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Unformatted text preview: Lecture 17
060311
Chapter 3 Basic Principle of Heredity
nd law: Independent Assortment
1.
2. Branch diagram to determine the phenotype
and genotype from the cross
3. Goodnessoffit chisquare test Mendels DihybridCross
Experiment
Cross involving two pairs of
contrasting traits F1:RrYy ConclusionfromDihybridCrossExperiment
Independent assortment
The inheritance of one trait does not influence the
inheritance of another trait in a dihybrid cross
Monohybrid cross: The F1 generation  dominant
phenotype: Round or Yellow.
Dihybrid cross: The F1  dominant phenotype for each
trait: Round and Yellow.
Two independent monohybrid crosses
During gamete formation, segregation of alleles at
independent or unlinked loci is independent of the
segregation of each other.
 The independent allele segregation and subsequent
random union of gametes at fertilization determine the
phenotypes observed. BranchDiagramfor
DihybridCross
ThePunnettSquare
canbeused,buta
branchdiagram
(Forked Line)is
ofteneasier
Dihybridcanbe
understoodas
twoindependent
monohybrid
crosses
Multiply
probabilitiesfor
eachcombination BranchDiagram/ForkedLine You will get the same results if you first list color and then shape TrihybridCross
Three pairs of contrasting traits:
A/a, B/b, C/c
Cross between AABBCC x aabbcc
Gametes:
ABC x abc
F1 hybrid:
AaBbCc
F1 Gametes (8 types): 2n
ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
Punnett square results in 64 boxes = cumbersome
Use the Branch Diagram/ForkedLine Method Branch Diagram for Trihybrid F i g : 3.10, K & C Practice:DihybridTestcross
Inthetestcross,
oneparentishomozygousrecessiveatbothloci
rryy(wrinkled,green).
Theotherparenthasdominantphenotyperoundandyellow.The
genotypecanbehomozygousdominantgeneonbothlocioronly
oneloci,orheterozygousonbothloci.
R_Y_=RRYY,RRYy,RrYY,RrYy(RoundandYellow)
ry
Practice:RrYyxrryy
Rr x rr =
Rr
rr Yy x yy=
Yy
yy
Yy
yy 1/4 RrYy 1/4 Rryy 1/4 rrYy
rryy 1/4 Practice:DihybridTestcross ObservedandPredictedRatioofProgeny
Genetic ratios are prediction based on Mendilian principles.
RrYy x rryy = ¼ RrYy: ¼rrYy: ¼ Rryy:1/4rryy
Predicted outcome 1 (round yellow):1(wrinkled
yellow):1(round green):1(wrinkle green).
The actual phenotypes we observed when we look at the
progenies from the cross may deviate from the expectation
 RrYy x rryy = 1000 progeny
 Expecting 250 round yellow, 250 wrinkle yellow, 250
round green, and 250 wrinkle green
 In fact, we may see 243: 262: 254: 239
The deviation between the expectation and actual
observation is due to chance
Example: toss coin 1000 times
486 heads and 514 tails are within reason.
Deviation is attributed to chance. EvaluatingFitnessofGeneticData
If other than chance causes the deviation, how can we tell?
Observed 25 round and 15 winkled seeds, would the
ratio still be 1:1?
How to evaluate the deviation between observation and
expectation is due to chance or not?
Maybe the inheritance of this character (seed shape) is
more complicated than was assumed
ChiSquare (X2) a statistic test to evaluate the fitness of
observed results to theoretical expectations TheGoodness
of
fitChi
SquareTest
Chisquare test evaluates the probability that the deviation between
the observed and the expected values is due to chance
Establish null hypothesis (Ho): yes due to chance
Ho: No real difference exists between the observed values and
expected values.
Calculate the ChiSquare value, which is measuring the deviation
between the observed and expected value
Used to find the probability that the deviation is due to chance
The lower the ChiSquare value is, the higher the possibility
that the deviation is due to chance, the Ho is truer
The higher the ChiSquare value is, the higher the likelihood
that the other factor other than chance alone causes the
deviation. Ho is false. We need to reject the null hypothesis: Ho CalculateChi
Square(X2)Value
ChiSquare (X2)
X2 ( O b s e r v e d E x p e c t e d )2
E xpected with degrees of freedom (df) = n1, where
n = the number of different phenotypic classes.
3:1 ratio, n = 2, so df = 2 1 = 1
9:3:3:1 ratio, n = 4, df = 3
Use the chisquare value and df to find the probability of
fitness (pvalue) using a chisquare table.
Use the 5% level, pvalue = 0.05, as a threshold for
determining whether to reject (p < 0.05) or accept (p
0.05) the null hypothesis.
chisquare table Accept Ho Reject Ho Chi
SquareAnalysis
Example
MonohybridCross,Purple&WhiteFlowers 3:1 E xpected
R a t io O b se r v e d
(o) E xpected
(e) ⇠ 105 ⇠ (150) = 112.5 1 /4 45 (150) = 37.5 (o
(o e) (o e )2 e )2
e 105112.5 = 7.5 (7.5)2 = 56.25 56.25/112.5 = 0.5 4537.5 = 7.5 (7.5)2 = 56.25 56.25/37.5 = 1.5 X 2 = 2.0 150 .1 < P < .5
2 (observed  expected) 2
expected n=2,df=1 Conclude:p>0.05,FailtoReject,theobservedDatafitstheexpected3:1ratio. Steps to perform a Chisquare test
1. Find following experimental data
1. Total number of offspring analyzed
2. Number of different classes of offspring
3. Number of offspring observed in each class 2. Calculate how many offspring would be expected in each
class if Ho is correct.
1. Multiply the expected ratio (portion) by the total number of offspring
analyzed in the experiment. 3. Calculate the Chisquare value.
4. Compute the degrees of freedom
5. Use the chisquare value and df to find the probability of
fitness (pvalue) using a chisquare table.
1. Use the 5% level, p = 0.05, as a threshold for determining whether to
reject (p < 0.05) or fail to reject (p 0.05) the null hypothesis. Chi
SquareAnalysis
Example
n=4,df=3
DihybridCross 9:3:3:1
E xpected
R a t io O b se r v e d
(o) E xpected
(e) 9/16 587 567 + 20 400 0.71 3 / 16 197 189 +8 64 0.34 3 / 16 168 189 21 441 2.33 1 / 16 56 63 7 49 0.78 1008 (o
(o e) (o e )2 e )2
e X 2 = 4.16
.1< P <.5 Conclude:p>0.05,FailtoReject,theobservedDatafitstheexpected9:3:3:1ratio. Chi
SquareAnalysis
Practice
Dihybrid test cross E x pected
R a t io O b se r v e d
(o) 1/2 1 262 1/2 258 1/2 E xpected
(e) df=? 243 1/2 1:1:1:1 (o
(o e) (o e )2 e )2
e 237 1000 X2 =
<P< ...
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 Summer '08
 GALLIO

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