Chi-Squared Genetics and Statistics A Tale of Two Hypotheses.

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Chi-Squared Genetics and Statistics A Tale of Two Hypotheses

Transcript of Chi-Squared Genetics and Statistics A Tale of Two Hypotheses.

Page 1: Chi-Squared Genetics and Statistics A Tale of Two Hypotheses.

Chi-SquaredGenetics and Statistics

A Tale of Two Hypotheses

Page 2: Chi-Squared Genetics and Statistics A Tale of Two Hypotheses.

Consider this story....

Two tigers at a zoo are bred together and they have four cubs. 

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Two of the four cubs are albino tigers.  Based on that, Kristin hypothesizes that both of the parents must be carrying a recessive gene for albinism.  The cross would look like:

 

A a  x  A a Who fell into the bleach?

At least they have a future in the circus.....

Don't hate me because I'm beautiful

If Kristin's hypothesis is accurate the punnett square would look like..

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Your friend, Emily is unconvinced.

If your hypothesis is correct, then only ONE of the four kittens should be an albino.

You are so dumb...you are really really dumb....

A a

A AA Aa

a Aa aa

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But isn't 1/4 pretty close to 2/4 ...maybe the difference is just due to chance....

Once I flipped a coin four times I got heads 3 times.  Sometimes it just happens that way.   Maybe you just got lucky and got an extra white kitten..

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The only way to solve this problem and the argument is to do a statistical analysis.

We call this type of analysis a CHI SQUARE The purpose is to determine whether the results are statistically significant. What are the odds that your tigers are Aa x Aa? Or could other factors be at work here?

I am so going to win this argument!

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Here's how to do a chi square.

Summed for all classes means that you  are looking at all the traits you observed - in this case, orange and white. 

Sum of all values for each “category”

Chi squared

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To apply the formula, plug in your "observed" and "expected" numbers....this will give you 

I do not like math!

*in this case, we have 2 categories, orange and white

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1.33?  Is that good or bad?  Who is right?  Who is wrong?  What time is it?

= 1.33

To determine if this number is good or not, you must look at a chi square chart. 

 "Degrees of freedom" is one less than the original number of classes you looked at, which was 2 (orange & white)

So we will look at the first row (DF = n-1 2-1=1)

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1.33 is between the 20% and 30% columns

Basically this means that the difference you observed between orange and white cubs can be expected to occur more than 20% of the time, just due to chance.

Scientists use 5% as the cut-off percent to reject a hypothesis.  Results are always better with a large sample size. 

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If you find that you have a "poor fit", that means that you probably need to reject the hypothesis.  In the tiger cub case, we did not have a poor fit. 

Well obviously, I was right.  You can run and tell that..

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Null Hypothesis

In statistics, the only way of supporting your hypothesis is to refute the null hypothesis, or opposite view. Rather than trying to prove your idea (the alternate hypothesis) right you must show that the null hypothesis is likely to be wrong – you have to ‘refute’ or ‘nullify’ the null hypothesis.

- In a sense, it is sometimes easier to prove the opposite wrong, though that only means your idea is more likely, not necessarily right.

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P value, or critical value Scientist generally accept p=.05 as the cutoff for accepting or rejecting a hypothesis. Basically, it shows the the likelihood of your expected or null hypothesis being by chance.

If X2 value is greater that critical p = statistically significant, meaning reject null hypothesis or reject expected, as it is unlikely to occur by chance given the parameters

If X2 value is less than critical p value: mean accept null hypothesis or expected/predicted outcome, as it can happen by chance more that 5 percent of the time.

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Poor fit.Emily thinks she gets it now.  So she looks at another case.  She breeds two black mice together and finds that over the course of 3 years, the parents produce 330 brown mice, and 810 black mice.   She hypothesizes that the parents are Bb (heterozygous).  How can she prove this with a chi square? 

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Online Chi Square Calculator at http://www.graphpad.com/quickcalcs/chisquared1.cfm

-- just plug in the observed and expected values

video review http://www.youtube.com/watch?v=WXPBoFDqNVkhttp://www.educreations.com/lesson/view/chi-square-in-genetics/3021987/?ref=link

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HW

Watch two video: (watch in order)

http://www.youtube.com/watch?v=WXPBoFDqNVk

http://www.educreations.com/lesson/view/chi-square-in-genetics/3021987/?ref=link

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Solving Chi Squared:

1. Underline key word or vocab words. 2. What type of problem is it?

- usually either genetics or other3. What is your expected?

- do you need to do a punnet square, what type?

- do you need a null hypothesis 4. Calculate percentages of total to get expected if needed5. Plug values into a chi squared equation

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An imaginary trait, bristles-with-split-ends, is hypothesized to be X-linked dominant. As before, the P females were homozygous wild type however this time the males had bristles-with-split-ends. The F1 84 males were all wild type while the 90 females all had split-ends. In addition, the data for the F2 generation revealed 26 wild type males, 35 wild type females, 29 split-end males and 40 split-end females. Does this data support or reject the hypothesis? Use chi square to prove your position.

QOD- how would you solve this problem?

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An imaginary trait, bristles-with-split-ends, is hypothesized to be X-linked dominant. As before, the P females were homozygous wild type however this time the males had bristles-with-split-ends. The F1 84 males were all wild type while the 90 females all had split-ends. In addition, the data for the F2 generation revealed 26 wild type males, 35 wild type females, 29 split-end males and 40 split-end females. Does this data support or reject the hypothesis? Use chi square to prove your position.

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The chi square for the bristles-with-spit-ends problem approximates 3.7 (depending on how your round your numbers) with three degrees of freedom. This gives a probability of approximately 30%, therefore, difference between what was expected and the data that had been collected was insignificant and the hypothesis is accepted.

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Percent

Value/ Total ex: 3/100= 3%

To find a percent of a total1. Turn the percent into a decimal2. Multiply by the total

Ex: what is 15 percent of 300? .15 * 300 = 45

Given a 1:2:1 ratio

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QOD: Explain what this image shows: what can you tell about the organism from the image? Include as much information as you can.

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Linkage Groups & Chromosome Maps

When Mendel crossed his F1 generation:  PpRr x PpRr, he got a 9:3:3:1 ratio.   He would have not seen this pattern if the alleles had been located on the same chromosome and inherited together.

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In eukaryotic species, each linearchromosome contains a long piece of DNA– A typical chromosome contains many hundredor even a few thousand different genes

The term linkage has two related meanings1. Two or more genes can be located on thesame chromosome2. Genes that are close together tend to betransmitted as a unit

Linkage and Crossing over

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A return to Drosophila....•Different alleles exist because any gene is subject to

mutation, or change, to a stable, heritable new form•Alleles can randomly mutate to become a different allele

depending on DNA sequence changes.•Wild type is a term used for the most common allele in the

population.•Other alleles, often called mutant alleles, may produce a

phenotype different from that of the wild-type allele.•An alternate form of designating alleles. Alleles that are wild

type are expressed with a +•Ex. Red eye color (w+) is dominant to white eye color (w).

The red eye is the wild type. Don't let this confuse you, its just a different way to express alleles.

Check out:  http://www.exploratorium.edu/exhibits/mutant_flies/mutant_flies.html

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Many genes are located on the same chromosome, and they do not assort independently, instead, they are inherited together Consider the following chromosome map of the fruit fly: All the alleles are located on chromosome 2 of the fruit fly, and are inherited together.

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Punnett Squares with Linkage Groups 

If a expected punnet square outcome is determined to be unlikely by a chi-square calculation, if may be due to linked genes

-as apposed to the predicted independent genes

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Chromosomes are called linkage groups– They contain a group off genes that are linked together

The number of linkage groups is the number of types of chromosomes of the species– For example, in humans 22 autosomal linkage groups An X chromosome linkage group

A Y chromosome linkage group

Genes that are far apart on the same chromosomecan independently assort from each other– This is due to crossing-over or recombination

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How Chromosome Maps Are DeterminedAlleles that are farther apart, like the gray body and the brown eye color allele are more likely to cross-over and exchange than ones closer together.

A B D

a b d

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Biologists use the percent of crossing over to determine the locus of alleles on a chromosome. 

The distance between alleles is measures in MAP UNITS, or MU. On the diagram the long wing allele is 13MU from the aristae allele. The image above is a linkage map because it shows the distance between the alleles

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