Transcript Chromosomes and Heredity

Sex Chromosomes
&
Probabilities and Chi Squared
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X-Linked Inheritance
• Special chromosomes determine sex in many organisms
• X and Y chromosomes = sex chromosomes which are non-identical but
share some genes
• Hemophilia is a classic example of human X-linked inheritance
• Males produce two different types of gametes: one containing X and
another Y chromosome (heterogametic)
• Females have two X chromosomes and produce only X-bearing
gametes
• In some organisms (birds, butterflies and some reptiles), females are
heterogametic
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Morgan studied inheritance patterns in Drosophila melanogaster and
observed that with some traits
Males X Females produced different ratios than when Females X Males
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Inheritence of Hemophilia A among the descendants of Queen Victoria.
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How is this trait
most likely
inherited?
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How is this trait
most likely
inherited?
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Predicting the liklihood of observing a specific outcome
in the progeny of a given cross
If a couple has three children, what’s the liklihood that they
have exactly one girl?
(the problem can be broken down using the multiplication and addition rules)
They could either have
(Boy AND Boy AND Girl) OR (Boy AND Girl AND Boy) OR (Girl AND Boy AND Boy)
So the probability (P) would be
P = (1/2 X 1/2 X 1/2)
P = (1/8)
P = 3(1/8)
P = 3/8
+ (1/2 X 1/2 X 1/2)
+ (1/8)
+ (1/2 X 1/2 X 1/2)
+ (1/8)
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The liklihood of observing specific outcomes in the progeny of a given
cross is predicted by the binomial probability
(p + q)n
In the previous example of a couple having three children,
where the probability of having a boy is p and the probability of having a girl is q…
the liklihood of observing each outcome is..
(p + q)3 = (p + q)(p2 + 2pq + q2) = p3 + 3p2q + 3pq2 + q3
Probability of
having exactly 1 girl
3p2q = 3(1/2)2(1/2) = 3/8
If the probability of possibility
A is p and the probability of
the alternative possibility B is
q, then the probability that, in
n trials, A is realized s times
and B is realized t times is
n! psqt
=
s!t!
3! (1/2)2(1/2)1=
2!1!
3/8
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Chi-Square Analysis
• The test of goodness of fit = test analyzes
whether observed data agree with theoretical
expectation
• A conventional measure of goodness of fit is a
value called chi-square, c2
• c2 = ∑( observed-expected)2 / expected
• A value of c2 = 0 means that the observed
numbers fit the expected numbers perfectly
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Chi-Square Analysis
•
Probability P that a worse fit (or one equally bad) would be obtained by
chance, assuming that the genetic hypothesis is true
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The critical values of P are conventionally chosen as 0.05 (the 5 percent
level) and 0.01 (the 1 percent level)
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Statistically significant refers to the magnitude of the difference between
the observed and the expected numbers
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To determine the P value corresponding to a calculated c2 we need the
number of degrees of freedom of the particular chi-square test
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The number of degrees of freedom equals the number of classes of data
minus 1
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Calculation of chi-square for a suspected
monohybrid cross … by example
OBSERVED
Experiment: We cross two pea plants that we suspect are
monohybrids for pea color. When we sampled 144 peas from the
progeny and got 99 yellow peas and 45 green peas
Hypothesis: (If) This is a monohybrid cross, (then) we should
observe a 3:1 ratio of yellow:green peas in the progeny. This means,
out of 144 peas, we should expect to observe 108 Yellow:36 Green.
G g
G
g
99
45
EXPECTED
G g
G
g
108
36
c2 = ∑(observed-expected)2/expected
= (observedYel-expectedYel)2/expectedYel + (observedGrn-expectedGrnl)2/expectedGrn
= (99 - 108)2 / 108 + (45 - 36)2 / 36
= 81/108 + 81/36
= 3.0
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We have two classes of data (either yellow or green)
so…our degrees of freedom = (# of classes) - 1
= 2 -1
=1
Giving a P value of ~0.08. So the amount of fluctuation that we observed was not
statistically different from what we expected to see for this sample. Indicating that
our hypothesis could be correct… and we a dealing with a trait inherited as a a
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monohybrid cross.
What would if the ratios stayed the same
when we sampled 1440 peas, instead of 144?
So we observed 990 yellow and 360 green
peas.
1st sample 144 peas
2nd sample 1440 peas
1080
?
?
450
360
?
?
1440
1440
990
?
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