Probability
and
The Product Rule
-
For
independent
both
occurring
⇐
For
what
a
=
:
events
(X
cross
are
Genetics
X
and Y )
between
the
and
odds
is
2
of
Y
( Aa)
heterozygous
n.÷
fm/aa-:
ty
probability
CP )
of
PW.rs/y)-
getting
↳
the
,
an
aa
individuals
individual ?
Pwbe=
£ £
'
=
¥
I
,
them
The
Rule
mutually
For
occur
⇐
/✗
For
what
or
a
=
exclusive
Y)
is
cross
is the
events
④
Y
and
X
,
the
probability
that
one
will
Plx)+P
between
probability
2
of
-0*4*-8
⑨
?⃝
-
Sum
aa
heterozygous
getting
①
④
③
1
1
dominant
a
A
1
A
a
( Aa)
individuals
genotype (AIAA) ?
allele (mom)
allele
allele
( mom)
( mom)
f- t.tt
+
,
,
1 A allele
+
t
t
=
1
a
1 A
allele
allele
(Dod )
( Dad )
( Dad)
Applying Probability
⇐
we
the
①
②
breed
#
whats
whats
of
the
the
2
Rules
dogs
B_bCc
pub
prob
.
to
with
genotype
puppies
of
of
Dihybrid
among
getting
a
getting
a
crosses
Bbcc
the
.
__
How
can
we
offspring ?
Bb
genotype
?
genotype ?
Cc
t.io#:: kd:--:o:2-y-2-4=12
¥
(prob
of
Bb)( prob
⇐ Ksi
⑤
)
of Cc
predict
-
Polydactyly
that
Assume
foot
-
-
expressed
is
a
marries
Having
second
child
has
child
will
P
p
=
=
have
six
fingers
with
woman
digits
extra
couple
The
a
an
with
man
a
has
is
extra
with
digits
normal
.
What
dominant
the
toes
six
digits
allele
and feet
hands
is
and
number of
and/or toes
fingers
on
each
.
.
,
but
probability
couple 's
the
that their
next
.÷÷÷÷t¥:*¥÷÷÷÷
normal
allele
¥
=L
the
chance
next
allele
÷
=
a
extra
polydactyly ?
polydactyly
-
hen
by
has
each hand
on
normal
a
caused
son
individual
when
-
pp
:
P
⑧
50%
50%
pp
polydactyly
normal
/
.
Tightly
If
what
a
curled
or
heterozygous
of
percentage
Curly
=
hair
wooly
curly
by
haired
person
mes
offspring
their
C
-
caused
is
s
would be
a
dominant
a
person
erected
to
dominant
-
straight
=
C
,
recessive
Ca
✗
cc
:¥¥8}
:
in
gene
with
have
humans
straight
.
hair ,
straight hair ?
Assume
that
produced
5
When
4
the
black
would
a
black
albino
and
guinea
offspring
was
⑤
⑧
=
BB
recessive
✗
bb
0b¥:b
an
albino
guinea
pig
a
second
were
black
produced
.
guinea
pig
,
What genetic
emanation
data ?
D= dominant
b
with
offspring
these
to
crossed with
pig
.
crossed
albino
3
apply
black
,
,
black
albino
::#
:|
:#
:* :#
Albinism
,
of
lack
2
recessive
gene
what
probability
is
.
pigmentation
parents
that
with
their
in
humans
normal
next
,
results
pigmentation
child
will be
from
have
wild
an
an
type ?
autosomal
albino child
The autosomal
normal
heterozygous
probability
for
gene
finger length
condition
that their
.
brachydactyly
Assume
is
first
married
child
that
to
will
a
,
short
female
a
man
have
with
fingers
with
normal
brachydacty.ly ?
,
is
dominant
brachydactyly
fingers
.
to
in
What is
the
the
A
certain
rare
a
deaf
type
autosomal
woman
,
congenital
of
recessive
could
gene
some
of
.
In
the
deafness
a
mating
children
in
humans
involving
have
is
a
normal
caused
deaf
man
hearing ?
by
a
and
How
with
many
the
kinds
genotype
of
gametes
Ppcc
TTRR
will
?
be
expected
from
an
individual
Two
F,
organisms
that
is
independently
,
AABBCCDDEE
self
-
fertilized
assorting
.
If
alleles
and
the
.
How
aabbccddee
capital
many
letters
are
,
mated
represent
different
to
produce
dominant
genotypes
will
an
,
occur
in
Fa ?
Assuming
the
cross
independent
AaBbCcDd
assortment
✗
,
what
Aabbccdd
will
proportion
have
of
the
offspring
aabbccdd
of
genotype?
Albinism
,
a)
b)
c)
child
albino
what
chat
what
is
is
is
of
recessive
autosomal
an
lack
the
the
the
pigmentation
gene
(a)
.
Two
in
humans
,
with
parents
from
results
normal
an
pigmentation
have
.
probability
probability
probability
that their
that
that
next
next
next
three
three
child
children
children
will
will
will
be
be
be
an
albino
albino
girl ?
?
wild-type ?