exam #1 w 2/11 at 7:30-9pm in bur 106 bonus posted

•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