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MENDELIAN GENETICS CHI SQUARE ANALYSIS€¦ · square chart you need to determine the degrees of...
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MENDELIAN GENETICS CHI SQUARE ANALYSIS
AP BIOLOGY
What is Chi-Squared?l In genetics, you can predict genotypes
based on probability (expected results)l Chi-squared is a form of statistical analysis
used to compare the actual results (observed) with the expected results
l NOTE: C2 is the name of the whole variable – you will never take the square root of it or solve for C
CHI SQUARE FORMULA:
NULL HYPOTHESIS:l The hypothesis is termed the null
hypothesis which states l That there is NO substantial
statistical deviation (difference) between observed values and the expected values.
l In other words, the results or differences that do exist between observed and expected are totally random and occurred by chance alone.
l If the null hypothesis is supported by data• The assumption is that mating is random and
normal gene segregation and independent assortment occurred.
• Note: this is the assumption in all genetic crosses! This is normal meiosis occurring and we would expect random segregation and independent assortment.
l If the null hypothesis is not supported by data• The deviation (difference) between what was
observed and what the expected values were is very far apart…something non-random must be occurring….
CHI SQUARE VALUE:
l In order to determine the probability using a chi square chart you need to determine the degrees of freedom (DF)
l DEGREES OF FREEDOM: is the number of phenotypic possibilities in your cross minus one. l DF = # of groups (phenotype classes) – 1
l Using the DF value, determine the probability or distribution using the Chi Square table
l If the level of significance read from the table is greater than 0.05 or 5% then the null hypothesis is accepted and the results are due to chance alone and are unbiased.
DF VALUE:
Step 1: Calculating C2
l First, determine what your expected and observed values are.
l Observed (Actual) values: That should be something you get from data that you measure
l Expected values: based on probabilityl Suggestion: make a table with the expected
and actual values
Step 1: Examplel Observed (actual) values: Suppose you
have 90 tongue rollers and 10 nonrollers(recessive)
l Expected: Suppose the parent genotypes were both Rr à using a punnett square, you would expect 75% tongue rollers, 25% nonrollers
l This translates to 75 tongue rollers, 25 nonrollers (since the population you are dealing with is 100 individuals)
Step 1: Examplel Table should look like this:
Expected Observed (Actual)
Tongue rollers 75 90
Nonrollers 25 10
Step 2: Calculating C2
l Use the formula to calculated C2
l For each different category (genotype or phenotype calculate (observed – expected)2 / expected
l Add up all of these values to determine C2
Step 2: Examplel Using the data from before:l Tongue rollers
(90 – 75)2 / 75 = 3l Nonrollers
(10 – 25)2 / 25 = 9l C2 = 3 + 9 = 12
Step 3: Determining Degrees of Freedom
l Degrees of freedom = # of categories – 1l Ex. For the example problem, there were two
categories (tongue rollers and nonrollers) àdegrees of freedom = 2 – 1
l Degrees of freedom = 1
Step 4: Critical Value l Using the degrees of freedom, determine
the critical value using the provided tablel Df = 1 à Critical value = 3.84
If X2 is greater than the critical value then reject the null hypothesis.
Step 5: Conclusionl If C2 > critical value…reject null…
there is a statistically significant difference between the actual and expected values.
l If C2 < critical value…accept null…there is a NOT statistically significant difference between the actual and expected values.
Step 5: Examplel C2 = 12 > 3.84à There is a statistically significant difference
between the observed and expected population
Tiger Examplel Two tigers at the zoo have 4 cubs.l Two of the four cubs are albino tigers.l Albinism is recessive, what are the genotypes
of their orange parents?l What is the expected number of albino tigers?
Apply Chi-Squared
Accept or reject null?
Chi-squared and Hardy Weinberg
l Review: If the observed (actual) and expected genotype frequencies are the same then a population is in Hardy Weinberg equilibrium
l But how close is close enough?l Use Chi-squared to figure it out!l If there isn’t a statistically significant difference
between the expected and actual frequencies, then it is in equilibrium
Ferret Example
Ferrets Expected Observed (Actual)
BB 74 78
Bb 72 65
bb 18 21
Example
l C2 CalculationBB: (78 – 74)2 / 74 = 0.21Bb: (72 – 65)2 / 72 = 0.68bb: (18 – 21)2 / 18 = 0.5C2 = 0.21 + 0.68 + 0.5 = 1.39
l Degrees of Freedom = 3 – 1 = 2l Critical value = 5.99l C2 < 5.99 à there is not a statistically
significant difference between expected and actual values à population DOES SEEM TO BE in Hardy Weinberg Equilibrium
EXAMPLE:DIHYBRID FRUIT FLY CROSS
x
Black body -eyeless
Wild type
F1: all wild type
F1 CROSS PRODUCED THE FOLLOWING OFFSPRING
5610 1881
1896 622
Wild type (normal body
and eyes)
Normal body -eyeless
Black body -eyeless
Black body –w/ eyes
ANALYSIS OF THE RESULTS:
l Once the total number of offspring in each class is counted, you have to determine the expected value for this dihybrid cross.
l What are the expected outcomes of this typical dihybrid cross? (9:3:3:1)Ø 9/16 should be wild typeØ 3/16 should be normal body eyelessØ 3/16 should be black body wild eyesØ 1/16 should be black body eyeless.
NOW CONDUCT THE ANALYSIS:Phenotype Observed Hypothesis
Wild 5610
Eyeless 1881
Black body 1896
Eyeless, black body 622
Total 10009
To compute the expected value multiply the expected 9/16:3/16:3/16:1/16 ratios by 10,009
CALCULATING EXPECTED VALUES:
l To calculate the expected value:l Multiply the total number of offspring times the
expected fraction for each phenotype classl TOTAL = 10,009
l Wild-type expected value: 9/16 x 10,009 = 5634l Eyeless expected value: 3/16 x 10,009 = 1878l Black body expected value: 3/16 x 10,009 = 1878l Black body & Eyeless expected value: 1/16 x
10,009 = 626
NOW CONDUCT THE ANALYSIS:
Phenotype Observed Hypothesis
Wild 5610 5634
Eyeless 1881 1878
Black body 1896 1878
Eyeless, black body 622 626
Total 10009
l Using the chi square formula compute the chi square total for this cross:Ø (5610 - 5630)2 / 5630 = 0.07Ø (1881 - 1877)2 / 1877 = 0.01Ø (1896 - 1877)2 / 877 = 0.20Ø (622 - 626)2 / 626 = 0.02Ø x2 = 0.30
l How many degrees of freedom? Ø 4 phenotype classes – 1 = 3 Ø Accept or reject null?
CALCULATING X2: