The
fore!mat of genes on a chromosomes – called a chromosome map - can
be determined by analysing how often the genes recombine away from
the parental genotypes to produce new combinations in the
offspring. Your task is to determine the chromosome map for 5
genes, A, B, C, D, E.
Example
of how to do it:
Consider two genes, Q and R.
All
parents have two copies of each gene (because they are
diploid)
Parent type 1 has the genotype of Qq and Rr
(therefore called “heterozygous” for these traits) and Parent type
2 has the genotype of qq and rr (therefore called “homozygous
recessive” for both traits)
If
parents with type 1 are crossed with parents of type 2 and the
following offspring are produced:
QqRr
= 78 offspring
qqrr
= 42 offspring
Qqrr
= 12 offspring
qqRr
= 6 offspring
You
will notice that the first two are the same as the parents and the
second two are different from the parents. These second two types
are known as “recombinant types”.
In
this example there are 18 recombinants out of a total of 138
offspring.
This
can be converted to a “cross over value” through the simple
formula:
Cross over value = number of recombinant
types/total number of offspring x 100
So
in this case the cross over value for Q and R is
18/138x100
Hence the cross over value equals approximately
13%
This
means that you can then say that genes Q and R are 13 map units
away from each other on a chromosome. This could look like the
following picture, but technically the Q and R could be anywhere,
so long as they are separated by 13 points:
Now
imagine we found another gene, “S”
If
we look at its recombination with “R” and find that it is 6, then
there would be two possible positions for “S”, as shown
below.
What
we would need to know, is the cross over value between S and Q. If
it equals 7, then the map is:
but
if the cross over value between S and Q equals 19 then the correct
map would be:
Now, for
you to solve:
There are 5 genes to consider: A, B, C, D and
E.
Cross over between A and B - Parents AaBb
crossed with aabb produce:
120
AaBb
171
aabb
5
Aabb
4
aaBb
Cross over between A and C - Parents AaCc
crossed with aacc produce:
84
AaCc
108
aacc
6
Aacc
2
aaCc
Cross over between A and D - Parents AaDd
crossed with aadd produce:
44
AaDd
45
aadd
1
Aadd
0
aaDd
Cross over between D and E - Parents DdEe
crossed with ddee produce:
128
DdEe
160
ddee
4
Ddee
8
ddEe
Cross over between C and D - Parents CcDd
crossed with ccdd produce:
165
CcDd
120
ccdd
5
Ccdd
10
ccDd
Cross over between B and E - Parents BbEe
crossed with bbee produce:
224
BbEe
168
bbee
5
Bbee
3
bbEe
Hints:
1)
Determine the cross over value
(recombination percentage) for each reported group
2)
Draw them on a chromosome
map
The
cache can be found at S35 11.WX1, E149 08.YZ5
Where:
W =
the cross over value between A and E
X =
the cross over value between B and D
Y =
the cross over value between B and C
Z =
the cross over value between C and E