in the table below.

a) metaphase I b) anaphase ! c) telophase II d) two chromosome arrangements equally probable at metaphase plate e) homologous chromosomes separate f) sister chromatids have separated g) chromosomes in haploid gametes combine in diploid zygote h) Y may sort with R or r; y may sort with R or r i) four types of gametes produced: YP, Yr, yR, and yr j) four phenotypes produced in F2 generation in 9:3:3:1 ratio

To try to explain this unusual data, you come up with two alternate hypotheses in addition to your original hypothesis of independent assortment.

Drag the labels to their appropriate locations in the table below. Labels may be used more than once. (Hint: First, figure out the predicted F1 gametes for each hypothesis; then construct a Punnett square to help you fill in the rest of the table.)

a) no b) BP, Bp, bP, bp in equal numbers c) mostly BP and bp d) 2 predominant phenotypes e) 2 predominant phenotypes f) 9 blue-pointy: 1 purple-rounded g) 3 blue-pointy: 1 purple-rounded h) no i) yes

Consider the following family history:
Bob has a genetic condition that affects his skin.
Bob’s wife, Eleanor, has normal skin. No one in Eleanor’s family has ever had the skin condition.
Bob and Eleanor have a large family. Of their eleven children, all six of their sons have normal skin, but all five of their daughters have the same skin condition as Bob.

Based on Bob and Eleanor’s family history, what inheritance pattern does the skin condition most likely follow?

– X-linked dominant
– Y-linked
– X-linked recessive
– autosomal dominant
– autosomal recessive

X-linked dominant

Predict the eye colors of F1 and F2 generations. (Assume that the F1 flies are allowed to interbreed to produce the F2 generation.)
Drag the correct label to the appropriate location in the table. Labels can be used once, more than once, or not at all.

Predicted eye color(s): all wild type all wild type all wild type 1/2 wild type 1/2 vermilion

Red-green color blindness is due to an X-linked recessive allele in humans. A widow’s peak (a hairline that comes to a peak in the middle of the forehead) is due to an autosomal dominant allele.
Consider the following family history:

A man with a widow’s peak and normal color vision marries a color-blind woman with a straight hairline.
The man’s father had a straight hairline, as did both of the woman’s parents.
Use the family history to make predictions about the couple’s children.

Drag the correct label to the appropriate location in the table. Not all labels will be used.

1) 1/4 2) 1/2 3) 0 4) 1

Humans are diploid and have 46 chromosomes (or two sets). How many sets of chromosomes are found in each human gamete?
– 1
– 2
– 3
– 4
– 5

1

Humans are diploid and have 46 chromosomes. How many chromosomes are found in each human gamete?
– 12
– 23
– 36
– 45
– 92

23

_____ is the process by which haploid gametes form a diploid zygote.
– Embryogenesis
– Meiosis
– Gastrulation
– Fertilization
– Mitosis

Fertilization

A particular diploid plant species has 48 chromosomes, or two sets. A mutation occurs and gametes with 48 chromosomes are produced. If self-fertilization occurs, the zygote will have _____ set(s) of chromosomes.
– 1
– 2
– 3
– 4
– 5

4

Which of these terms applies to an organism with extra sets of chromosomes?
– monosomy
– haploid
– trisomy
– polyploid
– diploid

polyploid

Mutant tetraploid plants _____.

– are usually sickly
– are able to interbreed with their parents
– have an odd number of chromosomes
– are unable to interbreed with a diploid plant
– unable to self-fertilize

are unable to interbreed with a diploid plant

Most polyploid plants arise as a result of _____.

– self-fertilization
– a mutation of gamete formation
– meiosis
– mitosis
– hybridization

hybridization