2/3/2012 Genetics

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Test 1 on February 13 over 2, 3,4, & 5

Gene Interaction

-Most phenotypic traits such as height, weight, and pigmentation are the result of many genes interacting.

-Mendel was very fortunate to have picked single gene influenced traits.

Two ZGene Interaction

(same results) 9:3:3:1

(diference: interaction of 2 genes influencing a single trait

R rose comb is dominant to r

P pea comb is dominant to p

R and P are codominant (walnut comb)

rrpp produce a single comb

Bateson and Punnet bred rose comb (RRpp) to a pea comb (rrPP). All the F1 offspring had a walnut comb (RrPp).

RP Rp rP rp
RP
RP
rP
rp

Glucose metabolism

-Nearly a dozen enzymes interacting in this single pathway resulting in a single phentypic product

-Each enzyme the result of the expression of one or multiple genes.

Epistasis (simplified example)

-Alleles of one gene masks the phenotypic effects of the alleles of another gene.

Can have two or more genes coding for different protein in an enzymatic pathway producing a single product.

Colorless precursor > Enzyme C > Colorless Intermediate > Enzyme P > Purple Pigment

-If one piece falls out, the whole process falls down

Epistatic Genes

C (enzyme C) dominant to c

P (enzyme P) dominant to p

cc or pp mask the P or C alleles producing white

CCpp (variety 1 white) x ccPP (variety 2 white) = CcPp all purple

F1 Cross

CP Cp cP cp
CP CCPP CCPp CcPP CcPp
Cp CCPp CCpp CcPp Ccpp
cP CcPP CcPp ccPP ccPp
cp CcPp Ccpp ccPp ccpp

Ch 5 p101-117

Linkage and Genetic Mapping in Eukaryotes

-Thousands of different genes, but relatively few chromosomes

-Genes located to close together on the same chromosome violates the law of independent assortment

Gene Mapping

-Any method used to determine the relative positions of genes on a chromosome and the distance between them.

-Gene mapping allows us to determine if genes are “linked”

“linkage” is used in two ways

-Two or more genes located on the same chromosome [i.e. linkage group]

-*Genes that are close together on the same chromosome that tend to be transmitted as a unit are also referred to as linked.

Crossing Over

Genes of the smae linkage group (if they are located far apart) can independently assort due to crossing over during meiosis I

Note: gene A & B assorted independently of C despite being on the same chromomsome

pastedGraphic.pdf

Haploid cells with new allele combination are called nonparentals or recombinants vs. those having same combination as parents called parentals or non-recombinants

The phenomenon of crossing over was discovered when crosses failed to follow simple Mendelian patterns of inheritance

Genetic Linkage Mapping

-The closer two genes are on a chromosome, the less likely they will become separated during a “crossing over” event

-Thus: *there is a correlation between the number of recombinant gametes and the distance between two genes on the same chromosome.*

Which genotype would you expect to find more often among gametes from this heterozygote individual?

A)RBh*

  1. Rbh

If you did a dihybrid crosses involving genes R&B and a separate dihybrid cross involving genes B&H, which cross would produce more recombinant offspring?

Expected ration in F2 generation of a dihybrid cross; if the alleles independently assorted

PL Pl pL pl
PL PPLL PPLl

Unexpected results

-Bateson and Punnet’s dihybrid cross in F2 Generation

Expected a 9:3:3:1 ratio, observed a 15.6:1.0:1.4:4.5 ration (offspring had a much greater proportion of the parental phenotypes)

Explanation: these two genes must not be assorting independently.  These two genes must be linked together (assorting as a unit)

But how did they know that the difference in the rations they observed and expected was not due to sampling error?

Chi Square Analysis

Hypothesis must be stated in the negative; hence, Null Hypothesis (H0).

There is no difference between the observed ratios and the expected ratios.

P=<5% reject     P=>5% fail to reject

Bateson and Punnet’s Offspring P<1% thus rejected H0

Genetic Linkage Map

The percentage of recombinant offsprings correlated to the distance between 2 genes

Map Distance

Map distance=Number of recombinant offspring/total number of offspring x 100

Units of distance are called map units (mu) or centiMorgans (cM)

One map unit is equivalent to a 1% recombination frequency

Figuring distance between A and B

AaBb x aabb

Crossing over in the case of this homozygous individual could not produce a recombinant gamete

see notebook

What would be the linear order of these three genes given the following map distances?

Map distance between Sh & bz = 2.0cM

Map distance between C & bz = 5.0cM

Map Distance between C & Sh = 3.0cM

Caution: as percentage of recombinant offspring approaches a value of 50%, this value becomes a progressively less accurate measurement of map distance.  This is due to multiple crosses in regions between genes that are far apart.

pastedGraphic_1.pdf

5 chiasmata in a grasshopper tetrad.

Posted on February 3, 2012, in Genetics. Bookmark the permalink. Leave a comment.

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