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Fig. 15-1
The location of a particular gene can be seen by tagging isolated chromosomes with a fluorescent dye that highlights the gene
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Fig. 15-4a
EXPERIMENT
PGeneration
GenerationF1 All offspring
had red eyes
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Fig. 15-4b
RESULTS
GenerationF2
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Fig. 15-4c
EggsF1
CONCLUSION
Generation
Generation
P XX
w
Sperm
XY
+
+
++ +
EggsSperm
+
++ +
+
GenerationF2
w
w
w
ww
w
w
w
ww
w
w
w
ww
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Fig. 15-7
(a) (b) (c)
XNXN XnY XNXn XNY XNXn XnY
YXnSpermYXNSpermYXnSperm
XNXnEggs XN
XN XNXn
XNY
XNY
Eggs XN
Xn
XNXN
XnXN
XNY
XnY
Eggs XN
Xn
XNXn
XnXn
XNY
XnY
The transmission of sex linked recessive genes
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• Sex-linked genes follow specific patterns of inheritance
• For a recessive sex-linked trait to be expressed– A female needs two copies of the allele– A male needs only one copy of the allele
• Sex-linked recessive disorders are much more common in males than in females
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
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Partial Linkage
• Linkage is different from sex linkage• Linked genes tend to be inherited together
because they are located near each other on the same chromosome. Results from genes being closely linked on the same chromosome
• Linked genes in genetic experiments deviate from the results expected from Mendel’s law of independent assortment.
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Dihybrid Testcross to
Detect Independent
Assort
YYRR X yyrr
YrRr X yyrr
YR Yr yRyr yr
yyRr
Yyrr
yyrr
YyRr
Eggs
Spermyr
YR
yr
Yr
yR
Phenotypic ratio1:1:1:1
Ratio of parental:Recombinant1:1
Dihybrid
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Fig. 15-9-1
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg b+ b+ vg+ vg+
Morgan 1912
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Fig. 15-9-2
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
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Fig. 15-9-3
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
Testcrossoffspring Eggs b+ vg+ b vg b+ vg b vg+
Black-normal
Gray-vestigial
Black-vestigial
Wild type(gray-normal)
b vg
Sperm
b+ b vg+ vg b b vg vg b+ b vg vg b b vg+ vg
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Fig. 15-9-4
EXPERIMENTP Generation (homozygous)
RESULTS
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
Testcrossoffspring Eggs b+ vg+ b vg b+ vg b vg+
Black-normal
Gray-vestigial
Black-vestigial
Wild type(gray-normal)
b vg
Sperm
b+ b vg+ vg b b vg vg b+ b vg vg b b vg+ vg
PREDICTED RATIOS
If genes are located on different chromosomes:
If genes are located on the same chromosome andparental alleles are always inherited together:
1
1
1
1
1 1
0 0
965 944 206 185
:
:
:
:
:
:
:
:
:
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Fig. 15-10Testcrossparents
Replicationof chromo-somes
Gray body, normal wings(F1 dihybrid)
Black body, vestigial wings(double mutant)
Replicationof chromo-somes
b+ vg+
b+ vg+
b+ vg+
b vg
b vg
b vg
b vg
b vg
b vg
b vgb vg
b vg
b+ vg+
b+ vg
b vg+
b vg
Recombinantchromosomes
Meiosis I and II
Meiosis I
Meiosis II
b vg+b+ vgb vgb+ vg+Eggs
Testcrossoffspring
965Wild type
(gray-normal)
944Black-
vestigial
206Gray-
vestigial
185Black-
normal
b+ vg+
b vg b vg
b vg b+ vg
b vg b vg
b vg+
Sperm
b vg
Parental-type offspring Recombinant offspring
Recombinationfrequency =
391 recombinants2,300 total offspring
100 = 17%
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20% recombination
A B
a b
With crossing overA B
a bParental
a B
A b
Recombinants
40%
40%
10%
10%
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Testing for Assortment/Linkage
1. Generate a dihybrid2. Testcross the dihybrid3. Compare the % of parental to recombinants
A. If 50% parental:50% recombinant – Independent Assortment
B. If more parental than recombinant – partial linkageC. If only parental and no recombinant – complete linkage
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• The discovery of linked genes and recombination due to crossing over led Alfred Strutevant to a method of constructing genetic maps
• He assumed the farther apart genes are , the higher the probability that a cross over will happen between them and therefore the higher the recombination frequency.
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The closer the two genes are on a chromosome the fewer recombinants
Minimum = 0% recombinants
The further two genes are on a chromosome the more recombinants
Maximum = 50% recombinants
Linkage therefore can be used as a measure of genetic distance on chromosome
1 Map Unit = 1 % recombination
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Gene Order on ChromosomeB – Vg 17 MUB – Cn 9 MUVg – Cn 9.5 MU
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Partial Linkage – two genes are so close on the same chromosome that recombination occurs less than 50% of the time.
Complete Linkage – two genes on the same chromosome so close that recombination cannot separate them.
Independent Assortment – two genes on different chromosomes or two genes on the same chromosome but far enough apart that recombinant occurs 50% of the time.
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Example ProblemIn Drosophila long wings is dominant to dumpy wings and round
eyes is dominant to star eyes. A dihybrid fly was generated by mating a long wing round eye fly with a dumpy wing star eye fly. This dihybrid fly was testcrossed and the following progeny were generated.
222 long wing round eye215 dumpy wing star eye33 long wing star eye30 dumpy wing round eye
a. Are these genes completely linked or partially linked?b. What is the genetic distance between these two genes?c. How would the results have differed if the genes
independently assorted?
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Exception to chromosomal Inheritance (Organellar Genes)
• The inheritance of traits controlled by genes present in the chloroplasts or mitochondria– Depends solely on the maternal parent because
the zygote’s cytoplasm comes from the egg– Maternal Inheritance
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Pedigree Symbols
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Nuclear vs Organellar
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Human GeneticsPedigree Analysis
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Autosomal vs Sex Linked
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Multifactorial Traits
• Heart disease• Personality• IQ
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Alterations of chromosome number or structure cause some genetic disorders.
• So far we’ve seen that the phenotype can be affected by small scale changes involving individual genes
• Random mutations are the source of all new alleles, which can lead to a new phenotype.
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Abnormal chromosome #: Aneuploidy
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Human Aneuploids
• Trisomy 21• Sex chromosome
– XO – turner syndrome– XXY – klinefelters– XYY
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Abnormal chromosome numbers
• Polyploidy: Common in plant• ~70 % of flowering plants, • Banana are triploid,• Wheat 6n• Strawberries 8n
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Alterations in chromosome structure
• Meiosis errors and damaging agents such as radiation can cause breakage of the chromosome
• four types of structural damage
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Chromosome Structure
reciprocal translocation between 9 and 22(Philadelphia Chromosome)
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Disorders caused by structurally altered chromosomes
• Cri du chat – deletion in chromosome 5 • Chronic myogenous leukemia
Normal chromosome 9
Normal chromosome 22
Reciprocaltranslocation Translocated chromosome 9
Translocated chromosome 22
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Fig. 15-8
X chromosomes
Early embryo:
Allele fororange fur
Allele forblack fur
Cell division andX chromosomeinactivationTwo cell
populationsin adult cat:
Active XActive X
Inactive X
Black fur Orange fur
Barr Body – inactive X visible in interphase nucleus
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Genomic imprinting• Def: a parental effect on gene
expression• Identical alleles may have
different effects on offspring, depending on whether they arrive in the zygote via the ovum or via the sperm.
• Fragile X syndrome: higher prevalence of disorder and retardation in males