PROBABILITY/STATISTICS IN INHERITANCE
We know that when two people who are both
heterozygous for a simple Mendelian autosomal gene alpha have a child, the probability
that the child will show the dominant phenotype is 3/4. Let's ask
a somewhat more complex question. If this couple has a total of four children,
what is the probability that 3 of the 4 will show the dominant phenotype? To
answer this, we will first derive the appropriate formula and then use it to
calculate the numerical answer. The same formula allows us to understand the
expected statistical distribution of the various possible phenotype patterns
in four-child ( or any other size) families in a large population.
1. Review: How can some simple overall probabilities be calculated
by combining the multiplication and addition "rules" we covered earlier?
Let's start with a very simple case: asking about gender probabilities in families
of three children.
What is the probability that all three children in a family will be the same
gender?
P(all female)= 1/2 x 1/2 x 1/2 = 1/8
P(all male ) = 1/2 x 1/2 x 1/2 = 1/8
P(all one gender) = P(all female) + P(all male) = 1/8 + 1/8 = 1/4
What is the probability that a three-child family is two girls and one boy?
Each possible birth order has P=1/8. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=1/8.
So, P(2G,1B)= 3/8 and P(1G,2B)= 3/8.
This allows us to write the overall gender probability distribution for families
of three children as follows:
1/8 will be three girls
3/8 will be two girls and one boy
3/8 will be one girl and two boys
1/8 will be three boys
Adding it all up, we have 1/8 + 3/8 + 3/8 + 1/8 = 1 (100%)
2. How can we understand and use "Pascal's triangle"
and the "general rule for repeated trials of events with constant probabilities",
formula 1 (page 161 in textbook)?
Consider the numerators of the fractions in the above three-child family gender
equation:
1 , 3 , 3 , 1. These numbers are the coefficients in the expansion of the term
(p + q) cubed. In general, the coefficients of any such binomial expansion { the
term (p+q) raised to any power} give the "number of ways" that something
can happen.
Figure 4.21, "Pascal's triangle", shows these coefficients for the expansion
of (p + q) raised to any power up to 10. The numbers in any row can be used
just as described above. For example, assume that over the next two decades
you have 6 kids. There are 64 possible gender birth orders, with 20 of these
resulting in you having three girls and three boys.
The terms p and q are the individual probabilities for a specific outcome from
a single "event". For "gender" calculations, the probabilities p and
q are equal, both = 1/2 (the equal probabilities of male and female births).
For "dominant : recessive phenotype" type of calculations, p and q will usually
not be equal. For a simple Mendelian inheritance from two heterozygotic parents,
p will = 3/4 (if AA and Aa give dominant phenotype) and q will
= 1/4 (aa gives recessive phenotype).
Generalizing this, we get to the formula on page 161 in your text that is "the general
rule for repeated trials of events with constant probabilities". The term (n!/s!t!)
is the number of possible ways (orders) of getting a certain net outcome ( "a
total of n; with s of one and t of the other"). This
number can either be calculated or taken directly from Pascal's triangle.
3. Sample problem: You and your mate are both heterozygous
for some simple Mendelian gene alpha (i.e., each of you has genotype
Aa, and both of you show the dominant phenotype) on chromosome #1. Over
the next decade, you proceed to have four children. What is the probability
that 3 of your children will show the dominant phenotype and one will show the
recessive phenotype? What are the probabilities of the other possible outcomes?
If we were looking at thousands of such families, we know that the overall ratio
of dominant to recessive phenotypes in the children would average out to 3:1,
as shown by a simple Punnett square. But for one couple having four children,
what's the probability, P(3D,1r)?
To calculate P(3D,1r), we use formula 1 for the case n=4, s=3, t=1, p=3/4,
q=1/4.
P(3D,1r)= 4!/3! x (3/4)cubed x (1/4) = 4 x (27/64) x 1/4 = .42 ( 42% )
By also calculating the other four possibilities, we can construct a graph that shows the statistical distribution you would expect
to see in a large population.
Problem S-4: "Heterozygous parents; have three children".
Modify the sample problem above to do the calculations for you and your mate (both Aa) having THREE children. Do the calculations for the probabilities that all three, two, one, or none of the three children will show the dominant phenotype from gene alpha. Construct the graph and compare the result with the "four children" graph done in class.