Exam #1 W 2/11 at 7:30-9pm in BUR 106 Bonus posted
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Transcript of Exam #1 W 2/11 at 7:30-9pm in BUR 106 Bonus posted
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•Exam #1 W 2/11 at 7:30-9pmin BUR 106
•Bonus posted
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Phenotype
Genotype
Fig 13.5
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Fig 13.5The inheritance of genes on different chromo-somes is independent.
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Y y
r R
Gene for seed color
Gene for seed shape
Approximate position of seed color and shape genes in peas
Chrom. 1/7 Chrom. 7/7
Fig 13.8
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Fig 13.8The inheritance of genes on different chromosomes is independent:independent assortment
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Fig 13.8
meiosis I
meiosis II
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Fig 13.8The inheritance of genes on different chromosomes is independent:independent assortment
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Fig 13.5
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Inheritance can be predicted by probability
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Probability of a 4= 1/6
Probability of two 4’s in a row=1/6x1/6=1/36
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Probability of 3 or 4 = 1/6+1/6= 1/3
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“and” multiply
“or” add
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Huntington’s Disease
D=disease
d=normal
Neurological disease, symptoms begin around 40 years old.
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Mom = dd Dad = Dd
d or d
D or d
Dd
Dd dd
ddpossible offspring50% Huntington’s50% Normal
Mom
Dad
Huntington’s Disease D=disease
d=normal
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Two different people:One with Huntington’s disease = Dd HhOne without Huntington’s disease = dd Hhmate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia?(Dd hh)
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Two people:One with Huntington’s disease = Dd HhOne without Huntington’s disease = dd Hhmate. What is the probability that their offspring will have
Huntington’s disease and sickle cell anemia? Dd hh
Probability of each outcome:
Probability of Dd (Ddxdd) = .5
Probability of hh (HhxHh) = .25
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Two people:One with Huntington’s disease = Dd HhOne without Huntington’s disease = dd Hhmate. What is the probability that their offspring will have
Huntington’s disease and sickle cell anemia? Dd hh
Probability of each outcome:
Probability of Dd (Ddxdd) = .5Probability of hh (HhxHh) = .25Multiply both probabilities .25 X.5 = 12.5%
chance Dd hh offspring
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Tracking two separate genes, for two separate traits, each with two alleles.
Ratio of 9:3:3:1
Fig 13.5
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Some crosses do not give the expected results
Fig 13.13
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CB 15.5
Heterozygous wild typegray w/ normal wings
b+ b vg+ vg
Homozygous wild typeblack w/vestigial wings
b b vg vg
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=25%
8%9%41%42%
Fig13.13
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Does this show recombination?
D/dM1/M2
d/dM1/M2
D/dM1/M2
d/dM2/M2
D/dM2/M2
d/dM2/M2
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Does this show recombination?
D/dM1/M2
d/dM1/M2
D/dM1/M2
d/dM2/M2
D/dM2/M2
d/dM2/M2
arental ecomb.
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=25%
8%9%41%42%
Fig13.13
Why fewer recombinants than parentals?
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These two genes are on the same chromosome
Fig 13.14
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These two genes are on the same chromosome,and close together.
Fig 13.14
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Homologouspair of chromosomes
Fig 13.15
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Fig 13.13
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By comparing recombination frequencies, a linkage map can be constructed
= ? m.u.
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By comparing recombination frequencies, a linkage map can be constructed
= 17 m.u.
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Linkage map of Drosophila chromosome 2
Fig 13.16
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Only 2 of the 4 chromosomes can cross-over.Fig 13.14
Recombinants
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Linkage map of Drosophila chromosome 2
Fig 13.16
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Yeast chromosome 3
physical distance linkage
map
Recombination is not completely random.
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Fig 13.19
A single gene with 2 alleles only has a few phenotypes
Traits coded for by multiple genes have a variety of phenotypes
Height of males at Conn. Ag. College in 1914
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Wheat color shows wide variation...Fig 13.20
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...and is coded for by three genes.Fig 13.20
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•Exam #1 W 2/11 at 7:30-9pmin BUR 106
•Bonus posted