Chi-square Statistical Analysis You and a sibling flip a coin to see who has to take out the trash. Your sibling

grows skeptical of the legitimacy of your coin because “tails” always seems to win.

You decide to test it out by flipping the coin 20 times.

Poss. Outcomes

TailsHeads

Obs.

137

Exp.

1010

(o-e)2/e

.9

.91.8c2 value

Chi-square Statistical AnalysisDetermines the probability that observed data could result from expected conditions1. For each phenotype, calculate

(Observed – Expected)2 / (Expected)2. Add up your figures. This is the chi-square value

3. Degree of freedom (df) = (# of phenotypes – 1)4. Find the probability in the table.

c2 = (o-e)2 eS

Fruit Fly Eye ColorP: Red-eyed female x White-eyed male F1: 100 Red-eyed flies F2: 75 Red-eyed, 25 white-eyedWrite out Punnett squares for both crosses.

75 25100

w+ = redw = white

Fruit Fly Eye ColorP: Red-eyed female x White-eyed male F1: 100 Red-eyed flies F2: 75 Red-eyed, 25 white-eyed

75 25

25 males

50 females, 25 males

Chi-square Statistical Analysis Our F1 cross produces 100 offspring If we assume eye color and gender are unlinked,

w+ = red w+w ♀ x w+w ♂w = white Then we expect… But we got…

Expected Phenotype

37.5 Red ♀12.5 White ♀37.5 Red ♂12.5 White ♂

Observed

500

2525

Calc.

4.212.54.2

12.5

33 c2 value

Only a 0.000027% probability

Therefore, eye color and gender are linked

X-Linkage

The gene with the white-eyed mutation is on the X-chromosome

Homozygous Dominant

Xw+Xw+

Hemizygous Recessive

XwY

Practice Problems1. Colorblindness is due to a recessive x-linked

allele. What are the chances of a normal male and a carrier female having a colorblind son as their first child?

2. Why are males more likely than females to have recessive x-linked traits?

Practice ProblemsIn sesame plants, the one-pod condition, (A) is dominant to the 3-pod condition (a), and normal leaf (B) is dominant to wrinkled leaf (b). An AaBb plant is testcrossed to produce the following offspring:

11 one-pod, normal12 one-pod, wrinkled7 three-pod, normal10 three-pod, wrinkled

Is this data likely if this is a case of independent assortment?

What about…11012070100?

What about…182317?

A

B

a

b

A

B

a

b

A

B

a

bA

B

a

b

A

B

a

b

A

B

a

b

A

B

a

b

A a

Bb

a

B

A

b

A

B

a

b

A

B

a

b

A

B

a

b

A a

Bb

Gametes

ABab

Parental

ABab

Recomb.

AbaB

AaBb

AaBb

A a

Bb

ORAbaB

Earlobes & Toes (Independent Assortment)P: FFTT x fftt

all FfTt(free earloes and 2nd toe longer)

Testcross of F1 individualsFfTt x fftt:

¼ Free, 2nd toe¼ Free, great toe¼ attached, 2nd toe¼ attached, great toe

Earlobes & Toes (Linked) P: FFTT x

fftt F1: FfTt F1 Testcross

FfTt x fftt

F

T

f

t

f

t

f

t

F

T

f

t

f

t

f

t

½ Att,Grt Toe

½ Free,2nd Toe

IndependentAssortment

¼ Free, 2nd

¼ Free, Grt¼ Att, 2nd

¼ Att, Grt

Earlobes & Toes (Linked + Recombination)

F

T

f

t

f

t

f

t

F

T

f

t

f

t

f

t

T

fF

t

F

t T

ff

t

f

t

Parental OffspringFfTt (free, 2nd)

fftt (attached, great)

Recombinant OffspringFftt (free, great)ffTt (attached, 2nd)

Recombination frequency = (# of recomb. offspring) (total offspring)

18 17 2 3

Genetic Recombination

Genes farther apart are more likely to cross over

Greater distance greater RF Enables us to map genes on a chromosome

Genetic Recombination b / c = 9% (9 map units or

centimorgans) v / c = 9.5% (9.5 mu) b / v = 17% (17 mu) According to our map, b/v = 18.5 mu

(not 17) Double crossover leads to lower than

expected recombinant frequencies

b

c9mu

v

v

OR

9.5mu

9.5mu

18.5mu

ReviewConstruct a linkage map given the following

information:a-b 7.2 %b-c 3.5 %a-c 3.9 %b-d 9.6 % c-d 6.3 %How many recombinant organisms would you

expect to find out of 1000 offspring from a cross between AaDd and aadd organisms?