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Lectures
posted online
after lecture.
Textbook
sections
and/or pages
posted a few
days prior to
each lecture.
CB 14.3
In many
instances there
is a unique
pattern of
inheritance.
Traits
disappear and
reappear in
new ratios.
CB 14.6
Genotype
Phenotype
Human blood types
AA or
AO
BB or
BO
AB
OO
CB tbl 14.2
Looking at the past:
If Frank has B blood type,
his Dad has A blood type,
And his Mom has B blood type…
Should Frank be worried?
possible
genotypes
Mom=B blood
Dad=A blood
BB or BO
AA or AO
Gametes all B / 50% B and all A / 50% A and
50% O
50% O
Frank can be BO
= B blood
…no worries
We can also
predict the
future
Fig 2.12
Inheritance of blood types
Mom = AB
Dad = AB
Inheritance of blood types
Mom = AB
Gametes:
A or B
Dad = AB
A or B
Inheritance of blood types
Mom = AB
Gametes:
A or B
A or B
Dad
A or B
A AA
Mom or
B AB
Dad = AB
AB
BB
Chance of each
phenotype for
each offspring
25% AA
50% AB
25% BB
Testcross:
determining
dominant/
recessive and
zygosity
CB
14.7
Sickle-cell anemia is caused by a point mutation
CB
5.21
Sickled and normal red blood cells
Sickle-Cell Anemia:
A dominant or recessive allele?
Mom = HS
Dad
H or S
H HH
Mom or
S HS
HS
SS
S=sickle-cell
H=normal
Dad = HS
possible offspring
75% Normal
25% Sickle-cell
Coincidence of malaria
and sickle-cell anemia
CB 23.13
Sickle-Cell Anemia:
A dominant or recessive allele?
Mom = HS
Dad
H or S
H HH
Mom or
S HS
HS
SS
S=sickle-cell
H=normal
Dad = HS
possible offspring
Oxygen transport:
75% Normal
25% Sickle-cell
Malaria resistance:
75% resistant
25% susceptible
Variation in Peas
Phenotype
Genotype
CB 14.8
The
inheritance
of genes on
different
chromosomes is
independent.
CB 14.8
Approximate position of seed color and shape genes
in peas
Y
y
Gene for seed color
r
Chrom. 1/7
R
Chrom. 7/7
Gene for
seed shape
The inheritance of genes on different
chromosomes is independent:
independent assortment
CB 15.2
CB 15.2
meiosis I
meiosis II
The
inheritance of
genes on
different
chromosomes
is
independent:
independent
assortment
CB 15.2
CB 14.8
CB 14.9
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
Dad = Dd
Dad
D or d
d Dd
Mom or
d Dd
D=disease
d=normal
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
Many traits are
coded for by
more than one
gene.
CB 14.11
Eye color: One trait controlled by multiple genes
Lectures
posted online
after lecture.
Textbook
sections
and/or pages
posted a few
days prior to
each lecture.