Homework Problems

Homework Problems
Homework Problems
BIOL 129-001 IMCDB
Spring 2009
Date to be SubmittedProblem Set
Mar 23 1
Mar 27 2
Apr 1 3
Apr13 4
Apr 17 5


Last updated 1/19/09

Problem Set 1

Homework Problems I
Homework Problems I

  1. You cross wild, gray-colored mice with white (albino) mice. In the first generation, all are gray. From many litters, you obtain in the F2 198 gray and 72 white mice.

    1. Propose a model to explain these results.
    2. On the basis of this model, diagram the cross and compare the observed results with those expected (e.g., use a Chi-square test).

  2. A woman has a rare abnormality of the eyelids called ptosis, which makes it impossible for her to open her eyes completely. The condition has been found to depend on a single dominant allele (P). The young woman's father had ptosis, but her mother had normal eyelids. Her father's mother had normal eyelids. (a) What are the probable genotypes of the woman, her father, and her mother? (b) What proportion of her children will be expected to have ptosis if she marries a man with normal eyelids?

  3. Short hair in rabbits (S) is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents of each cross.

    Parents
    		Offspring
    (a) short X short
    		4 short and 2 long
    (b) short X short
    		8 short
    (c) short X long
    		12 short
    (d) short X long
    		3 short and 1 long
    (e) long X long
    		2 long

  4. In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruits is crossed with a plant homozygous for cream fruits. The F1 are intercrossed to produce the F2.

    1. Give the genotypes and phenotypes of the parents, the F1, and the F2.
    2. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the orange parent.
    3. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the cream parent.

Problem Set 2

Homework Problems II

  1. A woman with detached earlobes (dominant trait) and a man with attached earlobes have a son with detached earlobes. Their son has a child with a woman with detached earlobes who has a mother with attached earlobes. What is the probability that they will have a son with attached earlobes?

  2. In snapdragons, red flower color (R) is incompletely dominant over white flower color (R1); the heterozygotes produce pink flowers. A red snapdragon is crossed with a white snapdragon, and the F1 are intercrossed to produce the F2.

    1. Give the genotypes and phenotypes of the F1 and the F2, along with the expected proportions.
    2. If the F1 are backcrossed to the white parent, what will the genotypes and phenotypes of the offspring be?
    3. If the F1 are backcrossed to the red parent, what will the genotypes and phenotypes of the offspring be?

  3. In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two traits assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2.

    1. What will be the phenotypic ratios in the F2?
    2. If an F1 plant is backcrossed with the bitter, yellow spotted parent, what phenotypes and proportions are expected in the offspring?
    3. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

  4. In sailfin mollies (fish), gold color is due to an allele (g) that is recessive to the allele for normal color (G). A gold fish is crossed with a normal fish. Among the offspring, 88 are normal and 82 are gold.

    1. What are the most likely genotypes of the parents in this cross?
    2. Assess the plausibility of your hypothesis by performing a Chi-square test.

Problem Set 3

Homework Problems III

  1. In silkmoths (Bombyx mori) red eyes (re and white-banded wing (wb) are encoded by mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are:

    PhenotypeNumber of Progeny
    wild-type eyes, wild-type wings418
    red eyes, wild-type wings  19
    wild-type eyes, white-banded wings  16
    red eyes, white-banded wings426

    1. What phenotype proportions would be expected if the genes for red eyes and white-banded wings were located on different chromosomes?

    2. What is the genetic distance between the genes for red eyes and white-banded wings?

  2. In tomatoes, tall (D) is dominant over dwarf (d) and smooth fruit (P) is dominant over pubescent fruit (p), which is covered with fine hairs. A farmer has two tall and smooth plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent plant and obtains the following progeny:



    		Progeny of


    Genotype
    		Plant A
    		Plant B
    Dd Pp
    		122
    		    2
    Dd pp
    		    6
    		  82
    dd Pp
    		    4
    		  82
    dd pp
    		124
    		    4

    1. What are the genotypes of plant A and plant B?

    2. Are the loci that determine height of the plant and pubescence linked? If so, what is the map distance between them?

  3. A cross between individuals with genotypes a+a b+b X aa bb produces the following progeny:

    GenotypeNumber of Progeny
    a+a b+b83
    a+a bb21
    aa b+b19
    aa bb77

    1. What is the evidence that indicates that the a and b loci are linked?

    2. What is the map distance between a and b?

Problem Set 4

Homework Problem Set 4

  1. A microbiologist discovers a new type II restriction endonuclease. When DNA is digested by this enzyme, fragments that average 1,048,500 bp in length are produced. What is the most likely number of base pairs in the recognition sequence of this enzyme?

  2. Two restriction endonucleases (EcoRI and BamHI) are used to cut a piece of linear DNA, singly and in pairwise combination into the following fragments (sizes are in kb). Determine the correct order of restriction sites, and draw the map, with the intervals between sites labeled.

    EcoRI - 2, 4, 16
    BamHI - 10, 12
    EcoRI and BamHI - 2, 4, 6, 10

  3. Two restriction endonucleases (HaeIII and PstI) are used to cut a piece of circular DNA, singly and in pairwise combination into the following fragments (sizes are in kb). Determine the correct order of restriction sites, and draw the map, with the intervals between sites labeled.

    HaeIII - 17
    PstI - 8, 9
    HaeIII and PstI - 2, 7, 8

Problem Set 5

Homework Problem Set 5

  1. Will restriction sites for an enzyme that has 4 bp in its restriction site be closer together, farther apart, or similarly spaced, on average, compared with those of an enzyme that has 6 bp in its recognition site? Explain your answer.

  2. Three restriction endonucleases (EcoRI, BamHI, HinfI) are used to cut a piece of linear DNA, singly or in pairwise combination into the following fragments. Determine the correct order of restriction sites, and draw the map, with the intervals between sites labelled.

    EcoRI - 17, 15, 7

    BamHI - 26, 13

    HinfI - 34, 5

    EcoRI and BamHI - 17, 9, 7, 6

    EcoRI and HinfI - 15, 12, 7, 5

    BamHI and HinfI - 5, 13, 21

  3. Three restriction endonucleases (HaeIII, PstI, SauIII) are used to cut a piece of circular DNA, singly or in pairwise combination into the following fragments. Determine the correct order of restriction sites, and draw the map, with the intervals between sites labelled.

    HaeIII - 5.5, 4.5

    PstI - 9, 1

    SauIII - 5.4, 4.6

    HaeIII and PstI - 4.5, 2.5, 2, 1

    HaeIII and SauIII - 4, 3.1, 1.5, 1.4

    PstI and SauIII - 4.6, 3.9, 1, 0.5

Chi Square Test

Chi-squared Test

Testing the Goodness of Fit of Genetic Ratios

If one has a perfectly balanced coin, one can calculate (by definition using probability theory) that the probability of the coin falling heads up is 0.50 and the probability of the coin falling with tails up is also 0.50. Therefore, if one flips the coin 100 times the expected frequency is 50 heads and 50 tails. However, it is also likely that in a sample of 100 coin flips one will observe 45 heads and 55 tails, or 52 heads and 48 tails, or 41 heads and 59 tails, or other combinations differing from 50 heads and 50 tails. Once in a while the observation will be exactly 50 heads and 50 tails, but most of the time you will observe these deviations called sampling error. These are called sampling error because the 0.50 heads and 0.50 tails will only be obtained from an infinite number of flips (an impossibility of course) and any lesser number is only a sample of the infinite number. Because samples are smaller than infinity, they usually deviate from the exact expectations derived from calculations.

It is of great importance that you understand that there is an exact probability which is known by statisticians that chance alone will cause sampling errors of any particular magnitude in a sample of a certain size. As the size of the sample increases, the probability of getting a sampling error of a certain percentage decreases. To illustrate, let us suppose that one were to flip a perfectly balanced coin 200 times. It is known ahead of time that there is a certain probability of obtaining 100 heads and 100 tails. It is also known that there is an exact probability, one which is considerably smaller, that you would get a sampling error of 10 (10%) from a result of 100 heads and 100 tails, i.e., 110 heads and 90 tails or 90 heads and 110 tails. Finally, if one flips the coin 2000 times, there is a very much smaller probability of getting a sampling error as large as 10% from the expected result of 1000 heads and 1000 tails.

The problem that one faces when one obtains a ratio of 60 black and 40 white offspring from a mating which one calculated should yield 50% black and 50% white can be stated very simply. It is the problem of deciding between two alternative hypotheses: (1) The genetic mechanism is one which really produces a 1:1 ratio and the deviation is due to chance (sampling error) or (2) The genetic mechanism is not one which really produces a 1:1 ratio (not necessarily a 6:4 ratio -- perhaps a 3:1 ratio). In order that your decision is not a mere guess, you should apply statistical techniques which will test goodness of fit of your observed results to your expected ratio (in this case 1:1). This test is the Chi-square test (formula presented below). You will arrive at a value for probability (p). You must know (a) what this probability refers to and (b) how to apply it to the making of your decision.

The probability which you calculate is exactly this: The probable frequency with which a deviation as great as, or greater than, the observed one will appear as a result of chance alone if another similar trial or another similar sample is taken. For example, if the value of the probability which you get from your calculation is 0.50 then there is a 50% probability that another similar trial will produce results which, because of chance alone, deviates as much or more than yours from the 1:1 ratio being tested. Another way of saying it is that one out of every two trials would be expected to show as much deviation because of sampling errors caused by chance alone. If this were true, obviously your deviation from a 1:1 ratio is not significant. It does not in any way support the hypothesis that the genetic mechanism is one which produces a 1:1 ratio; but, on the other hand, if there is a certain explanation which predicts that the mating should produce a 1:1 ratio, the deviation from 1:1 ratio places no suspicion upon that explanation because the deviation can be readily attributed to sampling errors.

However, if the value of the probability which you calculate is 0.05, then only 5% of trials like yours will deviate from the expected ratio as much as yours and because of chance alone or sampling error. You either actually have gotten a deviation that is expected to occur by chance alone only 1 out of 20 times or you have some genetic mechanism which is producing something other than a 1:1 ratio. If you reject the 1:1 ratio hypothesis at the 5% level, then you risk making an error of ejecting the true 1:1 producing genetic mechanism 5% of the time or 1 out of every 20 trials or samples. Rather arbitrarily, statisticians accept this level of potential error and call this the first level of significance. Probabilities of the value of 5% or lower are significant; that is, you better not ignore those 20 to 1 odds against your deviation being due to chance alone and you should refuse to accept your observations as an indication of a 1:1 ratio. If no acceptable alternative hypothesis gives a ratio with a high goodness of fit, then you should conduct more trials to collect additional observations to see if they show a better fit to the 1:1 ratio. (After all, maybe you hit the 1 out of 20 times that a 1:1 ratio is accompanied by that much deviation because of sampling error alone.) A value for probability of 0.01 means that only 1 out of 100 times another trial would show the amount of deviation you obtained because of chance alone. This is the second level of significance. Your observations should certainly not be considered a 1:1 ratio. Here again, more observations could be made or perhaps another acceptable hypothesis will be found to have a high goodness of fit.

Note that this statistical test does not prove nor disprove anything. It simply determines the probabilities that deviations from the expected are or are not due to chance. This helps you decide that the explanation upon which the expectations were based is probably correct or not correct. Also note that you do not generally calculate the probability (p), instead you calculate a Chi square value and compare that value to a table of values (presented below) assigned to different p values. Computer programs will often calculate a specific p value associated with each chi square value, however.

Degrees of Freedom means the number values you need to know in order to know all of the values. For example, let's assume there are 100 total individuals in a population and you want to know how many of those individuals have the dominant phenotype and how many have the recessive phenotype. If you count 75 dominant individuals, then you know that there are 25 recessive individuals. Therefore, there is one degree of freedom. If a population has three phenotypes, then you need to know how many individuals there are in total plus how many of phenotypes 1 and 2 to determine the number of phenotype 3. In general then, the degrees of freedom is equal to N-1, where N is equal to the number of classes (e.g., phenotypes or genotypes).

Chi-squared Formula




Table of X2 Values

Degrees of Freedom Probability, p
  0.99 0.95 0.05 0.01 0.001
1 0.000 0.004 3.84 6.64 10.83
2 0.020 0.103 5.99 9.21 13.82
3 0.115 0.352 7.82 11.35 16.27
4 0.297 0.711 9.49 13.28 18.47
5 0.554 1.145 11.07 15.09 20.52
6 0.872 1.635 12.59 16.81 22.46
7 1.239 2.167 14.07 18.48 24.32
8 1.646 2.733 15.51 20.09 26.13
9 2.088 3.325 16.92 21.67 27.88
10 2.558 3.940 18.31 23.21 29.59
11 3.05 4.58 19.68 24.73 31.26
12 3.57 5.23 21.03 26.22 32.91
13 4.11 5.89 22.36 27.69 34.53
14 4.66 6.57 23.69 29.14 36.12
15 5.23 7.26 25.00 30.58 37.70
16 5.81 7.96 26.30 32.00 39.25
17 6.41 8.67 27.59 33.41 40.79
18 7.02 9.39 28.87 34.81 42.31
19 7.63 10.12 30.14 36.19 43.82
20 8.26 10.85 31.41 37.57 45.32
21 8.90 11.59 32.67 38.93 46.80
22 9.54 12.34 33.92 40.29 48.27
23 10.20 13.09 35.17 41.64 49.73
24 10.86 13.85 36.42 42.98 51.18
25 11.52 14.61 37.65 44.31 52.62
26 12.20 15.38 38.89 45.64 54.05
27 12.88 16.15 40.11 46.96 55.48
28 13.57 16.93 41.34 48.28 56.89
29 14.26 17.71 42.56 49.59 58.30
30 14.95 18.49 43.77 50.89 59.70
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