Lecture 12 Gene Linkage and Genetic Mapping 2 PDF

Preview

Summary

Gene linkage and Genetic mapping...

Description

Lecture 12 Gene Linkage and Genetic Mapping

2/26/19

Genetic Linkage  Mendels principles of segregation and independent assortment are based on separation of homolog chromosome during anaphase I of meiosis.  Some genes may be located on the same chromosome there are inherited together. o Linked genes o Altered progeny ratio from the expected 9:2:2:1 segregation ratios. o Ratio differs if genes are on the same chromosome and how close those genes are on the same chromosome. (tightly linked means they are close).

Independent assortment we have 4 options (AB)(Ab)(aB)(ab) But for gene linkage we only have 2 options (AB) and (ab) Linkage symbolism: 3 ways to show linkage 1. Two lines: a. A____B _______ a. b This is conventional with independent assortment. Heterozygous for Gene A and B 2. One line: a. A_____B

a b 3. A Slash(/): a. aB/Ab Linkage Phases: Repulsion and Coupling - Coupling (cis): The recessive alleles of both genes are on one chromosome and the dominant alleles are on the corresponding homolog. - Repulsion (trans):A recessive allele of a gene and a dominant allele of another gene are on one chromosome and the corresponding dominant recessive alleles are on the homolog. - Complete linkage exists between two genes ONLY if located very close together on the same chromosome. - IF linkage exists and organisms heterozygous at both loci are mated an F2 phenotype ratio results that is unique. Ex: Brown eye (bw) and heavy wings (hv), normal type are red eyes (bw+) and thin wings (hv+) Parent 1: bw____hv+ Parent 2: bw+_____hv bw hv+ bw+ hv F1: bw_____hv+ bw+ hv F1 is in repulsion (trans) phase Selfing the F1: bw____hv+ X bw____hv+ bw+ hv bw+ hv F2: bw_____hv+ bw hv+

bw____hv+ bw+ hv

bw____hv+ bw+ hv.

bw+____hv bw+ hv

1:2:1 genotypic and phenotypic ratios ¼ - brown eyes and thing veins `/2 red eyes and thin veins (wild type) ¼ red eyes and heavy veins These ratios are characteristics of complete linkage. Linkage Group: the set of genes present together in a chromosome. One linkage group corresponds to a pair of homolog chromosomes. Humans -> 23 pairs of chromosomes 23 linkage groups.

Recombination

 It’s Not likely that two genes will be so close on a chromosome that they will demonstrate complete linkage.  Typically crosses involving two genes on a chromosome will result in a percentage of offspring produced form recombinant gametes. o Recombination: the process by which offspring derive a combination of genes different from that of either parent o Results from crossing over o Recombinant individuals is related to the distance between the two genes along a chromosome. Two parental gametes Two recombinant gametes AaBb A____B AB. a b Ab AB aB ab ab Recombination Frequency (RF): the percent of total progeny that re recombinants. RF% = (recombinants / total progeny) *100. RF is not affected by the linkage phase (coupling or repulsion) Linkage map: shows linear ordr and relative distance of genes along a chromosome No branches or discontinuities Linear structure of DNA Ex: the order of three genes on a chromosome: Determine the recombination frequency for each two gene combinations:  Yellow, white 0.5%  White, Miniature 34.5%  Yellow Miniature 35.4%  Gene combination with a low recombination frequency are closer than gneese with higher recombination frequency.  Gene order: yellow-white-miniature Map Distance - Map Unit: the distance between two genes that result in 1% RF o One map unit is also called a centimorgan (cM) o 1cM = 1% RCF -= 1 map unit o So the distances for yellow – white – miniature = Y->m = 35.4 and w> m = 34.5 so distance between y and w is .5 Genetic distance and Physical Distance

- Geentic map indicated the recombination frequency between genes. - It is not necessarily the actual physical distance between these genes on the chromosome. - Genetic distance between genes is generally correlated with physical distance on the chromosome. ***Genetic Distance - Experimental recombination frequencies between two genes are never greater than 50% Exam Question!! AaBb when we reach 50% we cannot tell which genes are independent or closely linked together. Max amount of observable recombination is 50% Ex. 2 genes are separated by 20 map units. Each gene is 10 map units and crossing over does not occur 80% of the time. - A RF near 50% suggests 1. The two genes are on different chromosome 2. The two genes lie far apart on the same chromosome - Crosses between two geens lying very far apart on a chromosome may show no linkage at all. Genetic Distances and Multiple crossing over. - If a double crossover involving the same chromatid, no recombination can be detected between two genes and distance would be underestimated. - Double crossing over actually reduces number of recombinance.

A

B Double crossing over creates the exact same as parents AB or ab.

a

b

Multiple crossovers - Even number : no recombination - Odd numbers: recombination - Need to follow 3 genes to study a double exchange. Three-Point Mapping  Analysis of three linked genes. o Determiantion of gene order

o Based on recombination frequency. (use linkage symbolism) ABC x abc abc abc 3 requirements for mapping 1. One parent must be heterozygous at all loci 2. The gametic genotypes produced must be apparent from observing the phenotypes of the offspring. 3. Sufficient number of offspring must be produced Ex. three point cross in corn - Growth habit: Normal (lz+) vs lazy (lz) - Leaf surface: normal (gl+) vs glossy (gl) - Endosperm: normal (su+) vs sugary (su) Objective: determine the order and genetic distance among three genes. - Parent 1: heterozygous for all loci (lz+,gl+,su+/lz,gl,su) - Parent 2 Homozygous recessive (lz,gl,su/lz,gl,su) - Out of total number of 740: o Normal for all traits (lz+ gl+ su+) = 286 o Lazy, normal, normal, (lz gl+ su+) =33 o Recessive for all genes (lz gl su) = 272 most similar to normal trait. - Parental genotypes are represented by the two classes with the highest number of progeny. - These are non- recombinant individuals o No crossing over occurred. o Normal for the three traits – lz+ gl+ su+/lz gl su Determining gene order - RF between lz and gl loci (ignore su) o Parental types for these two genes were: lz+ gl+ lz gl o Lz+ gl: 59+40=99 o Lz gl+: 33+44=77 o Go back to table and look at these new recombiantions. Looking for all lz+ and gl adding together and the same with lz gl+ o RF = (99+77)/740 = .2378 = 23.78% - Now RF between lz and su loci (ignore gl) - Recombinant types are o Lz+ su: 4+40=44 o Lz su+ 33+2=35 o RF = (44+35)/740 = .1067 or 10.67% - Lastly gl and su loci (ignore lz) o gl+ su: 4+44=48

-

-

-

  

 

o gl su+: 59+2=61 o RF = (48+61)/740= .1473 or 14.73% o Genetic distance between gl and su is 14.73 cM Determining Gene Order: o Since lz and gl has the largest RF value they are the furthest apart, and su must be between them. o Lz-su+su-gl=lz-gl -> 10.67+14.73=25.4 lz_________su_________gl ---10.67cM-- ---14.73cM-------------23.78cM------------The calculated distance (23.78cM) is shorter than the added one (25.4 cM) This is due to double crossing over that occurs between two genes. Results in less observed recombination suggesting that genes are closer together than they actually are. The probability of two crossovers occurring in a region is much lower than the probability of a single event. Double recombination genotypes o Represented by the classes with the fewest number of progeny. Use to determine order. o Since recombination of the su gene represents the least abundant genotype it must lie between lz and gl o The order of the genes must be lz su gl or gl su lz Double recombinant classes in our example slide 26: the 2 and 4 is from double crossing over. Double cross overs represented by fewest progeny and the gene out of place is the one that is in the middle of the actual order. The one out of place is the Gene in the middle...