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Diploid Mapping. Linked Genes: Genes that are part of the SAME chromosome are said to be LINKED. Genes that are linked will not independently assort and.

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Presentation on theme: "Diploid Mapping. Linked Genes: Genes that are part of the SAME chromosome are said to be LINKED. Genes that are linked will not independently assort and."— Presentation transcript:

1 Diploid Mapping

2 Linked Genes: Genes that are part of the SAME chromosome are said to be LINKED. Genes that are linked will not independently assort and segregate, since these Mendelian laws apply only to genes located on different chromosomes (unlinked genes).

3 Unlinked Genes: 2 unlinked genes will yield 4 possible gametes in equal proportions because of independent assortment. Example: AB AB ab ab Meiosis A A a a B b B b (1/4) (1/4) (1/4) (1/4)

4 3 Ways to Notate Linked Genes: A B or A B or A B

5 Linked Genes: (No Cross-Overs) 2 linked genes will only yield 2 possible gametes because they did not independently assort. Example: A Ba b Meiosis A B A B a b a b (1/2) (1/2)

6 Genes that reside on the SAME chromosome can undergo Recombination during the crossing-over event that occurs during Prophase I of Meiosis I. 1 Cross over = Single Cross Over (SCO) A BAB AB Parental Type Gametes SCO Meiosis aB ab Cross-Over Gametes abAbabAb abab

7 2 Cross over = Double Cross Over (DCO) ABC (Parental) A B C aBc A B C (DCO) AbCAbC DCO Meiosis Ab c (SCO 1) a b c aB C a b c ABcABc (SCO 2) abCabC abc (Parental)

8 Max. # of Cross-overs: 2 Linked Genes 1 Cross-over (SCO) 3 Linked Genes2 Cross-overs (DCO)

9 Test Cross: Heterozygote (F1) X Homozygote Recessive Purpose:  To Determine Linkage  If genes are linked, allows mapping of these genes. 2 Genes = 2 point test cross 3 Genes = 3 point test cross

10 Mapping Procedure: 1. Prepare Trihybrid 2. Testcross to Homozygous Recessive 3. Count Progeny and Place in Classes or Categories: a.) Parentals – highest numbers b.) SCO1 c.) SCO2intermediate numbers d.) DCO – lowest numbers 4. Determine Order of Genes 5. Determine Distances between Genes: 1 st to Middle 2 nd to Last 1 st to Last 6. Determine the Expected Number of DCOs 7. Calculate the Coefficient of Coincidence (C) 8. Calculate Interference (I)

11 Interference (I): If: I = 1.0  Interference is Complete: No DCOs occurred I = + number  (+) interference: Fewer Observed DCOs than Expected DCOs I = - number  (-) interference: More Observed DCOs than Expected DCOs

12 More on Interference:   (+) interference is most observed   A cross-over event in one region of a chromosome inhibits a second cross-over in neighboring regions of the chromosome   The distance between genes is directly proportional to the frequency of cross-over.   The closer genes are to one another, the more (+) interference is observed (the lower the C number). This may be due to mechanical stress that is imposed on chromatids during cross-over (one chiasma inhibits formation of a second chiasma).

13 2 Point Test Cross: Prepare Dihybrid Cross: P1: P LXpl P = Purple L = LongP Lpl l = Round p = Red (Purple, Long) (Red, Round) 2 point test cross: P LXpl p lpl ( Heterozygote F1) ( Homozygous Recessive) PhenotypesNumber purple, long……………………………………..387 red, round……………………………………….413 P red, long…………………………………………..96 purple, round…………………………………..104 SCO 1000 Distance: p to l = X 100 = 20% or 20 map units (mu) 1000 Chromosome Map: p 20 mu l

14 3 Point Test Cross: P1: + + +Xe k ste = ebony body + + +e k st st = scarlet colored eyes k = kidney shaped eyes 3 point test cross: + + +Xe kst e k ste kst PhenotypesNumber wild type…………………………………………370 ebony, scarlet, kidney………………………360 P ebony, kidney…………………….……………104 scarlet………………………………..…………….98 SCO1 ebony………………………………….…………..28 kidney, scarlet……………………….……….…32 SCO2 kidney…………………………………..…………..5 ebony, scarlet…………………………..………..3 DCO 1000

15 Distance: NOTE: 1 of the 3 combination is the summation of the other 2 combinations. e to k = X 100 = 6.8% or 6.8 mu 1000 k to st = X 100 = 21% or 21 mu 1000 e to st = X 100 = 27.8 % or 27.8 mu 1000 Chromosome Map: (Order of Genes & Distances) e 6.8muk 21must | 27.8mu| Expected Number of DCOs: (freq. of SCO1 X freq. of SCO2 X Total # of Progeny) = X 0.21 X 1000 = Coefficient of Coincidence (C): (Actual number of DCOs ÷ Expected # of DCOs) C = 8 ÷ C = 0.56 Interference (I): (1.0 – Coefficient of Coincidence) I = 1.0 – 0.56 I = 0.44 I = (+) interference  Fewer DCOs than Expected

16 Wild Type Test Cross Progeny Black,Vestigial, Brown

17 Calculate the distances between these 3 genes, draw a chromosome map, determine the coefficient of coincidence, and interference for BOTH class total and grand total (all genetics lab sections). NOTE: bl = black body, vg = vestigial wings, br = brown eyes Parentals SCO1 SCO2 DCO

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