1. Genetic Mapping
November 15, 2017

2. Mitosis Somatic cells Programmed stage
Any non-sex cell
Germ-line cells = Gametes = Sex cells
Programmed stage
One progenitor cell -> 2 genetically identical cells
2n  2 daughter cells that are both 2n
(Diploid cells)
2n  4n  2n
n  2n  n

3. Meiosis Specialized diploid cells call meiocytes
(sex cells progenitors)
Two sequential cell divisions take place
Meiosis I and Meiosis II
2n (Diploid)  n + n + n + n (4 haploid cells)
2n  4n  2n + 2n  n + n + n + n

4. Key stages of meiosis and mitosis

6. 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

7. 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

8. 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

9. 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

10. What does this mean? 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

11. Calculate Recombination frequency
Which ones are the recombinants?
=305
(305/2839) x 100 = 10.7% recombination frequency
When genes are close together on the same chromosome, they do not assort independently. They produce a recombinant frequency of less than 50%.

12. 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

13. 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?

14. 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

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

16. Crossing over Crossing over rarely occurs between sister chromatids
You must account for all 4 genotypes, if crossing over occurred at the two chromosome stage, we only see 2 genotypes

17. 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)

18. 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

19. 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

20. 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
5000
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

21. 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.

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
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. Centromere mapping with Linear Tetrads
Fungi with linear tetrads can be mapped using centromere mapping
In the simplest form, centromere mapping considers a gene locus and asks how far it is from its centromere
Reminder – A meiocyte produces a linear array of eight ascospores called an octad
The octad is a result of four products from meiosis followed by a postmeiotic mitosis.

31. Haploid meiosis followed by postmeiotic mitosis event

32. With Mendelian rules, there will be 4-A and 4-a
So let’s consider two haploid fungi with different alleles at one locus A x a
With Mendelian rules, there will be 4-A and 4-a
But how will they be arranged?
No crossing over it will be AAAAaaaa
But with crossing over we will see a different arrangement

33. WHAT?
42/300 = 14% Therefore, the A/a locus is 14 m.u. from the centromere right ? WRONG… BUT we can use this number to determine m.u.

34. Remember what m.u. are….
M.U. are defined by the percentage of recombinant chromatids from meiosis
In haploids, there is half the number of chromatid so we need to divide the 14% by 2 and then we get the answer 7 m.u.

35. Last thing to cover in chapter 4
Molecular Mechanism for crossing over

36. Molecular Mechanism for crossing over
How can two large DNA molecules exchange segments with a precision so exact that no nucleotides are gained or lost?
Crossing over is initiated by a double-stranded break in the DNA of a chromatid at meiosis.

37. Step 1: both strands of a chromatid break in the same location
Step 2: DNA is eroded at the 5’ end of each broken strand leaving both 3’ ends single stranded

38. Step 3: Single strand invades the DNA of the other chromatid
Enters the center of the helix and base pairs with its homologous sequence and the invading strand displaces the other strand
Step 4: The invading strand uses the adjacent sequence as a template for new polymerization,
Separation of the resident strands of the helix and we also see replication of the other single strand end to fill the gap left by the invading strand

39. Step 5: Complete double strand crossover

40. Holliday Junction

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

42. Mapping with Molecular Markers
Molecular markers – sequences of DNA that differ between two homologous chromosomes
Two types
Single nucleotide polymorphisms (SNP)
Pronounced “Snips”
Simple sequence length polymorphisms (SSLP)

43. Single Nucleotide Polymorphisms SNPs
-a DNA nucleotide variation occurring in the genome between two members of the same species

44. SNPs
Comparing sequences of individual genomes reveals about 99.9% similar and 0.1% SNP
If humans have 3 billion base pairs, how many bases are different between you and the person to your left?

45. Occurrance in Humans -1 in 1000 bases, 0.1%
-3 to 10 million SNPs in the
human genome

46. More on SNPs Location of SNPs
Located both in genes and not in genes
In exons and introns
The sequences of genes between Wild type vs. mutant allele are examples of SNPs
Most SNPs do not have a phenotype
Two ways to detect SNPs
Sequence a segment of DNA
Restriction fragment length polymorphism (RFLPs)

47. Restriction Fragment Length Polymorphism (RFLP)
-Restriction Enzymes

48. RFLPs
Restriction fragment length polymorphism (RFLPs)

49. SSLPs Sometimes called variable number tandem repeats (VNTRs)
Repetitive DNA fragments
Minisatellites – tandem repeats of units 15 to 100 nucleotides in length
In humans, these can be between 1 and 5 kb in length
Microsatellites – shorter tandem repeats
5′ C-A-C-A-C-A-C-A-C-A-C-A-C-A-C-A 3′
3′ G-T-G-T-G-T-G-T-G-T-G-T-G-T-G-T 5′

50. SSLPs