Mapping Eukaryote Chromosomes by Recombination Chapter 4 February 21, 2017

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

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 0.132 and ct-cv is 0.064 0.132 x 0.064 = 0.0084 (.84%) 1448 x 0.0084 = 12 We only saw 8. Therefore, there was interference

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%

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

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.

Haploid meiosis followed by postmeiotic mitosis event

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

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.

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.

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

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

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

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

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

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.

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

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

Step 5: Complete double strand crossover

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

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 0.132 and ct-cv is 0.064 0.132 x 0.064 = 0.0084 (.84%) 1448 x 0.0084 = 12 We only saw 8. Therefore, there was interference

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%

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)

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?

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)

RFLPs Who is your daddy?

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′

SSLPs