Download So consider the cross ABC x abc where all three genes are linked

So consider the cross ABC x abc where all three genes are linked. We'll be able to
determine when crossovers occur between each pair of genes. Let's draw up each of these
possibilities assuming there are two crossover's occurring .
So here's a diagramatic representation of what the chromosomes might looked like paired
at pachytene of meiosis. One homolog has two chromatids with the dominant alleles of
three genes, while the other has the recessives.
There are various ways in which two chiasmata can occur as follows:
A
B
C
A
B
C
a
b
c
a
b
c
Two-stranded double crossover. Here only two crossovers occur and both involve the
same two non-sister chromotids.
A
B
C
A
B
C
a
b
c
a
b
c
So here the products of meiosis that emerage are as follows:
ABC (parental)
AbC (double recomb)
aBc (double recomb)
abc (parental)
Three-stranded double crossover (at least one type of this).
A
B
C
A
B
C
a
b
c
a
b
c
So here the products of meiosis are as follows:
ABc (single recomb between B and C)
AbC (double recomb)
aBC (single recomb between A and B)
abc (parental or non-recombinant)
Four stranded double crossover.
A
B
C
A
B
C
a
b
c
a
b
c
So here we end up with the following meiotic products:
Abc (single recomb between A and B)
ABc (single recomb between B and C)
abC (single recomb between B and C)
aBC (single recomb between A and B)
There are no parentals and there are no double recombinants.
Note that which chromatids are involved in double crossing over is random and so
typically if you examined many meioses you would see varying numbers of each of these
kinds of double crossovers.
Imagine you had the following placement of double crossing over.
What would you see in terms of meiotic products?
A
B
C
A
B
C
a
b
c
a
b
c
So here you'd obtain the following meiotic products:
ABC (parental)
ABC (parental)
abc (parental)
abc (parental)
Has recombination occurred here?
Yes! But you wouldn't be able to tell with the particular genes you've been following.
Alfred Sturtevant, a student of Morgan made the first genetic maps using recombination
frequencies.
The percent recombination was used as a distance between genes and so 1%
recombination is sometimes called 1 map unit, or equivalently 1 cM (centiMorgan).
So recall in Morgan's study of linkage between vestigial wings and eye colour (red vs
purple)
Crossed rr vv X R R VV
F1 all
Rr Vv
Testcross:
Female
RrVv
x
Male
rrvv
1/4 RrVv : 1/4 Rrvv : 1/4 rr Vv : 1/4 rrvv
But he got:
1359
151
154
1195
r = number of recombinants / total number of offspring examined
r = 151+154 / (1359+151+154+1195)
r = 0.107
It is common to present this a percent so just multiply by 100% this gives :
10.7% recombination or we could say the genes pr and vg are 10.7 cM apart or 10.7 map
units (m.u.) apart from one another. (we now use the word locus to refer to the gene
positions generally). (Ie the pr locus or the vg locus).
We often will draw a genetic map (in this case a simple one) as follows
R
10.7
V
Note that if a genetic map is available we can also move in the opposite direction.
That is, if we know the distance between two genes, we can predict the frequency of
various progeny we might expect to see in various crosses.
So imagine you had the following genetic map:
A
B
20cM
If you performed a testcross as follows:
AB/ab x ab/ab
You'd expect there to be a total of 20% recombinants and
100%- 20% parental or non-recombs = 80%.
Now recall that there are two kinds of each recombinant so each should be at a percent of
20%/2 or 10%.
Likewise there are two parentals so you'd expect 80%/2 = 40% of each.
You'd get the following progeny genotypes:
Expect
AB/ab (parental or non recomb)
40%
Ab/ab (recomb)
10%
aB/ab (recomb)
10%
ab/ab (parental or non recomb)
40%
Another example:
Imagine you had the following genetic map.
E
40cM
F
What is the expected percent of each type of recombinant in a testcross?
Well, there'd be 40% of recombinants in total so you'd expect 40%/2 = 20% of each.
For nonrecombinants, you'd expect a total of 100%-40% = 60% nonrecombs or there
should be 60%/2 or 30% or each type of nonrecombinant.
Now, thus far in mapping a recombination we've been considering only testcrosses so
we've really only needed to worry about recominantion in one parent (the double
heterozygote). But imagine for the map above and genes E and F we made an F2 cross.
That is you cross EF/ef x EF/ef
Given the map above predict the frequency of each genotype.
And assuming dominance whereby E dominant over the e allele and
F is dominant over the f-allele.
Do this for next day and I'll take up the result.
Answer:
First determine gametic proportions.
Percent of recombinant gametes (Ef and eF) is 40% we expect 40%/2 or 20% of each.
As a proportion this is 0.2
Percent of parentals gametes (EF and ef) is 100%-40%=60%, we expect 60%/2 or 30% of
each. As a proportion this is 0.3 of each.
(note we assume below equal recombination in male and female which isn't always so).
Now just set up a Punnett square as follows:
0.3 EF
0.2 Ef
0.2 eF
0.3 ef
0.3 EF
0.09 EF/EF
0.06 Ef/EF
0.06 eF/EF
0.09 ef/EF
0.2 Ef
0.06 EF/Ef
0.04 Ef/Ef
0.04 eF/Ef
0.06 ef/Ef
0.2 eF
0.06 EF/eF
0.04 Ef/eF
0.04 eF/eF
0.06 ef/eF
0.3 ef
0.09 EF/ef
0.06 Ef/ef
0.06 eF/ef
0.09 ef/ef
Now add up proportions of each genotype:
EF/EF 0.09
EF/Ef 0.06 + 0.06 = 0.12
EF/eF 0.06 + 0.06 = 0.12
EF/ef and Ef/eF 0.09 + 0.04 + 0.04 + 0.09 = 0.26 (note two kinds double het, cis/trans)
Ef/Ef 0.04
Ef/ef 0.06 + 0.06 = 0.12
eF/eF = 0.04
ef/eF 0.06 +0.06 = 0.12
ef/ef 0.09
Make sure the proportions all add up to one (or you've made a mistake).
Now imagine the E is dominant to e, and F dominant to f. Write the expected phenotypic
proportions:
E-F- = 0.09 + 0.12 + 0.12 + 0.26 = 0.59
E-ff = 0.04 + 0.12 = 0.16
eeF- = 0.04 + 0.12 = 0.16
eeff = 0.09
So instead of the 9 : 3 : 3 : 1 ratio expected with independent assortment, with a
40cM map,
9.4: 2.6 : 2.6 : 1.4.
20cM map
10.6: 1.4 : 1.4 : 2.6
10cM map,
11.2 : 0.8 : 0.8 : 3.2
1 cM map,
11.9 : .08 : .08 : 3.9
0 cM map
3: 0
: 0: 1
So there is an excess of some phenotypic classes, especially those that have more nonrecombinant genotypes, while a deficiency of those containing recombinants.
In a more general way, you could write out the equations for each genotype in an F2
cross as above.
So if "r" is the porportion of recombinants, then you'd have the following gametic output:
The proportion of each recombinant gametes would be r/2
The proportion of nonrecombinant is 1-r so that
Each nonrecomb gamete occurs (1-r)/2 of the time.
So using these gamete proportion fill in the Punnett square as above.
Then determine the propoportion of each genotype in terms of r.
Then if you had dominance, determine the proportion of each phenotype in terms of r.
With these equations in terms of r, if you had done an F2 cross you can now in principle
estimate the recombination frequency using those equations and the observed proportions
of progeny from your F2 cross. With dominance, you'd have four equations (one for each
phenotype). You'd need to use a statistical method known as maximum likelihood to
obtain the best estimate of the recombination proprotion r. Although, the equations and
procedures for estimating the recombination proportions for this and other crosses are
well-known and have been detailed in the literature.
Three-point testcross
To begin to map more genes, we need to construct genotypes that are segregating for
more than 2 linked genes. Here we consider the use of three point testcrosses.
Beginning with two parental lines AABBCC x aabbcc we generate and F1 and then a
testcross:
We wish to discover if the genes are linked?
If linked, how far apart are they
If linked what is there order.
AaBbCc x aabbcc
You observe the following numbers of progeny (note I"ve left out the gamete
contribution from the tester.
OBS
AB
Recombinant for gene pair
AC
BC
ABC
175
a b c
175
ABc
25
R
R
abC
25
R
R
AbC
50
R
R
a Bc
50
R
R
A bc
5
R
R
a BC
5
R
R
Total
510
110
60
150
Now let's work out the recombination percent for each gene pair.
A & B loci = 100% x 110/510 = 21.5 cM
A & C loci = 100% x 60/510 = 11.8 cM
B & C loci = 100% x 150/510 = 29.4 cM
So all the loci are linked since the recombination fraction is less than 50% for each.
Map the loci.
Well we can see that the greater distance is between loci B & C so these must be farthest
apart on the chromsome. So A must lie between the B & C loci but closer to the C locus
since this has a smaller distance.
C
11.8cM
A
21.5cM
B
33.3cM
Note that our estimated distance between the B & C locis is 29.4 cM yet if you add up the
distances between C & A and A & B you obtain 11.8 + 21.5 = 33.3cM which is greater
than the estimate we obtained when just considering the C & B loci alone.
Why is there this discrepancy in distances obtained in the two ways?
Recall that what we wish to do is to use the number (proportion) of crossovers that occur
between gene loci as an estimate of distance between those loci. Now that we've done the
mapping and know the gene order to be C -- A -- B we can go back and look at our
summary table of data.
What we now can see is that there are two genotypes that represent double recombination
events. That is progeny that are Abc and aBC. So the problem is that when we
determined the distance betwee that C and B loci, we didn't count those progeny at all,
but we should have actually counted each of them twice as they each involved two
recombination events.
So for the distance between B & C we should have added:
# BC recombs 25 + 25 + 50 + 50 + 2 x 5 + 2 x 5 = 170 (not 150 as before)
% recomb BC = 100% x 170/510 = 33.3% or 33.3cM.
So we can either ensure that once we know the gene order, we take into account such
double recombination events, and/or don't worry about it but note that the correct map
distance between the two loci that are further apart is obtain by just adding up the
intervening distances as we did intially above.
Another example three-point testcross
Cross RRssTT x rrSStt gives F1 RsT / rSt
Testcross RsT / rSt x rst / rst
Recombinant for gene pair
OBS
RS
RT
ST
RsT
270
rSt
270
Rst
10
R
R
rST
10
R
R
RSt
90
R
R
rsT
90
R
R
RST
130
R
R
rst
130
R
R
Total
1000
440
200
280
Now let's work out the recombination percent for each gene pair.
R & S loci = 100% x 440/1000 = 44.0 cM
R & T loci = 100% x 200/1000 = 20.0 cM
T & S loci = 100% x 280/1000 = 28.0 cM
So all the loci are linked since the recombination fraction is less than 50% for each.
Since R&S have greatest distance they must be at the ends and T must be inside closer to
the R locus.
R
20 cM
T
28 cM
48 cM
S
So again here again we can see that the recombination percent if we just consider the two
loci R & S came out to 44cM, yet this underestimates the map distance which is best
obtained by the sum of 20 + 28 = 48 cM. (Note that the map distance can be greater than
50cM even though you would not obtain such an estimate is you just consider two loci.)
For three-point testcrosses, it is often possible to deduce the gene order by inspection of
the numbers of progeny. Typically you will observe numbers like the following:
Two equally high frequency classes representing parental genotypes;
Two at intermediate frequency representing single recombinants for one gene pair
Two at another intermediate freq respresenting the single gene pair recombinants
Two at low frequency representing double recombinants.
For a three point cross there are only 3 possible gene orders.
The order that gives the low frequency double recombinant class, is the correct gene
order.