General Genetic
Bio 221 Lab 6
Mendel’s Second Law
Law of Independent Assortment (The "Second Law")
The Law of Independent Assortment, also known as "Inheritance
Law", states that alleles of different genes assort and distribute
independently of one another , during gamete formation.
Mendel’s Law animation:
http://www.sumanasinc.com/webcontent/animations/content/independentassortment.
html
http://www.springer.com/cda/content/document/cda_downloaddocu
ment/1002s.swf?SGWID=0-0-45-753210-0
Dihybrid-hybrid cross
1- Test cross (Ratio 1:1:1:1)
2-Cross (Ratio = 9:3:3:1)
Row
purple
sweet (RR YY)
Purple wrinkled
( RR yy)
Yellow sweet
(rrYY)
Yellow wrinkled
(rryy)
/ Divided
on small N.
/ Divided
on small N.
/ Divided
on small N.
/ Divided
on small N.
1
2
3
4
5
Total
Experimental
The “Second Law in Drosophila Fly
Meting between Parents ( 1Wild male ++ with 2female of vestigial wings (vg) and
black body color , then wait ) :
(F1 = Wild type ,
F2=9: wild type: 3 Vg type :3 e type: 1 vge )
++(wild)
vg
e
vge
CHI SQUARE TEST(X2)
1- The chi square test:
•
Chi-square (X2) is a statistical test commonly used to compare observed data with
data we would expect to obtain according to a specific hypothesis.
•
The chi-square test is always testing what scientists call the null hypothesis, which
states that there is no significant difference between the expected and observed
result.
or the difference between your observed result and expected is probably due to
random chance.
2- Determining the chi-square value
H0 :( Null hypothesis) There is no significant difference between observed and
expected
H1 :(Alternative hypothesis ) There is a significant difference .
3- Degrees of Freedom (df)
•
Is a measure of how many values can vary in a final statistical calculation.
• It calculates as= the number of categories – 1
• EX: there are 2 classes of offspring : Purple and white .Thus, degrees of freedom =
2-1 =1 .
4- Chi square formula:
•
Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square:
1. State the hypothesis being tested and the predicted results.
2- Calculate X2 using the formula. Complete all calculations to three significant digits.
3- Use the chi-square distribution table to determine significance of the value.
4- Determine degrees of freedom and locate the value in the appropriate column.
5- Locate the value closest to your calculated X2 on that degrees of freedom df row.
6- Move up the column to determine the p value.
7- State your conclusion in terms of your hypothesis.
•
•
If the p value for the calculated X2 s p > 0.05, accept your hypothesis. 'The deviation
is small enough that chance alone accounts for it. A p value of 0.6, for example,
means that there is a 60% probability that any deviation from expected is due to
chance only.
If the p value for the calculated X2 is p < 0.05, reject your hypothesis, and conclude
that some factor other than chance is operating for the deviation to be so great. For
example, a p value of 0.01 means that there is only a 1% chance that this deviation is
due to chance alone. Therefore, other factors must be involved.
Mono-hybrid cross in corn
1- Test cross (Ratio= 1:1)
2-Cross (Ratio =3:1 )
Row
Black
(Purple )
(RR)
Yellow (rr)
Obs.
1
2
3
4
5
Total
Note:
95/ 33=3
33/33=1
N.
1- RR
2- rr
/
Divided
by small
N.
/ Divided on
small N.
Chi-square calculation
Apply X2 on first law of Drosophila fly
•
++
vg
Breeding between Parent
( 1Wild male ++ with 2vestagial wings
female)
F1 = Wild type ,
F2= 3 wild type : 1 Vg type
Total
Chi-square calculation
Obs.
1- ++
2- vg
N.
degrees of freedom =2-1=1
•
•
•
•
•
Question
Suppose you counted the second generation of flowerd plant and you find
that 95 of the flowers are red and 33 of them are white .
1-95 flowers red
2- 33 flowers white
Test your observation by doing the Chi-squre test and decide your
conclusion .
Phenotype Observed
s
(o)
flowers
95
red
flowers
33
white
Total
128
Expected
(e)
3/4x 128 = 96
d=o-e
-1
1
0.01
1/4x 128 = 32
1
1
0.03
0
d2
d2/e
= 0.04
Degrees of freedom: 2 – 1= 1
X2 =
0.04
Conclusion: ( P < 0.05 ) so there is no significant differences between
observed and expected frequencies and we accept the null hypothesis
according to the ratio 3:1
•
How to calculate of expected in monohybrid ?
• 1- monohybrid test cross 1:1
Expected : ½ ½
• 2- Monohybrid cross 3:1
Expected : ¾
¼
Dihybrid-hybrid cross in corn
1- Test cross (Ratio 1:1:1:1)
2-Cross (Ratio = 9:3:3:1)
Row
purple
sweet
(RR YY)
Purple wrinkled Yellow sweet
( RR yy)
(rrYY)
Yellow wrinkled
(rryy)
1
2
3
4
5
Total
Obs.
1-
RRYY
2-
RRyy
3-
rrYY
4-
rryy
N.
Chi-square calculation
•
How to calculate of expected in Dihybrid ?
•
•
1- Dihybrid test cross 1:1 :1:1
Expected : ¼ ¼ ¼ ¼
•
•
2- Dihybrid cross 9:3:3:1
Expected : 9/16 3/16 3/16
1/16
Apply chi square test on the “Second
Law in Drosophila Fly
•
Breeding between Parent
( 1Wild male ++ with 2vestagial wings
female for )
(F1 = Wild type , F2= 9 ,3 ,3 , 1)
Obs.
1-
++
2-
e
3-
vg
4-
vg-e
N.
Chi-square calculation
degrees of freedom=4-1=3
Question
In pea plant pod shape can be swollen or pinched and the seed color can be either green or
yellow. A plant with swollen yellow is crossed with a pinched green one and after self crossing
we have the following data:
1-360 swollen yellow,
2-120 swollen green,
3-120 pinched yellow
4-40 pinched green.
Test the observation by doing the Chi-square test and decide your conclusion.
Phenotypes
d=o-e
d2
d2/e
Observed
Expected
(o)
(e)
swollen yellow
360
9/16x640= 360
360-360=0
0
0
swollen green
120
3/16x640=120
120-120=0
0
0
pinched yellow
120
3/16x640=120
120-120=0
0
0
pinched green
40
1/16x640=40
40-40=0
0
0
Total
640
=0
Degrees of freedom: 4 – 1= 3
Tabled = 7.815
Conclusion: X2= 0
so there is no significant differences between observed and
expected frequencies and we accept the null hypothesis
according to the ratio 9: 3: 3: 1.
•
•
•
•
•
•
•
•
•
•
Question :
In pea plant pod shape :
*Can be swollen or pinched
* And seed color can be either green or yellow .
A plant with swollen yellow is crossed with a pinched green one and after
self crossing we have the following data :
1- Swollen yellow =295
2-Swollen green = 108
3-Pinched yellow =101
4- Pinched green =32
Test the observation by doing the chi-square test and decide your
conclusion.