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Unformatted text preview: Biology 313
Lecture 5
Sept 1, 2010 Corn detasseling
• http://www.livinghistoryfarm.org/farmingint
he30s/crops_03.html Predicting the outcomes
of genetic crosses
Punnett square Another method for determining the
outcome of a genetic cross is to use
rules of probability
probability
Probability: likelihood of the occurrence
Probability:
of a certain event
of Rules of Probability
Rules
If
• Sum rule: If two events are mutually exclusive, the
probability that either of two events will occur is the
either
sum of their separate probabilities
p(A or B) = p(A) + p(B)
p(A
If
• Product rule: If two events are independent, the
probability that the two events will both occur is the
product of their separate probabilities
p(A and B) = p(A) * p(B)
p(A In a family of four children,
what is the probability that
all children are females?
The probability that a female child
is born is independent of whether
or not that child has any sisters,
therefore the product law applies. In a family of four children,
what is the probability that
all children are females?
The probability of each child being
female is ½. Therefore the
probability of four children being
all female is (1/2)4 or 1/16 The branching diagram
The
Consider the cross: Tt x Tt
Gametes from Parent 1 = T and t
Gametes
Gametes from Parent 2 = T and t
What is the expected ratio of progeny
genotypes? TT, Tt, tt
genotypes?
Can use the product rule Gametes
parent 1 Gametes
parent 2 Progeny
genotypes 1/2 T = 1/4 Tt
= 1/4 tT 1/2 t 1/2 T 1/2 t
1/2 T 1/2 = 1/4 TT = 1/4 t Conclusion: 1 TT : 2 Tt : 1 tt tt Rules of probability in Mendelian crosses:
Rules
•
• Given the cross: Rr X Rr
Given
A. What is the probability the first three offspring are all
A.
R phenotype?
B.
• B. What is the probability the first three offspring are all
rr genotype?
Answer:
Genotype
Phenotype
= 1/4 RR, 1/2 Rr, 1/4 rr
R = 1/2 + 1/4 = 3/4, r = 1/4
• Rr x Rr
• Each offspring is independent, so the product rule is used for
combinations.
• A. all R Phenotype: 3/4 x 3/4 x 3/4 = 27/64
• B. all rr Genotype: 1/4 x 1/4 x 1/4 = 1/64 • C. With five offspring, what is the probability that 3 have the R
phenotype and 2 have the r phenotype? The Binomial Expansion
• binomial = (p + q)n
binomial
•
•
•
• p = probability of R
q = probability of r
p + q = 1 I.e. they are mutually exclusive
I.e.
n = number of trials (e.g., number of offspring) Expand
• Expand the binomial:
•
• • (p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 Each term is the probability of one combination of five offspring.
The probability of 3R and 2r = 10 (3/4)3(1/4)2 = 0.264
0.264 • P = n! psqt
s!t!
P = probability, n = total #, s = # first class, t = # second class
probability,
P= 5! (3/4)3 (1/4)2 = 5x4x3x2x1 (3/4)3 (1/4)2 = 10 (3/4)3(1/4)2 = 0.264
5!
0.264
3!2!
3x2x1x2x1
3!2! Mendel’s Monastery
Brno, Czech Republic F2 P F1 P Selfed F2 Dihybrid Crosses Reveal the Principle of
Independent Assortment Discovery of the Principle of Independent
Assortment came from Dihybrid Crosses
(crosses that involve two characteristics) Mendel Dihybrid Cross
Assumption #1: Traits are on
different chromosomes Assumption #2:
There is
Independent
Assortment of alleles
at Anaphase I Assumption #2:
There is
Independent
Assortment of alleles
at Anaphase I Can diagram results
using Punnett
Square or via
Branching (later) Thought Question
Assume Mendel did NOT obtain a 9:3:3:1
ratio in the F2.
Would you conclude that there is no such
thing as the 2nd Law? An example of the 2nd
Law from Chapter 2
(Figure 217) : random
distribution of
chromosomes at
Anaphase I produces
genetic variation
m = maternal
p = paternal How predict progeny?
• Branching method: an easier, faster way
to predict progeny of crosses than
Punnett Square method, especially for
more than two traits • Independent assortment applies so can
apply the multiplication rule Branching method for Dihybrid Cross
Consider each gene
(chromosome)
separately Yy x Yy Can predict genotypes and
phenotypes
phenotypes • Given the cross:
Rr Yy Tt X rr Yy tt
rr
What
• What is the probability of an
offspring with an rr YY tt
genotype?
genotype? Given the cross: Rr Yy Tt X rr Yy tt
Given
rr
What is the probability of an offspring with an
rr YY tt genotype?
Answer:
Genotype •
•
• Rr x rr
wrinkled
Yy x Yy
Tt x tt = 1/2 Rr, 1/2 rr
= 1/4 YY, 1/2Yy, 1/4 yy
= 1/2 Tt and 1/2 tt Phenotype = 1/2 round 1/2
= 3/4 yellow, 1/4 green
= 1/2 tall, 1/2 short Each event is independent, so the product rule is
used for combinations.
Genotype probability is 1/2 rr x 1/4 YY x 1/2 tt =
1/16 How to work a genetics problem?
How
See p. 60 (Good tips) Goodnessoffit chisquare test
When two (or more!) genes segregate
independently, do we expect to see
exactly the ratios predicted by rules of
probability? NO !!!!!!!!
Why is there deviation between an
observed outcome and the expected
outcome? chisquare test
2 steps:
1. Determine χ2 value
χ2 = Σ (observedexpected)2/expected
2. Use the calculation of χ2 to determine a
probability (pvalue)
Degrees of freedom (df) = one less than number of
independent variables (for us, this is one less than the
number of progeny classes) Mendel’s testcross: Rr Yy x rr yy
Mendel’s
• Hypothesis: Mendel’s F1 progeny are expected to have a
Hypothesis:
1:1:1:1 ratio.
1:1:1:1
(null hypothesis = no difference between observed and expected results) • χ2 test:
genotype
Rr Yy
Rr yy
Rr
rr Yy
rr yy
Total observed
observed
55 expected
expected
51.75 (obsexp)2/exp
(obsexp)
0.204
0.204 51
49
52 51.75
51.75
51.75 0.011
0.146
0.001 207 207.00 0.362 = χ 2
0.362 Determine pvalue
Determine
Degrees of freedom (df) = 4 1 = 3
Degrees
df = one less than the number of progeny
classes
classes
Go to χ 2 distribution table
Go Accept null
hypothesis: chance
explains the deviation
of the observed from
the expected Reject null
hypothesis:
something else
explains the deviation
of the observed from
the expected pvalue > 0.1 (i.e., a deviation between observed and
pvalue
deviation
expected due to chance is likely, thus accept the null
likely thus
hypothesis
hypothesis 0.362 ...
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 Spring '11
 Rodermel
 Biology, Genetics

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