1. Mapping Eukaryote Chromosomes by Recombination
Chapter 4
February 21, 2017

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

3. 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

4. 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%

5. 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

6. 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.

7. Haploid meiosis followed by postmeiotic mitosis event

8. 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

9. 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.

10. 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.

11. Good….because now we get to come back the your old friend….
To much math yet???
Good….because now we get to come back the your old friend….
The chi-square test

12. Use chi-square test to infer linkage
A/a . B/b x a/a . b/b
Assume there are 200 progeny

13. REJECT the hypothesis of no linkage
Fail to Rejct/
ACCEPT the Hypothesis
REJECT the hypothesis of no linkage
4 phenotypes -1 = 3 degrees of freedom
9.88

14. YOU WILL NOT BE TESTED ON THIS SECTION
SKIP section 4.6 and 4.7
Accounting for unseen Mapping crossovers
YOU WILL NOT BE TESTED ON THIS SECTION

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

16. 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.

17. 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

18. 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

19. Step 5: Complete double strand crossover

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

21. 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

22. 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%

23. 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)

24. 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?

25. 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)

26. RFLPs
Who is your daddy?

27. 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′

28. SSLPs