χ in the AP Biology Curriculum - Mrs. Monarch's...

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χ2 in the AP Biology Curriculum

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Statistics, Analysis, and AP Biology

χ2 Comparing Means

Type of Data Discreet Data or Counting

Data

Continous Data or Measuring Data

Calculated on the AP Exam YES NO

Graphing NO YES

Other Information

Needs Understanding of Degrees of Freedom and

Null Hypothesis

Needs Understanding of Standard Deviation and

Standard Error of the Mean and Normal

Distribution

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AP Biology Curricula that Incorporates χ2

•  Genetic Problems •  Hardy-Weinberg Equilibrium Problems •  Behavior Lab •  Mitosis Lab

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Statistics, Analysis, and AP Biology

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Student’s Pitfalls in Using χ2

•  Students lack an understanding of the “null hypothesis”.

•  Students have difficulties in determining what is expected.

•  Students have difficulties in working with fractions of different denominators.

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The Null Hypothesis

The null hypothesis refers to a general statement or default position that there is no relationship between two measured phenomena, OR a type of hypothesis used in statistics that proposes that no statistical significance exists in a set of given observations. The null hypothesis attempts to show that no variation exists between variables. If the null hypothesis is rejected or the data does not support it, then one can conclude there is statistical significance between variables.

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Activities for Null Hypothesis

•  Statistics and the Null Hypothesis

Materials Needed o Bag of Skittles o Bag of M&M’s o Calculators o Small Cups

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Easy χ2 Problem

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.

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Easy χ2 Problem

C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically different from your prediction?

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Easy χ2 Problem

C. If the offspring consisted of eight yellow peas and twelve green peas, would this be statistically different from your prediction?

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Easy χ2 Problem

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?

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2003 AP Biology Question 1

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. The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring.

Wild Type

Male Wild Type

Female White-eyed

Male White-eyed

Female Brown-eyed

Female

F1 Generation 0 45 55 0 1

Wild Type

Male Wild Type Female

White-eyed Male

White-eyed Female

Brown-eyed Female

F2 Generation 23 31 22 24 0

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(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.

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Question Using Different Denominators

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.

Purple, Smooth

Purple Wrinkled

Yellow Smooth

Yellow Wrinkled

F2 Generation 375 133 118 47

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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.

Purple, Smooth

Purple Wrinkled

Yellow Smooth

Yellow Wrinkled

F2 Generation 375 133 118 47

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Purple, Smooth

Purple Wrinkled

Yellow Smooth

Yellow Wrinkled

F2 Generation 375 133 118 47

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Behavior Problem Using χ2 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)

Side with Potato

Side with Cotton

0 25 25 10 21 29 20 17 33 30 14 36

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Behavior Problem Using χ2 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. Time

(minutes) Side with

Potato Side with Cotton

0 25 25 10 21 29 20 17 33 30 14 36

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Mitosis Using χ2 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

Root tip treated with caffeine

Phase Count Phase Count

Interphase 212 Interphase 370

Mitosis 72 Mitosis 51

Total Cell Count 284 Total Cell

Count 421

% Cells in Mitosis % Cell in

Mitosis

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Mitosis Using χ2

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.

Root tip treated with caffeine

# cells in mitosis

# of cells in interphase

Observed

Expected Expected Difference

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Hardy-Weinberg Equilibrium Problem Using χ2

In a field of flowers, there were three phenotypic colors for a particular species, and they were 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: a. Determine the allelic frequency for the P and P’ alleles.

Purple Flowers

PP

Violet Flowers

PP'

White Flowers

P'P'

Number 243 440 36

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Number Counts # P Alleles # P' Alleles

Purple Flowers PP 243 X

Violet Flowers PP' 440

White Flowers P'P" 36 X

Grand Total Alleles

Total

% Total

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a. Determine if this field was in Hardy-Weinberg equilibrium using the chi-square test.

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