Gene Lecture 8 Linkage

Fundamental Genetics Lecture 8

Linkage and Chromosome Mapping in Eukaryotes John Donnie A. Ramos, Ph.D. Dept. of Biological Sciences College of Science University of Santo Tomas

Linked Genes ‰ Genes located in the same chromosomes ‰ Initiated by Thomas Morgan and Alfred Sturtevant ‰ Will not segregate independently ‰ Affected by crossing over

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Linkage vs Independent Assortment

Linked Genes in Drosophila ‰ Red eyes (bw+) dominant to mutant brown eyes (bw) ‰ Thin wing veins (hv+) dominant to mutant heavy wing veins (hv)

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X-Linked Genes in Drosophila Cross A ‰ Gray body (y+) dominant to mutant yellow body (y) ‰ Red eyes (w+) dominant to mutant white eyes (w) Cross B ‰ Red eyes (w+) dominant to mutant white eyes (w) ‰ Normal wings (m+) dominant to mutant miniature wings (m) ‰ Due to linkage and crossing over that occurred during meiosis ‰ Distance between linked genes is related degree of crossing over

Chromosome Mapping ‰ Determining the distances between genes and the order of sequence in a chromosome ‰ Uses the frequency of crossing ‰ The shorter the distance between linked genes, the lower the frequency of crossing-over. ‰ The longer the distance, the higher the frequency of crossing over to occur.

‰ Frequencies of crossing over: 1. Yellow, white 2. White, Miniature 3. Yellow, miniature

00.5 % 34.0 % 35.4 %

‰ 1% of crossing over = 1 map unit (centimorgan, cM)

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Distance Affects Crossover

Single Crossover

‰ 50% are recombinant gametes ‰ 50 % non-crossover gametes ‰ If distance between genes is more than 50 map units, ~100 % crossing over will occur.

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Multiple Crossover

‰ Occurrence of more than one crossovers between non-sister chromatids. ‰ Produces double crossover (DCO) gametes ‰ If the probability of crossover between A and B is 20% (0.20) and the probility of crossover between B and C is 30% (0.30), the frequency of DCO is 6 % (0.06) ‰ Product Law: (0.2)(0.3)=0.06

Three-Point Mapping ‰ The percentage of crossing over could be used to map genes in a chromosome ‰ Three criteria needed for successful mapping: ‰Genotypes of organisms producing the crossover gametes must be heterozygous for all gene loci ‰Cross must be constructed so that the genotypes of gametes could be determined based on the phenotypes of the offspring. ‰Large number of offspring must be produced

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Traits considered: 1. Body color Gray(y+) dominant to yellow (y) 2. Eye color Red eyes (w+) dominant to white (w) 3. Eye shape Normal (ec+) dominant to echinus (ec)

10,000

Determining Gene Sequence Steps: ‰ Determine 3 possible orders ‰ w-y-ec (y at the middle) ‰ y-ec- w (ec at the middle) ‰ y-w-ec (w at the middle)

‰ Perform a theoretical double cross over ‰ Compare the theoretical DCO with actual DCO (least no.) ‰ Perform theoretical NCOI and NCOII and compare with data White echinus eyes, gray body Red normal eyes, yellow body Yellow body, normal white eyes Gray body, echinus red eyes Yellow body, echinus red eyes Gray body, white normal eyes

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Unknown Gene Sequence

Total=1109

Unknown Gene Sequence

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Not all crossovers can be detected

Degree of inaccuracy increases with increasing distance between linked genes

Observed vs Expected DCO ‰ Observed DCO = double cross-over that actually occurred ‰ Example: (44 + 42)/1109 = 0.078

‰ Expected DCO = theoretical double crossovers ‰ Product of all the SCOI and SCOII ‰ Example: (82+79+44+42) / 1109 = 0.223 (200+195+44+42) / 1109 = 0.434 DCO exp= (0.223)(0.434) = 0.097

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Coefficient of Coincidence and Interference ‰ Coefficient of Coincidence (C) ‰The measure of actual DCOs that occurred ‰C = Observed DCO / Expected DCO = 0.078/0.097 = 0.804 or 80.4%

Interference (I)

‰

phenomenon when a crossover event in one region of a chromosome inhibits a second event to occur in a nearby region) I = 1-C = 1-0.804 = 0.196 or 19.6% Interpretation: 19.6% of expected DCO did not occur or only 80.4% of expected DCO was observed

Problem 1 A stock of corn homozygous for the recessive linked genes colorless (c), shrunken (sh), and waxy (wx) was crossed to a stock of homozygous for the dominant wild type alleles of the genes (+ + +). The F1 plants were then backcrossed to the homozygous recessive stock. The F2 results were as follows: Phenotype

Number

Phenotype

+++

17,959

+ + wx

4,455

c sh wx

17,699

c sh +

4,654

+ sh wx

509

+ sh +

20

c++

524

c + wx

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a.

Determine the distance between the c and sh

b.

Determine the distance between the sh and wx

c.

Determine the distance between c and wx

d.

Give the coefficient of coincidence

e.

Compute for the interference

Number

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Problem 1: Solution c sh wx

+ + +

x

c sh wx

+ + +

c sh wx + + +

NCO SCOI SCOII DCO

+ + + c sh wx + sh wx c + + + + wx c sh + + sh + c + wx Total

= = = = = = = = =

c sh wx

x

17,959 17,699 509 524 4,455 4,654 20 12 45,832

c sh wx

35,658 = 77.80 % 1,033 = 02.25 % 9,109 = 19.87 % 32

= 00.07 %

Problem 1: Solution

Distance between c and sh = (509 + 524 + 20 +12) / 45,832 = 0.0232 or 2.32 % Distance between sh and wx = (4466 + 4654 + 20 + 12) / 45832 = 0.1994 or 19.94 % Distance between c and wx = 2.32 + 19.94 = 22.26

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Problem 1: Solution c

sh

wx

2.32 mu

19.94 mu 22.26 mu

C = (0.0007) / (0.0232)(0.1984) = 0.1521 or 15.21% I = 1-C = 1-0.1521 = 0.8479 or 84.79 %

Problem 2 In a variety of tomato plant, the mutant genes o (oblate fruit), h (hairy fruit), and c (compound inflorescence) are all located in chromosome 2. These genes are recessive to their wild type alleles round fruit, hairless and single inflorescence, respectively. A testcross mating of an F1 heterozygote for all three genes resulted in the following phenotypes: Phenotypes

Number

Phenotypes

+++

73

++c

Number 348

+h+

2

+hc

96

o++

110

o+c

2

oh+

306

ohc

63

a. Determine the sequence of the 3 genes in chromosome 2 b. Give genotypes of the homozygous parents (P1) used in making the F1 heterozygote. c. Compute for the map distances between the genes d. Give the coefficient of coincidence and interference

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Problem 2: Solution +++ ohc

NCO:

ohc

x

Inference from given data:

ohc

‰ Sequence of genes is not correct

o h + = 306

‰ One chromosome contains 2 wild type alleles while the homolog contains the 3rd wild type allele

+ + c = 348 DCO:

+h+ = 2 o+c = 2

Three possible orders of the genes involved: ‰o h c

oh+ ++c +ch o++ +oh c++

Find a sequence that will satisfy both NCO and DCO

‰o c h ‰h o c

Satisfies NCO but not DCO Satisfies DCO but not NCO Satisfies both NCO and DCO

Problem 2: Solution +oh c++

NCO: DCO: SCO I: SCOII:

x

oh+ ++c +h+ o+c

ohc

Try if the sequence can satisfy the SCOs

ohc

(same as + o h) (same as c + +) (same as + + h) (same as c o +)

+++ o h c (same as c o h) o + + (same as + o +) + h c (same as c + h)

= 306 = 348 = 2 = 2 = = = =

73 63 110 96

654 = 0.654 or 65.4% 4 = 0.004 or 0.4% 136 = 0.136 or 13.6% 206 = 0.206 or 20.6%

Total = 1,000

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Problem 1: Solution

Distance between c and o = (73 + 63 + 2 + 2) / 1,000 = 0.140 or 14 % / cM Distance between o and h = (110 + 96 + 20 + 12) / 1000 = 0.210 or 21 % / cM Distance between c and wx = 14 + 21 = 35 cM

Problem 1: Solution c

o

h

14 cM

21.0 cM 35 cM

C = (0.004) / (0.14)(0.21) = 0.1361 or 13.61% I = 1-C = 1-0.1361 = 0.8639 or 86.39 %

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