BISC202 Lecture 16 Trihybrid Cross Mapping, interference, and X-linked genes PDF

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Lecture notes for BISC 202: Genetics in the Fall 2016 Semester...

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BISC202 Lecture 16: Trihybrid Cross Mapping, interference, and X-linked genes Mapping 3 Genes with a Trihybrid Testcross ● Rather than perform 3 dihybrid test crosses, it is easier to map 3 genes with one TRIHYBRID testcross ○ ex.) ABC/abc x abc/abc ● ex.) if the mapping distance between 2 genes is 4.5 mu ○ The crossover frequency is 0.045 ○ The DOUBLE crossover frequency is 0.045*0.045 = 0.00203 (use decimals or fractions and NOT %) ● Step 1: sort the progeny from most to least common (reciprocal classes, which refers to exact opposite genotypes, are organized together) ○ (NCO 1) A B C: 592 ○ (NCO 2) a b c: 580 ○ (SCO 1) A b c: 94 ○ (SCO 2) a B C: 89 ○ (SCO 3) a B c: 45 ○ (SCO 4) A b C: 40 ○ (DCO 1) a b C: 5 ○ (DCO 2) A B c: 3 ● Step 2: identify the parental (non-crossover; NCO) classes ○ The genotype that resembles the parents ○ In this case the top 2 genotypes ● Step 3: identify the double crossover (DCO) classes ○ The genotype with the least frequency, since DCO is very rare ○ In this case the bottom 2 genotypes ● Step 4: identify the single crossover (SCO) classes ○ These are the 2 pairs of classes with intermediate frequency ○ In this case the 4 genotypes in the middle ● Step 5: figure out the proper gene order ○ DCOs only exchange the middle gene ○ Line up the parentals (NCOs) and DCOs and see which gene was swapped, that must be the middle gene ○ In this case compare NCO 2 and DCO 1 ■ Since the “c” allele was swapped for the “C” allele, the loci for the C gene must be in the middle ■ As a result the correct gene order would be: A  C B or b c a ● Step 6: rewrite the F2 genotypes (tester alleles are omitted) in correct order ○ (NCO 1) A C B: 592 ○ (NCO 2) a c b: 580 ○ (SCO 1) A c b: 94 ○ (SCO 2) a C B: 89 ○ (SCO 3) a c B: 45 ○ (SCO 4) A C b: 40 ○ (DCO 1) a C b: 5 ○ (DCO 2) A c B: 3

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Step 7: identify where crossing over occurred for each class ○ NCO has crossing over at neither interval ○ SCO (1/2) has crossing over at the first interval (i) ○ SCO (3/4) has crossing over at the second interval (ii) ○ DCO has crossing over at the both intervals (i, ii) Step 8: calculate “r” for each interval ○ Interval (i) = (94+89+5+3) / (592+580+94+89+45+40+5+3) ○ Interval (i) = 13.2 mu ○ Interval (ii) = (45+40+5+3) / (592+580+94+89+45+40+5+3) ○ Interval (ii) = 6.4 mu ○ NOTICE how the DCO frequencies are used in calculating for the recombination frequency in both intervals ○ Add these map distances to the linear map How would you recognize whether only 2 of the 3 genes are linked? ○ 1: If there are 4 equally common or uncommon classes (since common classes will have parental combinations for 2 genes but the 3rd one will vary) ○ 2: The map distances with unlinked genes could be 50% but that is possible with distantly-linked genes How would you recognize when 3 genes are on separate chromosomes? ○ 8 equally common classes (this means every class is equally possible and they are independently assorted)

Making Genetic/Linkage Maps ● By mapping a few linked genes at a time, it is possible to map every gene found on a chromosome ○ This is known as a linkage group ● Some gene pairs can be more than 50 mu apart ○ These will show an “r” value of 50% and were mapped by adding the smaller (less than 50 mu) and more accurate intervals between them ● Also, if we mapped one of each non-homologous chromosome, we can make a complete genetic  map ○ ex.) Drosophila fruit flies are 2n = 8 so the linkage map consists of 4 different linkage groups Positive and Negative Interference ● Interference: It is possible that a crossover can inhibit another crossover happening nearby resulting in fewer DCOs than expected ○ Formula: ■ 1 - (Observed  # DCOs) (Expected # DCOs) ○ Observed # DCO can be taken from data ○ Expected # DCO is found by taking the recombination frequency between each interval and multiplying by the total number of genes ● Positive interference (between 0 and 1) indicates that DCOs were suppressed as crossovers interfered with nearby crossovers

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If Interference = 0, we would see ALL 12 DCOs (as expected) as there would be no interference ○ If interference = 1, we would see NO DCOs, due to 100% interference Negative interference refers to more DCOs being present than expected, meaning crossovers are promoting more crossovers elsewhere ○ This is mostly seen in prokaryotes (SCOs make circular chromosomes linear and linear DNA is degraded by prokaryotic cells)

Another Trihybrid Testcross Example ● Imagine a cross between (ignoring tester alleles): “A•B•C x a•b•c” with the following results: ○ 1 ABC: 245 ○ 2 abc: 240 ○ 3 ABc: 243 ○ 4 abC: 239 ○ 5 AbC: 10 ○ 6 aBc: 9 ○ 7 Abc: 11 ○ 8 aBC: 9 ● These results show that one gene is unlinked from the other 2 (C) because when “C” varies, the progeny #’s are not affected significantly ● To calculate “r” for “A” and “B”, take the bottom 4 progeny (recombinants) divide by the total progeny (via formula) ● To calculate “r” for “A” and “C”, take 3,4,7,8 (recombinants) and divide by the total progeny (via formula) ○ From this data, it should be ~50 mu (same for B and C), suggesting that the C gene is located on a different chromosome ● CONCLUSIONS: ○ When the data shows 4 equally common and 4 equally uncommon classes, it is likely that 1 gene is unlinked from the other 2 (3rd gene varies without an effect) ○ The map distances with the unlinked genes should be close to 50% HOWEVER it is possible that the 2 genes are just very far apart from each other Mapping 2 X-linked Genes ● ex.) Drosophila fruit flies have XX = females and XY = males ○ w+ = red eyes and w = white eyes ○ m+ = normal and m = small wings ● Since recessive tester is male, it will only pass down its recessive X to females and Y to males ○ Therefore the phenotype of the offspring reflects what the genotype of gametes from the dihybrid mother was ● P: wm/wm (female) x w+m+/Y (male) ● F1: w+m+/ wm x wm/Y ● F2 shown below in the table

Dihybrid Gamete

Females

Males

Parental or Recombinant?

w+ m+

439

352

P

wm

218

237

R

w+m

235

210

R

wm

359

391

P

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Mapping distance calculation between the 2 genes: ○ Total = 2447 ○ 218+237+235+210 = 900 ○ (900/2441) x 100% = 37% ○ 37 mu...