Transcript 2/4

•Exam #1 W 2/11 at 7:30-9pm
in BUR 106
•Bonus posted
Phenotype
Genotype
Fig 13.5
The
inheritance
of genes on
different
chromosomes is
independent.
Fig 13.5
Approximate position of seed color and shape genes
in peas
Y
y
Gene for seed color
Fig 13.8
r
Chrom. 1/7
R
Chrom. 7/7
Gene for
seed shape
The inheritance of genes on different
chromosomes is independent:
independent assortment
Fig 13.8
Fig 13.8
meiosis I
meiosis II
The
inheritance of
genes on
different
chromosomes
is
independent:
independent
assortment
Fig 13.8
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.
Huntington’s Disease
Mom = dd
D=disease
d=normal
Dad = Dd
Dad
D or d
d Dd
Mom or
d Dd
dd
dd
possible offspring
50% Huntington’s
50% Normal
Two different people:
One with Huntington’s disease = Dd Hh
One without Huntington’s disease = dd Hh
mate. 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 Hh
One without Huntington’s disease = dd Hh
mate. What is the probability that their offspring will have
Huntington’s disease and sickle cell anemia?
Probability of each outcome:
Probability of Dd (Ddxdd) = .5
Probability of hh (HhxHh) = .25
Dd hh
Two people:
One with Huntington’s disease = Dd Hh
One without Huntington’s disease = dd Hh
mate. 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
Multiply both probabilities .25 X.5 = 12.5%
chance Dd hh offspring
Fig 13.5
Tracking two
separate
genes, for
two separate
traits, each
with two
alleles.
Ratio of
9:3:3:1
Fig 13.13
Some crosses
do not give
the expected
results
CB 15.5
Heterozygous wild type Homozygous wild type
gray w/ normal wings black w/vestigial wings
b+ b vg+ vg
b b vg vg
Fig
13.13
=25%
42% 41%
9%
8%
Does this show recombination?
D/d
M1/M2
d/d
M1/M2
D/d
M1/M2
d/d
M2/M2
D/d
M2/M2
d/d
M2/M2
Does this show recombination?
D/d
d/d
D/d
M1/M2
M1/M2
M1/M2
arental
ecomb.
d/d
M2/M2
D/d
M2/M2
d/d
M2/M2
Why fewer recombinants than parentals?
Fig
13.13
=25%
42% 41%
9%
8%
Fig 13.14
These two genes are on the same chromosome
Fig 13.14
These two genes are on the same chromosome,
and close together.
Fig 13.15
Homologous
pair of chromosomes
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.
Fig 13.16
Linkage map of
Drosophila
chromosome 2
Only 2 of the 4 chromosomes can cross-over.
Fig 13.14
Recombinants
Fig 13.16
Linkage map of
Drosophila
chromosome 2
Recombination is not
completely random.
physical
distance
Yeast chromosome 3
linkage
map
A single gene with
2 alleles only has a
few phenotypes
Traits coded for by
multiple genes have
a variety of
Fig 13.19
phenotypes
Height of males at Conn. Ag. College in 1914
Wheat color shows wide variation...
Fig 13.20
Fig 13.20
...and is coded for by three genes.
•Exam #1 W 2/11 at 7:30-9pm
in BUR 106
•Bonus posted