1.1.flylab-fall 2010 copy - intro.bio.rpi.eduintro.bio.rpi.edu/lab/ppt_presentations/7.flylab...

Genetics Simulation

Fly

Lab

Pre-Lab Exercise:

Making Predictions And Analyzing

Data From Genetic Crosses

•! Analyze results of crosses in corn plants

•! Monohybrid

seed color (purple and yellow seeds)

•! Dihybrid

seed color and shape (round and wrinkled)

•! Statistical Analysis - Chi Square

Monohybrid Cross

Prediction: Purple seed allele is dominant.

Cross between true-breeding purple seed and yellow seed corn plants

Record results for F1 and F2 generations

P = allele for purple seed color

p = allele for yellow seed color

P: PP x pp

Monohybrid Cross: true-breeding purple & yellow seed-bearing corn plants

Expected results:

F1: all Pp (genotype)

all purple seeds (phenotype)

F2: 1 PP and 2 Pp - 3/4 purple seeds

1 pp - 1/4 yellow seeds

Purple Yellow

Number of seeds

85 35

Observed results for F2:

Monohybrid Cross

Chi Square (!2)Test

•! Analyzes results of genetics crosses

•! Compares the numbers observed with

those predicted (or expected)

•! Hypothesis:

There is no difference between the

observed numbers and the expected

ones.

•!Called the null hypothesis (Ho)

Chi Square (!2)Test

!2 = " ( observed – expected )2

expected

Calculation of Chi Square

Class 1 (purple) Class 2 (yellow)

Observed 85 35

Expected 90 30

Observed – Expected -5 +5

(Observed – Expected) 2 25 25

(Observed – Expected) 2

Expected

25/90 25/30

!2 =" (observed – expected) 2 = 0.28 + 0.83 = 1.11

expected

Class 1 (purple) Class 2 (yellow)

Observed 1213 387

Expected 1200 400

Observed – Expected 13 -13

(Observed – Expected)2 169 169

(Observed – Expected)2

Expected 169/1200 169/400

!2 = 0.1408 + 0.4225 = 0.5633

Table 3. Table of Chi Square Values

d.f. p=0.99 p=0.95 p=0.80 p=0.50 p=0.30 p=0.20 p=0.10 p=0.05 p=0.02 p=0.01

1 0.00016 0.0039 0.064 0.455 1.074 1.642 2.706 3.841 5.412 6.635

2 0.0201 0.103 0.446 1.386 2.408 3.219 4.605 5.991 7.824 9.210

3 0.115 0.352 1.005 2.366 3.665 4.642 6.251 7.815 9.837 11.341

4 0.297 0.711 1.649 3.357 4.878 5.989 7.779 9.488 11.668 13.277

5 0.554 1.245 2.343 4.351 6.064 7.289 9.236 11.007 13.388 15.086

Critical value of Chi Square corresponding

to a probability of 5% with 1 d.f.

For first example, !2 = 1.11

•!P-value of between 0.20 and 0.30

•!Do NOT reject the null hypothesis.

For second example, !2 = 0.5633

•!P-value of between 0.50 and 0.80

•!Do NOT reject the null hypothesis.

The difference between observed and expected values was close enough that it could have be different by chance alone.

Both !2 values

Dihybrid Cross

Cross true-breeding corn with purple smooth seeds

and yellow wrinkled seeds corn plants

Record results for F1 and F2 generations

R = allele for smooth seeds

r = allele for wrinkled seeds

P: PPRR x pprr

F1: all purple smooth seeds (PpRr)

Prediction: Smooth seed allele is dominant.

Dihybrid Cross:

True-breeding corn plants with purple smooth seeds mated with yellow wrinkled seeds

Dihybrid Cross

Cross F1 plants:

What is the expected ratio of phenotypes from

cross of PpRr x PpRr?

Purple, round: 9

Purple, wrinkled: 3 Yellow, round: 3

Yellow, wrinkled: 1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Phenotypic ratio for

dihybrid cross:

9:3:3:1

Dihybrid Cross Table 4. Observed phenotype results in offspring (F2)

of a cross between parents that are heterozygous for

seed color and texture:

PpRr x PpRr

purple-

round

purple-

wrinkled

yellow-

round

yellow-

wrinkled

896 317 331 107

If you are going to calculate Chi square, how do you

determine the numbers of expected seeds for each class?

Sum all of the individuals counted. Then, take 1/16th, 3/16th

and 9/16th of the sum to determine the numbers for each class.

Results of Dihybrid Cross:

Class 1 Class2 Class 3 Class 4

Observed 896 317 331 107

Expected 929 310 310 103

Observed – Expected -33 +7 +21 +4

(Observed – Expected)2 1089 49 441 16

(Observed – Expected)2

Expected 1089/929 49/310 441/310 0/103

!2 = 1.1722 + 0.1581 + 1.4226 + 0.1553 = 2.9082

Table 3. Table of Chi Square Values

d.f. p=0.99 p=0.95 p=0.80 p=0.50 p=0.30 p=0.20 p=0.10 p=0.05 p=0.02 p=0.01

1 0.00016 0.0039 0.064 0.455 1.074 1.642 2.706 3.841 5.412 6.635

2 0.0201 0.103 0.446 1.386 2.408 3.219 4.605 5.991 7.824 9.210

3 0.115 0.352 1.005 2.366 3.665 4.642 6.251 7.815 9.837 11.341

4 0.297 0.711 1.649 3.357 4.878 5.989 7.779 9.488 11.668 13.277

5 0.554 1.245 2.343 4.351 6.064 7.289 9.236 11.007 13.388 15.086

Critical value corresponding to a

probability of 5% with 3 d.f.

!2 Analysis of Dihybrid Cross

•! P-value of between 0.30 and 0.50

•! Do NOT reject the null hypothesis (Ho)

•! The difference between observed and

expected values was close enough that it could

have been due to chance.

Getting To Know FlyLab

Monohybrid cross:

wild type female and sepia eye male

•! F1 offspring; F2 offspring

•! Statistical analysis with Chi square

•!3:1 ratio

•!4:1 ratio

•!Opportunity to play the role of geneticist

•!Way to learn principles of genetic inheritance

I. Monohybrid cross:

P: se+se+ ! x se se "

F1

I. Monohybrid cross:

F1: se+se ! x se+se "

F2: se+ se

se+ se+ se+ se+ se

se se+ se se se

Inheritance

in Humans

Sex Chromosomes in Drosophila

First Exercise:

Learn About X-linked Inheritance Determine the mode of inheritance from patterns of

phenotypes in flies:

•! Mate a wild type female with a white eye male

•! Develop hypothesis about the mode of inheritance of the

white eye allele

•! Complete Punnett squares for parental and F1 generations based on your hypothesis; give expected proportions of

phenotypes generated

•! Use FlyLab for the same crosses and compare results

•! Do Chi square test.

•! Should you accept your hypothesis?

Using FlyLab To Learn About

Inheritance Patterns The mode of inheritance of the white eye allele can be one of

the following:

–! Autosomal recessive

–! Autosomal dominant

–! X-linked recessive

–! X-linked dominant

If you hypothesize autosomal inheritance, use:

–! w+ (wild allele)

–! w (white eye allele)

(FlyLab uses symbol w for phenotype of white eye fly)

If you hypothesize X-linked inheritance, use:

–! X w+ (wild allele on X chromosome)

–! Xw (white eye allele on X chromosome)

–! Y male chromosome – does not carry allele

Determining Mode of Inheritance

from Phenotype Patterns Examine cross of a female fly with wild-type eye color and a male fly with white eye color

•! If allele for white eye color is autosomal recessive, what phenotypes and ratios would you get for F1 and F2?

•! If the allele for white eye color is autosomal dominant, what

phenotypes and ratios would you get for F1 and F2?

•! If allele for white eye color is X-linked recessive, what phenotypes

and ratios would you get for F1 and F2?

•! If allele for white eye color is X-linked dominant, what phenotypes

and ratios would you get for F1 and F2?


from Phenotype Patterns Examine cross of a female fly with wild-type eye color and a male fly with white eye color

•! If allele for white eye color is autosomal recessive, what phenotypes and ratios would you get for F1 and F2?

F1: all wild type flies

F2: ! wild type flies and " white eye flies

•! If the allele for white eye color is autosomal dominant, what

phenotypes and ratios would you get for F1 and F2?

F1: all white eye flies

F2: ! white eye flies and " wild type flies


from Phenotype Patterns •! Examine cross of a female fly with wild-type eye color

and a male fly with white eye color

•! If allele for white eye color is X-linked recessive, what phenotypes and ratios would you get for F1 and F2?

F1: all wild type flies – # wild females and # wild males

F2: ! wild type flies and " white eye flies

However, # are wild type females, " are wild type males and " are white eye males

•!Would you get the same results for the F1 and F2 flies, if

you did the reciprocal cross?


from Phenotype Patterns •! Reciprocal cross (X-linked recessive): female fly with

white eye eye color and a male fly with wild-type color

•! If allele for white eye color is X-linked dominant, what phenotypes and ratios would you get for F1 and F2?

F1: # wild type females and 1/2 white eye males

F2: " white eye females and 1/4 wild type females

" white eye males and 1/4 wild type males

•!Would you get the same results for the F1 and F2flies, if



from Phenotype Patterns •! Examine cross of a female fly with wild-type eye color

and a male fly with white eye color

•! If allele for white eye color is X-linked dominant, what phenotypes and ratios would you get for F1 and F2?

F1: # white eye females and # wild type males

F2: " white eye females and 1/4 wild type females

" white eye males and 1/4 wild type males

•!Would you get the same results for the F1 and F2 flies, if


X-linked genes:

A unique pattern of inheritance

The white eye color mutant is X-linked recessive

X chromosome carries many genes unrelated to sex

White eye color is one of them and is X-linked in Drosophila

Mode of Inheritance for Whit

eyets: X-Linked Recessive

For X-linked inheritance

•!Use X and Y

•!Show allele as superscript

•!w+ for wild type allele

•!w for white allele

•!Female – wild type eyes

•!has two red-eyed alleles and red eyes (XW+XW+)

•!Male – white eyes

•!has only single allele for eye color (on X chromosome)

•!has a white-eyed allele (XWY)

Allele for white-eyes-on X chromosome only, not on Y

Guide to Determining Modes of Inheritance

•!FlyLab lets you see only the proportion of phenotypes; no

information on genotypes

•!Guide lets you compare proportions of phenotypes generated by FlyLab with the proportions you predicted

based on your hypothesis

•!Example: Autosomal Recessive

•! For F1 generation expect 1:1 ratio of wild type males to

wild type females

•! For F2 generation expect 3:3:1:1 ratio of 3 wild males to 3 wild females to 1 mutant male to 1 mutant female

•! Reciprocal cross gives same results

•! Use Fly Lab to test your predictions - Chi Square

Modes of Inheritance •!Determine of Modes of Inheritance for three mutant traits in Drosophila (assigned)

•!Each one of the mutants has one of four (4) possible modes of inheritance:

•! Autosomal recessive

•! Autosomal dominant

•! X-linked recessive

•! X-linked dominant

Use Guide to Determining Modes of Inheritance to help you determine the proportions of phenotypes you would expect for each mode of inheritance

Modes of Inheritance

•! Determine the mode of inheritance for one of

three mutants assigned to you.

•! Complete Analysis/Prediction sheet; do both

cross and reciprocal cross

•! Add supporting data - lab notes from FlyLab

•! Hand these in before you leave.

•! Complete two other mutant traits for post-lab

assignment. These will be collected next lab.

Prediction

Analysis

Sheets

Include Chi square analysis for F2 generations for both original cross and reciprocal cross with your FlyLab notes

Prediction

Analysis

Sheets

Genetic Stutter Increases Risk of ALS

•! Guilty gene: Ataxin-2, plays a role in another neurodegenerative

disease called spinocerebellar ataxia type 2 (SCA2)

•! In patients with SCA2, the Ataxin-2 gene stutters on a sequence of

three letters: CAG, which together tell the cell to make the amino acid

glutamine. In patients with the disease, this sequence can repeat more

than 34 times, misdirecting the cell to manufacture Ataxin-2 with too

much glutamine.

•! A scan of DNA from 980 people with neither ALS nor SCA revealed that

in most individuals, the Ataxin-2 gene has 22 or 23 CAG repeats

•! in almost five percent of samples from more than 900 patients with ALS, the gene carried between 27 and 33 CAG repeats. So the stutter in

•! Ataxin-2 associated with ALS patients is longer than typically found,

although shorter than what causes the human disease SCA2

•! .http://www.hhmi.org/news/bonini20100826.html

Before Leaving . . .

Hand in:

1.! Prediction /Analysis sheet (cross and reciprocal

cross) for one of your mutant traits

2. Include printouts of your lab notes for one mutant

trait (cross and reciprocal cross).

For Post-Lab:

•! Determine the modes of inheritance for the other

two mutants assigned you

For next week: Complete pre-lab

Due for lab:

•!Remaining two (2) mutants from FlyLab with Lab Notes

•!Quiz in lab on Genetics & Meiosis Demystified

Due Friday, Oct 8th:

•!Drosophila Genetics Lab Report

•!Lab Notebook

Turn in items in WALKER 6213 between

1PM-5PM

1.1.flylab-fall 2010 copy - intro.bio.rpi.eduintro.bio.rpi.edu/lab/ppt_presentations/7.flylab...

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