1. No category

Chi-squared Problems
1. Homozygous green peas were crossed with homozygous yellow peas. All the offspring were
phenotypically yellow. A test cross was performed with one of the offspring from the F1.
A. Using Y as the symbol for pea color, what are the genotypes of the parents used in this cross?
B. If the F2 generation consisted of 20 offspring, predict how many will be green and how many
will be yellow.
C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically
different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the
results of the cross. Show all your work. What is the best explanation for the difference between
what was expected and what was observed?
D. If there were 200 offspring and they consisted of 80 yellow peas and 120 green peas, would this
be statistically different from your prediction? Use a Chi-squared test on the F2 generation data
to analyze the results of the cross. Show all your work.
2. In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant
allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female
white-eyed fruit fly produced the following offspring.
F1
Generation
Wild Type
Male
Wild Type
Female
Whiteeyed Male
White-eyed
Female
Brown-eyed
Female
0
45
55
0
1
The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the
following offspring.
F2
Generation
Wild Type
Male
Wild Type
Female
Whiteeyed Male
White-eyed
Female
Brown-eyed
Female
23
31
22
24
0
A. Determine the genotypes of the original parents (P generation) and explain your reasoning. You
may use punnett squares to enhance your description, but the results from the punnett squares
must be discussed in your answer.
B. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental
genotypes. Show all your work and explain the importance of your final answer.
C. The brown-eyed female in the F1 generation resulted from a mutational change. Explain what a
mutation is, and discuss two types of mutations that might have produced the brown-eyed female
in the F1 generation.
3. In certain strains of corn, the phenotype for color can be yellow or purple. Also the texture of the
kernels can be “plump” (smooth) or the kernels can be wrinkled. The cross between a homozygous
purple and wrinkled strain of corn was crossed with a homozygous yellow and smooth strain of corn. The
offspring from this cross produced were all smooth and purple. The F1 offspring were crossed with one
another and the phenotypes of the kernels from several ears of corn were counted and recorded in the data
table below for the F2 generation.
F2
Generation
Purple,
Smooth
Purple
Wrinkled
Yellow
Smooth
Yellow
Wrinkled
375
133
118
47
A. What is the null hypothesis for this investigation?
B. Use a Chi-squared test on the F2 generation data to analyze the results of the cross to determine if
the results supports that these genes are unlinked. Show all your work and explain the
importance of your final answer.
4. In an investigation of pill bug behavior, a covered choice chamber
is used to test whether the distribution of pill bugs is affected by the
presence of a sliced potato. A slice of potato is placed at one end of
the chamber and dry cotton is placed at the other end. To test the pill
bugs preference for the potato 25 pill bugs are placed on each side of
the choice chamber.
The positions of pill bugs are observed and recorded every 10 minutes
for 30 minutes.
Time
(minutes)
0
10
20
30
Side with
Potato
25
21
17
14
Side with
Cotton
25
29
33
36
A. What is the null hypothesis for this investigation?
B. Perform a chi-square test on the data for the last data collected at the 30-minute time point in the
investigation. Explain whether your hypothesis is supported by the chi-square test and justify
your explanation.
5. Onion roots were treated with 1 M solution of caffeine, and other onion roots were just submerged in
water with no treatment. After three days, four samples of the roots were removed and a root tip squash
was done. The tips were observed under the microscope. The total number cells observed were counted
and the number of cells undergoing of mitosis were also counted.
Root tip with no
treatment
Phase
Count
Root tip treated
with caffeine
Phase
Count
Interphase
212
Interphase
370
Mitosis
72
Mitosis
51
Total Cell
Count
284
Total Cell
Count
421
% Cell in
Mitosis
% Cells in
Mitosis
A. What is the null hypothesis for this investigation?
B. Perform a chi-square test on the data for the data collected caffeine treatment t in the
investigation. Explain whether your hypothesis is supported by the chi-square test and justify
your explanation.
6. In a field of flowers, there were three phenotypic colors for a particular species purple, violet and
white. It was determined that the color was controlled by a single autosomal gene that exhibited partial
dominance. The following counts were made:
Number
Purple
Flowers
PP
243
Violet
Flowers
PP'
440
A. Determine the allelic frequency for the P and P’ alleles.
White
Flowers
P'P'
36
Number
Purple
Flowers
PP
Violet
Flowers
PP'
White
Flowers
P'P"
Counts
#P
Alleles
X
243
440
36
# P'
Alleles
X
Grand
Total
Alleles
Total
Allelic
Frequency
B. Determine if this field was in Hardy-Weinberg equilibrium using the chi-square test.
Chi-squared Problems Key
1. Homozygous green peas were crossed with homozygous yellow peas. All the offspring were
phenotypically yellow. A test cross was performed with one of the offspring from the F1.
A. Using Y as the symbol for pea color, what are the genotypes of the parents used in this cross?
YY x yy
B. If the F2 generation consisted of 20 offspring, predict how many will be green and how many will
be yellow.
One half will be yellow and one half will green as the cross is between a green pea plant that is
heterozygous Yy and the other is homozygous for green yy. The results should be ten that are
green and ten that are yellow
y y
Y Yy Yy
y yy yy
C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically
different from your prediction? Use a Chi-squared test on the F2 generation data to analyze the
results of the cross. Show all your work. What is the best explanation for the difference between
what was expected and what was observed?
∑
(12-10)2 (8-10)2
+
=0.8 0.8 < 3.84 which means that the difference is not statistically
10
10
significant. The difference can be accounted for by the small sample size.
D. If there were 200 offspring and they consisted of 80 yellow peas and 120 green peas, would this
be statistically different from your prediction? Use a Chi-squared test on the F2 generation data to
analyze the results of the cross. Show all your work.
∑
(120-100)2 (80-100)2
+
=8 8 > 3.84 which means that the difference is statistically
100
100
significant. The results do not support that this cross is a simple Mendelian cross controlled by
one simple gene.
2. In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant
allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female
white-eyed fruit fly produced the following offspring.
F1
Generation
Wild Type
Male
Wild Type
Female
Whiteeyed Male
White-eyed
Female
Brown-eyed
Female
0
45
55
0
1
The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the
following offspring.
F2
Generation
Wild Type
Male
Wild Type
Female
Whiteeyed Male
White-eyed
Female
Brown-eyed
Female
23
31
22
24
0
A. Determine the genotypes of the original parents (P generation) and explain your reasoning. You
may use punnett squares to enhance your description, but the results from the punnett squares
must be discussed in your answer.
This is a sex linked cross with the white being recessive to wild type. The genotypes of the
parents involved in the P cross are XeXe x XEY. All the female offspring are phenotypically red
and all the male offspring are phenotypically white.
XE Y
Xe XEXe Xey
Xe XEXe Xey
B. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental
genotypes. Show all your work and explain the importance of your final answer.
The parental genotypes are XeXe x XEY and based on the punnett square, there should 25% of the
total for each phenotype.
Xe
Y
XE XEXe XEy
Xe XeXe Xey
Wild
Type
Male
χ2 =
Wild
White- White- BrownType
eyed
eyed
eyed
Female Male Female Female Total
F2
Generation
Observed
23
31
22
24
Expected
25
25
25
25
(23 − 25)2 (31 − 25)2 (22 − 25)2 (24 − 25)2
+
+
+
25
25
25
25
0
100
=2 2<7.82 the differences between what was expected and
what was observed is not statistically significant.
C. The brown-eyed female in the F1 generation resulted from a mutational change. Explain what a
mutation is, and discuss two types of mutations that might have produced the brown-eyed female
in the F1 generation.
This may be the result of a mutation in the production of the pigment such as a point mutation or it
may be the result of a mutation in the biochemical pathway responsible for the deposition of the
pigment.
3. In certain strains of corn, the phenotype for color can be yellow or purple. Also the texture of the
kernels can be “plump” (smooth) or the kernels can be wrinkled. The cross between a homozygous
purple and wrinkled strain of corn was crossed with a homozygous yellow and smooth strain of corn.
The offspring from this cross produced were all smooth and purple. The F1 offspring were crossed
with one another and the phenotypes of the kernels from several ears of corn were counted and
recorded in the data table below for the F2 generation.
F2
Generation
Purple,
Smooth
Purple
Wrinkled
Yellow
Smooth
Yellow
Wrinkled
375
133
118
47
A. What is the null hypothesis for this investigation?
There is no statistical difference between the results of this cross and the predicted 9:3:3:1 ratio
for a Mendelian dihybrid cross.
B. Use a Chi-squared test on the F2 generation data to analyze the results of the cross to determine if
the results supports that these genes are unlinked. Show all your work and explain the
importance of your final answer.
F2
Generation
Expected
Expected
Difference
χ2 =
Purple,
Smooth
Purple
Wrinkled
Yellow
Smooth
Yellow
Wrinkled
375
133
118
47
9/16X(673) 3/16X(673) 3/16X(673)
379
126
126
4
-7
8
(375 − 379)2 (133 − 126)2 (118 − 126)2 (47 − 42)2
+
+
+
379
126
126
42
Total
673
1/16
X(673)
42
-5
=1.5 1.5<7.82 the differences between what was
expected and what was observed is not statistically significant.
4. In an investigation of pill bug behavior, a covered choice chamber
is used to test whether the distribution of pill bugs is affected by the
presence of a sliced potato. A slice of potato is placed at one end of
the chamber and dry cotton is placed at the other end. To test the pill
bugs preference for the potato 25 pill bugs are placed on each side
of the choice chamber.
The positions of pill bugs are observed and recorded every 10 minutes for 30 minutes.
Time
(minutes)
0
10
20
30
Side with
Potato
25
21
17
14
Side with
Cotton
25
29
33
36
A. What is the null hypothesis for this investigation?
Providing pill bugs with the potato has no effect on the distribution of the pill bugs in the choice
chambers.
B. Perform a chi-square test on the data for the last data collected at the 30-minute time point in the
investigation. Explain whether your hypothesis is supported by the chi-square test and justify
your explanation.
χ
=
2
(14 − 25)2 (36 − 25)2 =9.68 9.68>3.84 The data supports that the difference between what
+
25
25
was observed and what was expected is statistically significant. The data supports that
providing pill bugs potato has an effect on their distribution.
5. Onion roots were treated with 1 M solution of caffeine, and other onion roots were just submerged in
water with no treatment. After three days, four samples of the roots were removed and a root tip squash
was done. The tips were observed under the microscope. The total number cells observed were counted
and the number of cells undergoing of mitosis were also counted.
Root tip with no
treatment
Phase
Count
Root tip treated
with caffeine
Phase
Count
Interphase
212
Interphase
370
Mitosis
72
Mitosis
51
Total Cell
Count
284
Total Cell
Count
421
% Cells in
Mitosis
% Cells in
Interphase
0.25
0.75
A. What is the null hypothesis for this investigation?
Caffeine has no effect on the mitotic division of cells in the root tip of an onion.
B. Perform a chi-square test on the data for the data collected caffeine treatment t in the
investigation. Explain whether your hypothesis is supported by the chi-square test and justify
your explanation.
χ
=
2
Root tip
treated with
caffeine
# cells in
mitosis
# of cells
in
interphase
Observed
51
370
Expected
Expected
Difference
421 x .25
105.25
-54.25
421 x .75
315.75
54.25
(51 − 105)2 (370 − 316)2 =37.28 37.28>3.84 which supports that the difference between what
+
105
316
was expected and what was observed is significant. This supports that a 1 M solution of caffeine has a
negative effect on the mitosis of onion root tips.
6. In a field of flowers, there were three phenotypic colors for a particular species purple, violet and
white. It was determined that the color was controlled by a single autosomal gene that exhibits partial
dominance. The following counts were made:
Number
Purple
Flowers
PP
243
Violet
Flowers
PP'
440
White
Flowers
P'P'
36
A. Determine the allelic frequency for the P and P’ alleles using the data provided in the table
shown.
Number
Purple
Flowers
PP
Violet
Flowers
PP'
White
Flowers
P'P"
Counts
#P
Alleles
# P'
Alleles
243
486
X
440
440
440
36
X
72
Grand
Total
Alleles
512
1438
Total
926
Allelic
Frequency
B. Determine if these flowers found in the field was in Hardy-Weinberg equilibrium by using the
chi-square test. If the flowers in the field is in Hardy-Weinber equ