Math 223 Homework 2 Solutions
Spring 1997
This project is designed to help you think about
vectors as non-graphical
objects. Remember, vectors are really just a way of organizing information
into a list. A vector
can have as many components as needed to specify one
element of the list.
Consider the genetics of a population. The following table gives information
about the relative frequency of four different alleles (variants of a gene)
as they occur in four different populations.
| Allele | A | B | C | D |
|---|
| A1 | 0.29 | 0.10 | 0.20 | 0.22 |
|---|
| A2 | 0.00 | 0.08 | 0.06 | 0.00 |
|---|
| B | 0.03 | 0.12 | 0.06 | 0.20 |
|---|
| O | 0.67 | 0.69 | 0.66 | 0.57 |
|---|
Thus, we can think of a population as a
vector with four components, each
component representing the frequency of one of the four alleles for that
population. For example, the vector for population A would be
A = (0.29, 0.00, 0.03, 0.67)
An anthropologist has just returned from the Island of Wak-Wak with genetic
information about the population of the island. He is interested in tracking
their racial history using the genetic data. The island "natives" have the
following genetic vector:
X = (0.15, 0.05, 0.15, 0.64)
1. Define the genetic distance between two populations as the angle between
the vectors which represent each population. Using the
dot product, calculate
the genetic distance between population X and each of the other four
populations.
The dot product of two vectors can be computed two ways. If you have
the components of the two vectors, the dot product can be found by
summing up the product of the x components of the two vectors with the
product of the y components and so on. An alternative is calculate the
dot product as the product of the magnitudes of the two vectors and the
cosine of the angle between the two vectors. Simply rearranging these
forms of the dot product lets one calculate the angle between two vectors.
We know the magnitude of X to be 0.676 from straightforward
computation. The following table shows the magnitudes of each population
the dot product of their genetic vectors with X and the genetic
distance between each population and X (the angle between the two.)
| A | B | C | D |
|---|
| Magnitude | 0.731 | 0.7120 | 0.6948 |
0.6429 |
|---|
| Dotted with X | 0.4768 | 0.4786 |
0.4644 | 0.4278 |
|---|
| cos(angle) | 0.9649 | 0.9944 | 0.9887 |
0.9844 |
|---|
Angle (deg)=
Genetic Distance | 15.2 |
6.1 | 8.6 | 10.1 |
|---|
2. Which race are the Wakos most closely related to? Are there any of the
four that you know cannot possibly be genetic ancestors for the Wakos?
Which ones? Why?
Clearly, population B is closest to the Wakos since the genetic
distance is only 6.1 degrees.
All of the
other populations are further (genetically speaking) from the Wakos.
Populations A and D both lack the A2 allele, while the Wakos
have a frequency of 0.05 for the A2 allele. This means that regardless of
genetic distance, the only way the Wakos could be ancestors of the A
or D populations is through mutation. Otherwise, they would not
possess the A2 allele.
3. Now make a more realistic assumption. Assume that the Wakos are
descended from an equal combination of two of the above populations. Make a
new table (which should have six columns) which gives the genetic information
on these combined races. For example, the combination of A and
C would be the vector (assuming equal contributions from both)
0.5 A + 0.5 C = (0.245, 0.03, 0.045, 0.665)
The following table shows the genetic vectors of the combined
populations formed by equal contributions of two pure populations.
For example, the populations labeled BC below was calculated by
adding half of the B population vector to half of the C
population vector.
| Allele | AB | AC | AD | BC | BD |
CD |
|---|
| A1 | 0.195 | 0.245 | 0.255 | 0.15 |
0.16 | 0.21 |
|---|
| A2 | 0.04 | 0.03 | 0.00 | 0.07 |
0.04 | 0.03 |
|---|
| B | 0.075 | 0.045 | 0.115 | 0.09 |
0.16 | 0.13 |
|---|
| O | 0.68 | 0.665 | 0.62 | 0.675 |
0.63 | 0.615 |
|---|
4. Which of these combinations can be ruled out as possible genetic
ancestors for the Wakos? Which of the remaining combinations is closest to
the Wakos?
Simply looking at the data in the table, we see that the population
labeled BD is almost identical to the Wakos. This is confirmed
by calculating the genetic distance between BD and X as in
the part one. The distance between these two is 1.6 degrees, which is
after calculation of the other distances, the closest.
Again we have a population, in this case AD, that lacks the A2
allele completely. Thus, AD can be discounted as a possible
ancestor for the Wakos. This is logical since neither A or D
could have been ancestors individually.