•Exam #1 W 2/11 at 7:30-9pmin BUR 106
•Bonus posted
Phenotype
Genotype
Fig 13.5
Fig 13.5The inheritance of genes on different chromo-somes is independent.
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
Fig 13.8The inheritance of genes on different chromosomes is independent:independent assortment
Fig 13.8
meiosis I
meiosis II
Fig 13.8The inheritance of genes on different chromosomes is independent:independent assortment
Fig 13.5
Inheritance can be predicted by probability
Probability of a 4= 1/6
Probability of two 4’s in a row=1/6x1/6=1/36
Probability of 3 or 4 = 1/6+1/6= 1/3
“and” multiply
“or” add
Huntington’s Disease
D=disease
d=normal
Neurological disease, symptoms begin around 40 years old.
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
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)
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
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
Tracking two separate genes, for two separate traits, each with two alleles.
Ratio of 9:3:3:1
Fig 13.5
Some crosses do not give the expected results
Fig 13.13
CB 15.5
Heterozygous wild typegray w/ normal wings
b+ b vg+ vg
Homozygous wild typeblack w/vestigial wings
b b vg vg
=25%
8%9%41%42%
Fig13.13
Does this show recombination?
D/dM1/M2
d/dM1/M2
D/dM1/M2
d/dM2/M2
D/dM2/M2
d/dM2/M2
Does this show recombination?
D/dM1/M2
d/dM1/M2
D/dM1/M2
d/dM2/M2
D/dM2/M2
d/dM2/M2
arental ecomb.
=25%
8%9%41%42%
Fig13.13
Why fewer recombinants than parentals?
These two genes are on the same chromosome
Fig 13.14
These two genes are on the same chromosome,and close together.
Fig 13.14
Homologouspair of chromosomes
Fig 13.15
Fig 13.13
By comparing recombination frequencies, a linkage map can be constructed
= ? m.u.
By comparing recombination frequencies, a linkage map can be constructed
= 17 m.u.
Linkage map of Drosophila chromosome 2
Fig 13.16
Only 2 of the 4 chromosomes can cross-over.Fig 13.14
Recombinants
Linkage map of Drosophila chromosome 2
Fig 13.16
Yeast chromosome 3
physical distance linkage
map
Recombination is not completely random.
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
Wheat color shows wide variation...Fig 13.20
...and is coded for by three genes.Fig 13.20
•Exam #1 W 2/11 at 7:30-9pmin BUR 106
•Bonus posted