Download 1 Chapter 3: Basic Principles of Heredity, Part 2

Mendel’s Genetic Conclusions
Chapter 3: Basic Principles of
Heredity, Part 2
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Plants possess two genetic factors for a character. --alleles
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The two genetic factors separate when a gamete is formed. --two
alleles segregate from each other during meiosis.
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Some factors are dominant over other factors. --The dominant and
recessive phenotypes are in a ratio of 3:1 in the F2 generation. The
genotypic ratio is 1:2:1.
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Separate factors assort independently of each other. --Alleles at
different loci assort independently of each other during meiosis.
Laws of Probability
Probability is defined as the likelihood of a particular event occuring
Always between 0 and 1
0 means the event will never happen
1 means the event will always happen
1/6 x 1/6 = 1/36
P = 0.95?
P = 0.05?
1/6 + 1/6 = 2/6 = 1/3
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Applying Probability to Genetic Crosses
Tt x Tt
1
2
1
TT
Tt
tT
tt
(T gamete and T gamete
(T gamete and t gamete
(t gamete and T gamete
(t gamete and t gamete
½x½=¼
½x½=¼
½x½=¼
½x½=¼
tall
tall
tall
short
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Punnett Square: The long way to figure out
probabilities of particular genotypes or phenotypes.
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A branch diagram in a dihybrid cross: An easier way to predict
phenotypic and genotypic ratios
Break a dihybrid cross into two
monohybrid crosses
(Break a trihybrid cross into three
monohybrid crosses!)
A branch diagram in a testcross
Using Probability directly in a Dihybrid Cross: my favorite way!
Consider a cross of RrYy x RrYy
What is the probability of getting an individual with round, green seeds?
(= the Ry phenotype)
R phenotype
y phenotype
¾
¼
= 3/16
What is the probability of getting an individual with the RrYy genotype?
Rr
Yy
½
½ = 1/4
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Using Probability in a Trihybrid Test Cross
Using Probability in a Trihybrid F1 Cross
Consider a cross of RrYyTt x RrYyTt
What is the probability of getting an individual with the RRYytt genotype?
RR
Yy
tt
¼
½
¼ = 1/32
1. Consider a cross of RrYyTt x RrYytt
(Round, Yellow, Tall plant x Round, Yellow, short plant)
What is the probability of getting an individual with the RYt phenotype?
Round
Yellow
short
R phenotype
Y phenotype
t phenotype
¾
¾
½
= 9/32
2. Consider a cross of RrYyTT x RrYytt
(Round, Yellow, Tall plant x Round, Yellow, short plant.
Same phenotypes as 1st problem)
What is the probability of getting an individual with the Ryt phenotype?
(Same phenotype as 1st problem)
Round
R phenotype
¾
Yellow
Y phenotype
¾
short
t phenotype
0
= 0
A way to determine whether observed
data fit the expected ratios.
What is the probability that the
numbers seen were due to chance?
Chi Square Test
A measure of the deviation of
observed values from expected values
Coin flip example
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white
purple
A larger chi-squareÆ
greater deviation from
the expected values
Looks like a monohybrid cross?
Expected ratio: 3:1
3/4:1/4
Do these observed numbers
deviate significantly from the
expected numbers?
P<0.5
The probability is high
that the deviation from
expected is merely
due to chance.
Total: 150
Expect 3/4x150=112.5
1/4x150=37.5
If the flowers can
only be purple,
how many degrees
of freedom are
there?
Thus we did not refute our hypothesis: It still
may be a monohybrid cross, simple
dominance.
The most common cutoff for P is 0.05
P>0.05: Deviation of observed values from the
expected values may be due to chance
P<0.05: Deviation of observed from expected is
probably not due to chance.
There is something going on that does not fit
our expectations (hypothesis).
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Additional example in chapter 3:
Domestic cats: Black is dominant over grey
Bb x Bb
Expect B kittens:b kittens = 3:1
Observe: 30 black 20 grey
Chi-square = 6.0
Æ check chi-square chart
P<0.025
= a less than 2.5% probability that the deviation from 3:1 is due to chance.
What might be the reason?
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