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Heredity, Gene Regulation, and Development I. Mendel's Contributions II. Meiosis and the Chromosomal Theory III. Allelic, Genic, and Environmental Interactions IV. Sex Determination and Sex Linkage V. Linkage A. Overview. Independent Assortment. A. A. b. B. a. a. B. b. AB. ab. - PowerPoint PPT Presentation
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Heredity, Gene Regulation, and Development
I. Mendel's ContributionsII. Meiosis and the Chromosomal TheoryIII. Allelic, Genic, and Environmental InteractionsIV. Sex Determination and Sex LinkageV. Linkage
A. Overview
A aA a
b BB b
AB ab Ab aB
Independent Assortment
V. Linkage
A. Overview
A aA a
b BB b
AB ab Ab aB
Independent Assortment
A a
B b
Linkage
AB ab
V. Linkage
A. OverviewLinkage
A a
B bAB ab
A a
B bAB ab
In Prophase I of Meiosis – Crossing-over
A a
b B
Ab aB
X
AABB aabbAB
AB
ab
ab
V. Linkage
A.OverviewB.Complete Linkage
Test Cross
- if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene.
X
AABB aabbAB
AB
ab
ab
AB abGametes
AB
abF1
V. Linkage
A.OverviewB.Complete Linkage
- if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene.
X
AB
ab
Gametes
AB
ab ab
ab
F1 x F1
ab
V. Linkage
A.OverviewB.Complete Linkage
- if genes are immediate neighbors, they are almost never separated by crossing over and are ‘always’ inherited together. The pattern mimics that of a single gene.
F1 x F1 X
AB
ab
Gametes
AB
ab ab
ab
ab
AaBb
aabb
1:1 ratio A:a
1:1 ratio B:b
1:1 ratio AB:abNOT 1:1:1:1
V. Linkage
A.OverviewB.Complete Linkage
Phenotypes
AB
ab
aB ?
Ab ?
C. Incomplete Linkage
B
a b
A b
a b
a
C. Incomplete Linkage
- So, since crossing-over is rare (in a particular region), most of the time it WON’T occur and the homologous chromosomes will be passed to gametes with these genes in their original combination…these gametes are the ‘parental types’ and they should be the most common types of gametes produced.
B
a b
b
a b
a
A B
a b
A
C. Incomplete Linkage
- But during Prophase I, homologous chromosomes can exchange pieces of DNA.
- This “Crossing over” creates new combinations of genes…
These are the ‘recombinant types’
B
a b
b
a b
a
A B
a b
A
a B
A b
C. Incomplete Linkage
As the other parent only contributed recessive alleles, the phenotype of the offspring is determined by the gamete received from the heterozygote…
B
a b
b
a b
a
A B
a b
A
a B
A b
gamete genotype phenotype
ab aabb ab
ab AaBb AB
ab aaBb aB
ab Aabb Ab
LOTS of these
FEW of these
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabbV. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross - determine expectations under the
hypothesis of independent assortment
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
B b Row Total
A 43 12
a 8 37
Col. Total
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross - determine expectations under the
hypothesis of independent assortment
Easy with a 2 x 2 contingency table
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
B b Row Total
A 43 12 55
a 8 37 45
Col. Total
51 49 100
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross - determine expectations under the
hypothesis of independent assortment
Easy with a 2 x 2 contingency table
Compute Row, Columns, and Grand Totals
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
B Exp. b Row Total
A 43 28 12 55
a 8 37 45
Col. Total
51 49 100
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross - determine expectations under the
hypothesis of independent assortment
Easy with a 2 x 2 contingency table
Compute Row, Column, and Grand Totals
E = (RT x CT)/GT
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total
51 49 100
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabbV. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- test cross - determine expectations under the
hypothesis of independent assortment
Easy with a 2 x 2 contingency table
Compute Row, Column, and Grand Totals
E = (RT x CT)/GT
Phenotype Obs Exp (o-e) (o-e)2/e
AB 43 28 15 8.04
Ab 12 27 -15 8.33
aB 8 23 -15 9.78
ab 37 22 15 10.23
X2 = 36.38
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total
51 49 100
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
- Chi-Square Test of Independence
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabbV. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
2. Detemining the arrangement of alleles in the F1 individual; which alleles are paired on each homolog?
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
A B
a b
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
2. Detemining the arrangement of alleles in the F1 individual; which alleles are paired on each homolog?
- most abundant types are ‘parental types’
A B
a b
a B
A b
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
2. Detemining the arrangement of alleles in the F1 individual; which alleles are paired on each homolog?
- most abundant types are ‘parental types’ - least abundant are products of crossing-over: ‘recombinant types’
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
A B
a b
20 map units
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage
1. Determining if the genes are linked, or are assorting independently:
2. Detemining the arrangement of alleles in the F1 individual; which alleles are paired on each homolog?
3. Determining the distance between loci:
Add the recombinant types and divide by total offspring; this is the percentage of recombinant types. Multiply by 100 (to clear the decimal) and this is the index of distance, in ‘map units’ or centiMorgans.
20/100 = 0.20 x100 = 20.0 centiMorgans
V. Linkage
A.OverviewB.Complete LinkageC.Incomplete Linkage D.Summary
- by studying the combined patterns of heredity among linked genes, linkage maps can be created that show the relative positions of genes on chromosomes.
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