Mapping Eukaryote Chromosomes by Recombination Chapter 4 February 14, 2017

χ2 = Σ (O – E)2/E for all classes 38 tall purple 38 tall white 37 short purple 37 short white 45 tall purple 35 tall white 40 short purple 30 short white χ2 = Σ (O – E)2/E for all classes

ACCEPT the Hypothesis REJECT the hypothesis 4 phenotypes -1 = 3 degrees of freedom

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

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

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

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

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+ 1339 pr · vg 1195 pr+ · vg 151 pr · vg+ 154 2839

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

Calculate Recombination frequency Which ones are the recombinants? 151 +154 =305 (305/2839) x 100 = 10.7% recombination frequency

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

Calculate the recombination frequency here? Identify the recombinants….. 157 + 146 = 303/2335 = .1297 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?

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

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

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

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)

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 .

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

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

So how do we figure map units… Let’s take a look back at our first problem with diagnostic linkages 151 +154 =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

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+ 2200 vg · yfg 2300 vg · yfg+ 200 vg+ · yfg 300 5000 vg yfg vg+ yfg+ Predict the genotype and Identify the recombinants Add the recombinants 200+300=500 Divide by the total 500/5000 = 0.10 x 100 = 10% = 10 m.u. = 10 cM

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.

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.

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

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

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…

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.

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…. 13.2 + 6.4 = 19.6 not 18.5

These are double recombination events and should have been counted…and actually counted twice since they involve two crossover events 45 + 40 + 89 + 94 + 3 + 3 + 5 + 5 = 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

Let’s practice Identify P genotype Look at 2 genes at a time + + + 74 B and C ~ 74+70+44+50=238 238/1000=.238*100 = 23.8m.u. A and B ~ 74+70+4+2=150 150/1000=.15*100 = 15 m.u. A anc C ~ 44 + 50+ 4+ 2 = 100 100/1000=.1 10 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?

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