1. Mapping Eukaryote Chromosomes by Recombination
Chapter 4
February 9, 2017

2. Genome Mapping What positions are genes located on the chromosomes?
Gene position is crucial information needed to build complex genotypes
Knowing the position of genes provides a way to discover structure and function
Understand evolutionary genetic mechanisms
Chromosome map – arrangement of genes on chromosomes is represented as a unidimensional diagram
Loci – (sing. Locus) location of the gene on a chromosome

3. Warm up question #1
A single mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to DNA replication. The mutation is not corrected, and the cell undergoes DNA replication and a normal meiosis producing four gametes. How many of these gametes will carry the mutation? A) 1 B) 2 C) 3 D) 4

4. Diagnostics of Linkage
Recombination maps are assembled 2 to 3 genes at a time
Two genes are linked when the loci of two genes are located on the same chromosome
Therefore, the Mendelian Genetics rules do not explain/apply to the progeny ratios

5. So what does that look like…
P: pr/pr ; vg/vg x pr+/pr+; vg+/vg+
Gametes: pr ; vg and pr+ ; vg+
F1 dihybrid: pr+/pr; vg+/vg
Perform a testcross to determine which genes are linked
WHAT IS A TESTCROSS?
Testcross: pr+/pr; vg+/vg and pr/pr; vg/vg
The tester parent contributes gametes carrying only recessive alleles, the phenotypes of the offspring directly reveal the alleles contributed by the games of the dihybrid parent
The analyst can concentrate on meiosis in one parent (the dihybrid)
Pr – purple pr + is red
Vg – vestigial (very small wings) vg+ normal wing length

6. Testcross: pr+/pr; vg+/vg and pr/pr; vg/vg
Following mendelian genetics, what ratio would you expect to see with this cross?
1:1:1:1
The test cross actually showed
We see the first two alleles combinations are in a greater majority, which indicts they are linked (or located on the same chromosome) AND PARENTAL CHROMOSOMES
pr+vg+
pr+vg
prvg+
prvg
pr; vg
pr+pr;vg+vg
pr+pr;vgvg
prpr;vg+vg
prpr;vgvg
pr+ · vg
pr · vg
pr+ · vg
pr · vg
2839

7. So how can we make any sense out of this?!?!?!?
Remember back to the parents…
P pr/pr ; vg/vg x pr+/pr+; vg+/vg+
F1 pr+/pr;vg+/vg
Therefore the F1 dihybrids must have been…
That explains the linked progeny phenotypes/genotypes we see

8. Calculate Recombination frequency
Which ones are the recombinants?
=305
(305/2839) x 100 = 10.7% recombination frequency

9. Let’s look at another… P pr/pr ; vg+/vg+ x pr+/pr+; vg/vg
Gametes pr+; vg and pr; vg+
F1 dihybrid is pr+/pr; vg+/vg
What would you use for a test cross?
What mendelian ratio would you expect?
1:1:1:1
What were the actually ratios for F2?
What do you think the F1 hybrids were?
pr/pr; vg/vg
pr+vg+
pr+vg
prvg+
prvg
pr; vg
pr+pr;vg+vg
pr+pr;vgvg
prpr;vg+vg
prpr;vgvg

10. Calculate the recombination frequency here?
Identify the recombinants…..
= 303/2335 = x 100= 12.97% or 13%
Even though it is the same gene as the one before we got a different answer (remember it was 10.7 before)….WHY?

11. Crossing over produces recombinants
When chromosomes align during meiosis they occasionally break and exchange parts with its homolog
The cross-shaped structure is called chiasma
Chiasmata (pl)
The new combination is called crossover products
WHAT STAGE OF MEIOSIS CAN WE OBSERVE CROSSING OVER?
Prophase I – Diplotene
Prophase I – Leptotene
Metaphase I
Metaphse II

12. Chiasmata
Key concept: A crossover is the breakage of two DNA molecules at the same position and their rejoining in two reciprocal recombinant combinations

13. Crossing over Crossing over does not occur between sister chromatids
You must account for all 4 genotypes, if crossing over occurred at the two chromosome stage, we only see 2 genotypes

14. Multiple crossovers can include more than two chromatids
Does crossovers occur between sister chromatids?
Yes, but they are rare and they do not produce new allele combinations (so there are not counted)

15. Linkage symbolism and terminology
Cis conformation – same (adjacent)
Trans conformation – different (opposite)
Alleles on the same homolog have no punctuation between them
A slash symbolizes two homologs
Genes known to be on separate chromosomes are separated by a semi colon ; -AB/ab ; CD/cd
Genes of unknown linkage are shown by a dot .

16. How would you write the following allele pairings
A is located on the same homolog as F AF
Bc/BC has an unknown linkage to D/d
Bc/BC . D/d
Plant C1473 is dihybrid for genes A and B which are in a cis conformation
AB/ab or ++/ab
Plant M2243 is dihybrid for genes M and P which are in a trans conformation
Mp/mP or +p/m+
Genes located on separate genes use a semi colon
Aa;Bb;Cc

17. Mapping by Recombinant Frequency
Frequency of recombinants produced by crossing over is key to chromosome mapping
The farther apart two genes are, the more likely a crossover will take place
Therefore, the proportion of recombinants is a clue to the distance separating two gene loci on a chromosome map

18. So how do we figure map units…
Let’s take a look back at our first problem with diagnostic linkages
=305
(305/2839) x 100 = 10.7% recombination frequency
Use the percent to generate a map and use the percentage as a linear distance between two genes
So 10.7% = 10.7 m.u. or cM
m.u. – map units
cM -centimorgans

19. Practice
We already know that the distance between vg and pr is 10.7 m.u. So what is the distance between vg and your favorite gene (yfg) with the following progeny numbers?
vg+ · yfg
vg · yfg
vg · yfg
vg+ · yfg
3000
vg yfg
vg yfg+
Predict the genotype and Identify the recombinants
Add the recombinants
=500
Divide by the total
500/5000 = 0.10 x 100 = 10% = 10 m.u. = 10 cM

20. How to diagram 20.7 m.u. OR 0.7 m.u. OR pr vg 10.7 m.u. yfg 10 m.u. pr
What is the map distance between pr and yfg?
20.7 m.u. OR 0.7 m.u.

21. Practice problem
If A and B had a recombination frequency of 5% and A and C had a recombination frequency of 10%. What are the possible map(s)? B C A
5 m.u.
5 m.u. OR A C B
10 m.u.
5 m.u.

22. Three-point testcross
Trihybrid – triple heterozygote
The next level of complexity…
Three-point testcross
Three-factor cross

23. How many gametes are possible for a trihybrid cross? 2 x 2 x 2 = 8
Do a branch diagram to check
v+·cv+·ct+
v+·cv+·ct
v+ ·cv·ct+
v ·cv+·ct+
v+ ·cv·ct
v ·cv·ct+
v·cv+·ct
v·cv·ct

24. So let’s analyze the loci two at a time v and cv loci How do we know the box are the recombinants? Look back the parental genotypes…

25. There are 268 of these recombinants and 1448 flies so…
There are 268 of these recombinants and 1448 flies so…. 268/1448 x 100 = 18.5 % or 18.5 m.u. or 18.5 cM
How about v and ct?
191… so 191/1448 x 100 = 13.2% or 13.2 m.u.
How about ct and cv?
93 …So 93/1448 x 100 = 6.4% or 6.4 m.u.

26. So what does the map look like?
What can the testcross be rewritten like?
v+ ct cv/v ct+ cv+ × v ct cv/v ct cv
Does anyone see something wrong???
Hint: v-cv distance versus v-ct and ct-cv
But WAIT A SECOND….
= 19.6 not 18.5

27. These are double recombination events and should have been counted…and actually counted twice since they involve two crossover events
= 284.
284/1448 x 100 = 19.6% or 19.6mu
In practice, we do not need to do this calculation because the sum of the two shorter
distances gives us the best estimate of overall distance

28. Let’s practice Identify P genotype Look at 2 genes at a time + + + 74
B and C ~ =238
238/1000=.238*100 = 23.8m.u.
A and B ~ =150
150/1000=.15*100 = 15 m.u.
A anc C ~ = 100
100/1000= m.u.
+ + + 74
a b c 70
a + + 44
+ b c 50
+ + c 4
a b + 2
a + c
368
+ b +
388
1000
15m.u.
10m.u. b a c
So what was the genotype of the parents?

29. Deducing gene order by inspection
Two at high frequency
Two at intermediate frequency
Two at a different intermediate frequency
Two rare

30. Identifying and calculating interference
Knowing the existence of double crossovers helps us determine if there is interference in the crossover events
Interference is the likely tendency that a crossover at one spot on a chromosome will decreases the likelihood of a crossover in a nearby spot
If there is no interference, we should be able to use the recombination data to calculate the number of double recombinants
The v-ct RF is and ct-cv is 0.064
0.132 x = (.84%)
1448 x = 12
We only saw 8.
Therefore, there was interference

31. How to calculate interference
Coefficient of coincidence (c.o.c.)
Observed/expected (ranges from 0-1)
In our example 8/12 or 66.7%
To figure interference 1 – c.o.c
1 - 8/12 = 4/12 = 1/3 =33%

32. So getting back to Mendelian genetics
What can we now determine about a 9:3:3:1 ratio of plants?
The plants are a result of monohybrid cross
The plants are a result of a dihybrid cross with genes on the same chromosome
The plants are a result of a dihybrid cross with genes on different chromosomes
The plants are a result of a trihybrid cross