20 Basic Genetics Notes
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Biology
Date
Nov 24, 2024
Type
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20
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2 0
.1
Genetic
Terms
--liliC:il!l!!::..-::=---
--
!l!!!!!!l!!C!!:---.1!~~-=====-~
A . Alleles
C'
They
are
different.forms
of
a
gene
which occupy
th
e
sa
me
poc;,
it
io
n on
hrm
/(Jl
ogou,
chromosomes.
Example:
eye colour gene
-
_/
_
-~~
-l
L--:~
'
brown eye
_l_}
t-----i
:
allele
'
i----,
other genes
B.
Phenotype
C,
IL
refers
to
t
he
ap
pearance of a
tr
ait
.
C.
Genotype
blue eye
allele
{
It
refe
rs
to
the
Kenef
ic
make-up
or
an
organism.
93
"'
t
,;
and
c
xe1
(
:iSC
'J
N
:'
,S
8H"i<
0\
-
/1·tG
/iS
I\ {
No
e,
;
Rn
94
ul/c
/1'
.,.
E.
H
ete
ro~.ygote
I
.
.
1
·
1
c
ul
ar
ch
aracteri
s
tic
co
nlai
n
1;,
two
dif{ert
I
,,
11
ot\'\
JCO
a pa1.
c
··
111
lt
1
:--
,
\\1
l1
r
~a
ni~rn
,,
l()\C
~ l
-
a
l/
rlc
s.
F.
Hy
brid
.
.
..
1
1
. d
I
by
crossi
na
two
in
di
v
idua
ls w
hich
are
ge
ne
ti
calh
\t
,,
an
1nd1,
1c
ua
p1
o u
cec
.
-·
c
•
diffe
rent
in
o
ne
or
more
ch
arac
te
ri
s
ti
cs.
G.
Do
m;nant
allele
11
,s
the
a
ll
ele whi
ch
c
an
ex
p
ress
it
se
lf
in
h
etero::ygo
11
s
co
ndit
io
n.
H.
Dominant
character
It
is
the
phenotype
w
hi
ch
can
s
ho
w
up
in
hete
ro:,ygo
us
co
nd
iti
on;
co
nt
ro
ll
ed
by
t
he
do
mjnant
a
llel
e.
I.
Recessive
allele
It
is the
alle
le w
hic
h
effect is
mask
ed
in
het
ero::,ygo
us
co
ndition, i.
e.
,
it
c
an
o
nl
y
ex
pr
ess
it
se
lf
in
honw
::
,ygo
us
co
nditi
on.
J.
Recessive
chara
ct
er
C
It
is
th
e phen
oty
pe w
hi
ch
can
only s
how
up
in
hom
crvoo
us
co
i
1
ct
·t·
II
!
~-
o
1 10
11
:
co
n
tro
ec
by
th
e
r
ec
e
ss
i
ve
allel
e.
K.
Monohybrid
inheritance
,
It
refe
r,
to
the s
tud
y
or
lh
e i
nheritan
ce
nf
a
single
pair
of contrast
in
g character
s.
L.
Dihybrid
inher
i
tance
C
The
inh
er
itance
nf
two
pai
r
::,
or
contra-.tin
g
char;1c
1cr
s.
Basic
Genet/e-i:;
20.2
Monohybrid Inheritan
ce
~~-~-------
A.
Exa,np
le: inheritanc
1·1
e
,~
111
111
0
11
eye
c
o/o11r
Gen
otype
: h
on,
oz.ygous
Phenotype:
brown
homozygous
brown
homozy
qou
'i
blue
0)
This trait is controlled by a pair
of
alleles (two
differentforms
of
eye colour gene):
brown allele and blue allele, and therefore has two contrasting
phenotypes
:
brown
eye
and blue eye.
~
Every
person has
two alleles
(
on homologous chromosomes) to determine his/her
eye
colour
and the two alleles are
received
from his/her parents,
one allele from
each parent.
~
If
both alleles are the
same,
the person
is
said to be
homozygous.
C:,
If
both alleles are brown aJleles, the person must have brown eyes.
C)
If
both alleles are blue alleles, the person must have blue eyes.
C)
If
the two alleles are
different
(i.e. one brown and one blue), the person is said to
be
heterozygous.
C)
In
this
ca
s
e,
only
one
of
them
can
express
itself and the other one will be
ma
s
ked
.
c:,
Si
nce
a
hete
ro
zygous
person
h
as
brown eyes , the brown allele must have
expressed it
self
a
nd
it is there
fo
re
the
dominant allele
and
brown eye colour is
the
dominant phenotype·
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---~="'"""---:
Jn
~-:,~n
,v
Not,
,nil
r~
xucr,es
• , .
11
..,cll
in
the
pr
e1.ic
n
cc
of the
bro
\':n
al
lele .
(
'1
11<'
h
l1
w
:d
kk
c
,1111wt
, xpll
'·
'
"
"'Q
i<
96
_ 1
1
,
1
..
1
·
1
17
,rcl
or
c
1h
c
r
eces~
i
vc
all
ele.
The blue
eye
cr,i,
'd
h
I
I
h,
1
11
' ,
II
II ', ,
c ,
,
<111
"
c
. ,
,u
r "
tf
·
.
h·
t
, .
11
,1
tl'
l"
occ
ur
s onlv
in
homo
z
yg
ou
,
cond1t
1on.
''
n·c
e"i\
t
\'
l'
p
t•
no
~
'p
t:
,1
~
,
-
U.
Men
del's
firs
t
I.o
w
-
The
Law
of
Segregation
C
~cndcl \
hn
.
·c
d
tn
g
c:<.
pcrimc
nt
'-t
on
ga
rd
en
pe
a:
0
'1
r:
<11
,
tuJi
cd : !lo
we
r
co
lo
ur
!two contrasting characters : r
ed
and
v.
hi
teJ
0
Pur
c-
hr
e~
d
in
g rcd-
ll
owcrcd
p
ea
pl
anb
we
re
cr
os
s-po
lli
nat
ed w
ith
pure-br
eedin
,;
wh
it
c-
fl
o
wc
rcd
pea
pl
a
nt
s,
'
a·
R
e
Su
lts
of
the
breeding
experiment:
Pa
rents:
Red
X
(
pur
e-
br
eeding)
l
c
ros
s-pollination
all
red
(hybrid)
se
lf
-
pol
I
ination
Red
Phenotypic
ratio:
3
b.
Explanation
using
genetic d
iagram
:
Defining
symbols:
White
(
pure-breeding)
White
Let 'R' re
pre
se
nt
the
dominant
allele
f
d
II
I
f
or
re
flower
and
' '
a e e
or
white
flower
r represent
th
•
·
e
recessive
Parents:
Red
(homozy gote)
RR
t
Ga
mete
s:
R
F
.
) •
X
Rr
(hete
ro
zyg
ote)
a
ll
red
White
(homozygote)
rr
t
r
Basic
Genet
F,:
Rr
X
Rr
/
~
/
~
G:
l{
r
I<
r
~r--::J
F2:
RR
Rr
Rr
rr
Genotypic
ratio:
2
Phen
otypic
ratio:
red
3
~
This
ratio is known as monohybrid ratio.
Significance
of
the ratio:
wh
it
e
0
It
indicates
the
proportion
(or
%)
of
offspring having a
part
ic ul
ar
phenotype
e.g. 3/4
or
75%
of
the
F
2
plants will produce red flower.
0
It indicates the
probability
of
appearance
of
a particular phenotype in
an offspring.
e.g.
the
chance
for an F
2
plant
to
produce red flower
is
75%.
C)
Mendel's first law
of
inheritance (in modern terms) -
for a pair
of
alleles. o
nl
y
ONE
can
be
passed to the offspring through
gamete .
C)
The
Mendel's
law can be applied as follows:
If
the two
phenotypes
of
the offspring from a certain cross are
in
the
3
:
l ratio ,
according to Mendel's law
of
inheritance, bo
th
parents must be
heteroz)
1
gous.
97
98
.
.J
E
erc
i
se
s
N
otes
at1•
1
x ·
/n
ten
s
ve
.
(cr
osses
(.
Tvpe
.\
~
•
111
exa
mpl
e :
~
I
1
)1
'''" "" '
•
. l
1
1
~
;l I
l
r
11
· '
l
,
/
7o
H'
l'
r
I
o
/tJIII
'
---'.'.:l~=~~J---
"F
-;;
ei
;:~:-----;7
7rr,
..
'...~-'~·~
(
l
',
\
11
l.'.
{
l l •
-
ro
•·
HK
-..
Rl{
RR
·,
Rr
RR
>-
n
R
r'
Rr
Rr
x,
rr
rr
X
TT
D. Te
st
c
ross
all
RR
RR
: Rr
=
J :
I
all
Rr
RR
: R
r :
IT
=
I : 2:
1
Rr
:
rr
=
I : I
alJ
rr
all r
ed
a
ll
red
a
!J
red
re
d :
white
=
3 : I
red : w
hj
te
=
J :
l
all
white
a.
to
identify
the
ge
notype
of
an
organism
with
a dominant character
Example:
To
iden
ti
fy
th
e genotype
of
a red-flowered pea plant.
(Given that flower colo
ur
is
co
ntro
ll
ed
by a pair of alleles
and
red
is
dominant.)
Solution:
Tes
t-
c
ro
ss
the plant
with
a
white-flowered
plant, if only
red-flowered
offspring
are
produc
ed
, then
it
is homozygous dominant; if some
white-flowered
offspring
are
produced,
then
it
is heterozygou
s.
b.
to
identify
the
dominant allele
I
phenotype
c:>
T
es
t
cro
ss
two
homozygous
individuals of
different
ph
enotypes.
c:>
All
their offspring
will
be
heterozygous.
c:>
In
heterozygous condition , the allele
expressed
I
phenotype
shown
by
the
off
\
prin
g is dominant.
E.
Pedigree analysis
~
e.g.
In
man,
th
e
abi
lit
y
to
roll
t
on
gue is determined
by
a pair nf a
ll
eles.
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10 -
---
□
2
6,
Key:
D
mak
tonguc-rullcr
D
rnalc non-
ro
l
le
r
Q
female tongue-
ro
ll
er
Q
female non-ro
ll
er
~~\
-
- ~
( ,
C:-
1
Informa
ti
on obta
in
ed from
th
e above pedigr
ee:
C:,
Tongue-rolling is domina
nt
.
C:,
Both individuals 1 and 2 are hete
ro
zygous.
F.
Solving
genetic
proble,ns involving monohybrid inheritance
a.
The
phenotypes
and
I
or
genotypes
of
the parents are
gi
ven,
C:,
deduce the phenotypes and / or genoty
pe
s of
th
e offsp
ri
ng and
th
e
ir
pheno
l~
pie
ratio and /
or
genotypic ratio;
C:,
predict
the
proportion
of
off
spring
th
at will have a c
ena
in phen
ol~
pe
genotype;
C:,
determine the probability that
an
offspring will have a ce
rt
ain phenotype
genotype.
Example:
A
white-flowered pea
pl
a
nt
is
se
lf-c
ro
ssed. W
ith
th
e help of a
ge
ne
tic
di~1gr~m.
predict
th
e p
ro
portion
of
th
e
F
1
pl
a
nt
s w
hi
ch
w
ill
p
rod
u
ce
red
fl
cn
wr
~.
(Gi
ve
n : Red is domina
nt
over white.)
Solution:
Le
t ,
R
·
rep
re
se
nt
the domina nt
al
le
le
for
red
fl
ower and ·r· represl'n l the
re
cessive a
ll
ele for
wh
ite
flo
wer.
99
-i:::!fd
d
E
e1c1se
s
I
nf
:lf1SIV(I
NSS
B1<
1
lt
1
Q)
-
t;::
N
otes
an
-x
White
IT
it
100
P:
Whit
e
rr
t
r
G:
rr
l
<'
•
\ '
a
ll
w
hit
e
.
. ·
oducino
r
ed
fl
ower
s
===
0
The
μro
p
or
tt
on
of
F,
pl
an
ts
p1
c
.d
d deduce the phenotypes
and
I
(Jr
.
es
are
provt e
'
b.
Tire
result
of
certam cr
oss
ge
notypes
o.f
the
parents.
CASE
1
h
t
0
·ven that
both
parents show the
dominant
Deduce
the
oe
notypes of t e paren s
0 1
0
,ffi
•
h
the
rec
essive character.
character
and
o
ne
/
some
of
the
OJJ
spring s
ow
Example
:
ln mice, black
fur
is dominant over brown fur. Two black mice were mated gi\
ing
3
brown and
1
black mice.
Assuming that
the
fur colour is
con
tr
olled
by
a single pair
of
alleles, deduce
th
e
genotypes
of
the
par
ents
.
Deduction:
The
brown
offspring must
be
homo
zy
gous recessive.
Therefore,
th
ey
must
haw
received
one
recessive allele
from each parent.
Since both of
th
e parents should have
a
dominant allele
for
being
black,
their
genotypes
mu
st
be
hetero
z.ygo
us.
CASE2
Deduce
the
ge
notypes
(and
phenotype
s)
of
the
parents
o-
i
ven
tt
1
at
tl
1
t
.
h
::o
c
e
wo
p enotypes
in
the
offspring
are
in
the
ratio
of
3:
1.
Exa
mple:
A
mai
ze
plant
wa
s
se
lf-pollinated
and
cobs
bearing both
pur\Jle a
11
c1 ,,
II
~
Jc
ow
corn
grains were
produce
d.
The number
of
purpl
e
grnins
and
ye
llo
w grains
0
11
one
of
the
c
ob~
were
53
and
18
re
sp
ec
ti
ve
l
y.
purple Drains
ye
ll
ow gra
in
s
\.
, 1
'
.·
/
••Q1t
l1
eo1
0
1111
0
illi11
8
~~
1~1~•~••~
!si
~•
0
1110
11111,~lf'
corn cob
Basic
Genetics
·h .,
11111i11~
rllw
the
()
r
·
.
I
. -
.
.
•
"'
LI
i/
i
(
CJ
0
111
,.,
co
111ml/
ed
hy
o .\III
J.!.
fr
/Ht
11
·
n/
(l//f'lf' \,
r/('d11r
r·
rhf'
gcnn(
Y
/>C
0nd
/Jh
enm
_r
pe
ql
th
e
por
e
11,
p/cmt.
De
duction:
The
purple grains
and
yellow grains
ar
e
in
a 3 : I ra
ti
o
(53
:
J
8
=
3 : I
J.
According to
Mendel's Law
of inheritance,
th
e pare
nt
mu
st be
heterozygous
and
pur
pl
e grain
is
th
e
dominant
character.
There
fo
re
th
e phenoty
pe
of the parent
mu
st be
purple grain
(i.e. the parent plant
had de
Yel
oped from
a
purple grain).
CASE3
Deduce the genotypes of the parents given that
the
phenoty
pe
s of the parents
an
d
the offspring are the same.*
Example:
A
green pea plant was self-pollinated and all offspring produce green leaves.
Ass
umin
g
that
le
af
colour is controlled
by
a pair
of
alleles, deduce the genotype
uf
rh
e
pea
plant
.
Deduction:
Since all the offspring have
th
e same phenotype as their parent.
th
e parent
pl
ant
must
be
homozygous.
c.
The phenotypes
of
parents and the phenotypes (and phenotypic ratio)
of
the
o
jf
,;;
pring are provided, deduce the dominant
I
recessive character for certain
phenotype.
CASE l
Deduce the
Jo
m
in
ant
/ r
eces~
i
ve
alle
le
of a character
given
th
at bo
th
pare
nt
s have
the
same phenotype
but u
ff
-;
pri ng
ar
e
of
different
ph
enotyp
es.
·ri . .
.
'
I'
I -
ro· 'l
JC
''
'] l'tri•c
nu
mber
or
orr~pr
in
° .
1\
,
1s
onlv true·
for
oq:.alll'-1n'-
w
·1
1c
1 u
rn
P
u
~
'
'
c-
"'
101
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es
an
d E
xe
rc
ises
NSS
Biology-
lnt
e
nstve
Nol
,
10~
Kram
pl
e I:
_
. " .
.
.
.
II
· I by
a
pair
ol allel
e~
.
Jn
n
b1
ced
1n
g
ex
pe
rim
ent
•
1
_
r
n
1
._
,
s
co
111
,
o cl
.
1
,
•.
f
ur
L'U
1.)
Llt
11
" ·
·
1
.
. ,
1
,
g
ave
birth
to
1hr
ee
gr
ey
and
fi
ve
bJ,
1
k
.,,
I
·
11
1d
11l
c
c
1nc1
L
, c
r·
1lf
('\
J':l
l'i
wer
e
1n
atc
1.
'
,
')
r,
I
.
I
ell
\
::-
.
•
.
, •
1
,-
ev
0
,-
h/
<1r
·k
('(J/0111 . c XJJ
mn _
\O
llr
an
swer
H
·
1,
;
(11
i.,
rhc·
r<'<'
c
'
SS/\
'<
'
('/1(/1
u,
f
ti
·
-~
·
lkduction:
Rlac
A
1:--
n:c~
:--
s
i\'
c
h
ec
ausL
':
)h
,th
parents
h
ave
gre
y
.fur.
th
ere
fore
e
ach
of
th
em
mu
5t
po
ssess
at
i
rnl/
on
e
gre
y
allele.
S
om1.:
of
the
ir
o
ff
s
pri
ng
ha
ve
black.fur
sh
ow
in
g
that
th
ey
mu
st ha
ve
recei
ved
w
lc1as1
one
a
ll
ele
for
bl
ack
f
ur
fr
om
e
ith
er
of
th
e parent
s.
At
le
a
st
one
of
the
pa
ren
ts
is
hetero
zygous.
in
hete
r
ozygo
us
co
nditi
on,
th
e
rece
ss
ive
all
ele is
m
as
k
ed.
Therefo
re
the
black
allele
is
recessive
.
Example
2:
ln
man.
the
ab
ilit
y
to
roll
tongue is determined
by
a pair of alleles.
D
ma
le t
ong
ue-
ro
ll
er
D
ma
le
non
-
ro
ll
er
Q
fema
le
tongue
-
ro
ll
er
Q
female
non
-ro
ll
er
Wi
th
rc
}N
ef/('e
I
ti
1
0
U'
0 Jn 1
1
e
/'Jcdi[lrec
..
,,1
11
·(·/
• ,
1
·
'"
• •
I I,\
f
I
l'
r/
>
·
or
tnr,hility to
ro
ll
10
11
>
.,,..,
.-
·
•
<
111111r1111
clwracr
e.
. .
.
.~
lf
t .
L.rp
l
mn
yo
11
r
c
111
s
11
·er
.
, •
the
ahtl
1(
r
Basic
Genetics
Deduction:
The
abi!i(y
to roll
1011
' "
)
. .
I
.
,I<
(
' "
l
0
11111
1.
111[
h L'l' ] ll '-;t':
Both
p,ircnts ;
1rc
tou
,.
, .
I
, . ,
.
~
11
'
,n
h
' ·' ·
tlwrl'll>1t
·
L'i
1L
·h
1
i1
·
1l1
c
111
m11
q
p
o<.,<.,
cs.,
u/
/co
st
one
to11g1w-rol/i,,,::
:i
lkk
.
Some
or
!heir
cllildre
11
'l .. ,
.
.
,
•
,
•
<
I L
notHo/ler
s
, how
111
1~
1l1
;
1t
1l1
cy
n,11
, 1
h;1vc
received
or
ha.
\/
0
11
c
alklc
for
llOJ
_
1
n
, .
.
.
.
.
1
Ul\ ,
..
LI
L
1oll111
g trnrn
c
ilh
er
1)1
Ili
c pcin·n
h.
At kast o
ne
of
the
parents
is
heterozygous.
In
hetcro
zy
,
ious
.
r ·
_
.
~
c-
'
co
nuti
o
11
,
only
th
e
dum1nc1nt
allele will
h
<:
n p r('
,w'
rl.
·r
l
1crclorc.:
the
tongue-rolling
allele
is
dominant.
CASE2
Deduce
th
e dominant / recess
iv
e a.
ll
ele / c
hara
ct
er
given that the parents have
d~fferent
ph
enotypes
but o
ff
s
prin
g
are
of
the
same
pheno(vpes.*
Example:
Wing length
of
fruit flies
is
controlled
by
a pair of alleles.
In
a
br
eeding experiment.
long-winged flies were crossed
with
short-winged
fli
es
. a
ll
the F
1
fli
es
,
vere
long-
winged .
Which
of
the
par
e
nts
shows the dominant character? Explain
ru11r
m1s1t
·er.
Deduction:
Some
offsprjng must have rece
iv
ed
a
long-winged
allele
and
a
short-winged
allek
from their
parents/
so
me offspring must
be
heterozygous.
A
ll
offspring have
long
wings.
Therefore.
th
e
long-winge<I
a
ll
ele must
be
dominant over the
short-winged
allek.
·
. -
-
-·
-·
·-
.
--
.
-
---
~
..
.
1
lun·
a Li
r<'C'
111
1111h
cr
11f
offsprin
g.
l
'11
, 1s
nnlv
1ru
i.:
lu
1
11rt•
.
a111
o.;111-;
,
...
111l
Ii
i.:
,11
1
P
1
(
1
t,
.
~
103
,
gy-
lnlensi,
e
Notes
and
E,crcises
.,.
,;J,:..---
--~-
20.3
Oihybrid
Inheritance
..-
~1
'8c::=
104
;1.
Mendel'
.~
.
,.
1
,c
1111
d
/aw
-
The
/
,,
aw
of
Ind
ependent
Assortment
{
In
m
odl•r
n
te
rm
s:
D1
m
11
~ ~
amc
k'
r,
,,ma
t ion.
e
ith
er
one
of
a
pair
of ,
iil
clcs
ma
y
com
bi
ne
rand
" i1
h
,
ithrr
011e
of
anoth
er
pa
ir
and
e
nt
er
th
e
sa
me
gamete eel
I.
°"''
i
OR
gametes
B.
J1
endel's
experiment
on
pea plant
a.
Re
s
ult
s
of
the
breedi
.
,
ng
expermzent:
pu
re br
ee
ding
X
homologous
chromosome pair
ro
und
and
ye
llow
see
ds
pure breeding
wrinkled and
green seeds
F1
s
ee
ds
ar
e all round . I
anc
ye
llow
(d
'h
.
I
I
ybnd)
+
se
lf
-pollination
F · 31S
2
-
.
round a
nd
ye
llow
IOI
wr
i
nk
led
a
nd
e
ll
ow
,
.
y
(new
comb
1na1i
on)
I 08
ro
unJ
a
nd
gr
ee
n
(
.
new comb
mnlion)
12
w
ri
11
klcd
and
gr
ee
n
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Basic Genet/
TliC'
pn
1p,1rt1t11
is.
lll. J
l1
-
I
l'
llnl
,,
,,
,,
..,
.,
1
dih,h
ri<
.1
•
··
•·
.
,
1p 1u
'<.
11
11
;
11
c d
111
ii
rn
110
,
,r
<J
:
J:
3 : I
-
the
•
.I
• '
10
.
h.
Explanation
11
.\·inx
R<'
11
eli<'
dia
~ra
111
:
I
l'I
:
I
} .
' IL'j)t°l.'sc nt s
the I
· .
.
L
um
111ant
allele
fur
rc
,ullll
.:;
ccd
r n
.'
]ll"l'
:--.
cnts
I
he
re , , . . ·
1
.
·
l
L:SS
I\
c a k le lor wrinkled
see
d
'\ represe
nt
s
th
e domi1nnt
a
ll
ele
f .
II
d
~
-
0 1
ye
ow see
y
represents
th
e
re
ce
ss
i
ve
·
1l
lelc
f .
d
•
c
0 1
green see
P:
ro
und. ye
ll
ow
seed
x
wrinkled. green seed
G:
R
RYY
rr
yy
t
RY
ry
RrYy
F,
seeds
are
all
round
and
yellow
(
dih
y
brid
s)
t
self-pollination
RRYY
RRYy
Rr
YY
round
&
yellow
round
&
y
ell
ow
ro
und
&
ye
ll
ow
RR
Yy
RRy
y
R
rYy
round
&
ye
J
low
rou
nd
&
green
round
&
ye
ll
ow
Rr
YY
Rr
Yy
rr
YY
R
rY
y
round
&
yell
ow
Rry
y
ro
u
nd
&
green
rrY
y
round
&
yellow
ro
und
&
ye
ll
ow
w
ri
nk
le
d & ye
ll
ow
wrinkled
&
yellow
RrY
y
Rryy
IT
Yy
rr
yy
round and ye
lJ
ow
round a
nd
green
wri
nk
led
&
yellow
wrinkl
ed
& gr
ee
n
Punnett square s
ho
w
in
g all po
ss
ib
le c
omb
ina
ti
o
ns<~
{
ga
met
es
to
form F
1
105
;:_;::zd
c'e~
,··d
E~erc
ses
i.)
----
RE
\1
\RI\.\ :
J'
ff
crcnl
ch
rom
osome
s.
Uni~
true
!0
1
ge
ne" ho
rn
c on
I
20.4
I
nherita
nce
in
Humans
106
A.
Multiple alleles
C
Th
ey a
n.
.:
a
se
ri
es
of
3
or
more
alleles
(alternative form
s)
of
a
gene which
occ
ur
at
t
he
s
am
e
locus on homologous chromo
so
mes.
C;
On
ly 2
of
them
can
be
present together
in
a
cell.
C
Exa
mpl
e:
human
ABO
blood
gro
up
s
/
1
-
domi
n
ant
all
e
le
r
es
pon
s
ibl
e
for
produc
in
g a
nti
gen
A
on red blood ce
ll
s.
t'1
-
dom
in
a
nt
all
e
le
r
es
ponsible
for
produ
ci
ng antigen B on red blood cell
s.
,
--
recessive alIde
produ
c
in
g
neith
er antigen .
/
1
t.1
/
11
d
.
d . .
.
b
all
arc
co omuiant
an , 1s
recess
iv
e
to
oth.
0
(;e1101,·pe
!"!°
:
s
in
ce
th
e
two
alleles arc
equally
dominant , they can express
th
emselves
at
th
e
sa
me
ti111
e
in
heterozyg
ous
co
ndition.
0
Human
bloot.l
gro
up
ge
not
yp
es
:
A
1:3
AB
0
Genotype
\'
t'rt
.
t
'i
t11
J'i
,
t i
r'/'1
ii
A
B
both
no
ne
t;t
•,
1
l1k11
,,
i
: ,'
II
•
/
1
,1
1 I
/" t
11
I
(
Bns1c
Gonoftc.,
r
•
l\lo
r1n
t1
r1
on
Blood
group
A
An
tl
qo
11
[l
Blood group B
Antigen A
Blood group AB
Blood group
0
B.
Sex
deten11i11ation
( ' Sex
is
:-1
genetically determined trait.
C:
1
Membe
rs
of
all homologous chromosome pairs are identical
in
s
iz
e and sh
ape:
exec-pr
the
sex chromosomes
which are important for the deter
min
ation of se
x.
A
ll
other chromosomes are known
as
autosomes.
C)
In
humans, the sex chromosomes
in
male
are
not
alike
in
size and shape
whil
e
in
femaJc, the sex chromosomes
are
identicaJ
in
size
and
shape.
C
XX-XY system:
in
humans ,
0
th
e female possesses two X chromosomes while
the
male
on
e X and
on
e Y
0
X
and Y are
of
different sizes; Y being shorter,
contains
very
few genes
0
at meiosis, half
of
the sperms
wilJ
contain X
and
half Y
Q
maleness
is
determined
by
Y
chromosome which controJs the differentia
ti
on of
th
e testis which
in
turn affects
the
dev
elopment of the genital organs and male
characteristics .
107
=-:it!
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IU
I
.,
\C"fL
f...)
\
.,..,.'l
vgy
-
,r
11
cr
11'"'
11
C'
/'ll l l
/C.)
,;JI
-
QC
U>vC
,
108
1,
,,
11
«I
61
4
5
2
3
3\&l~HHU
lh
6
7
l\ U
AA
13
14
15
U
)UC
19
20
8
9
11
Xl
21
22
10
11
12
1°C
II
H
16
17 18
(
t-
XY
))
,i
2
31
&I
6
7
n
A6
AA
13
14
15
U
XJ
19
20
11
3
it
u
8
9
11
XI
21
22
«i
H
4
5
n
ax
lh
10
11
12
re
ix
11
16
17
18
,,
xx
Ch
romosomes of male
Chromosomes of female
c:,
The chance of a child being a
boy
or
a girl
is
50%
.
P:
Father
XY
/~
G:
X
C.
Sex-linkage
in
man
xx
0
irl
0
y
X
XY
boy
Mother
xx
!
X
~
Genes carried
on
the
sex
chromosomes
are
said
to
be
sex-linked.
In
human
male.
th
e
re
is a portion of
the
X chromosome
for
which
there
is
NO
homologous region
of
th
e
y
chromosome.
Genes carried
on
th
e non-homologous por
ti
on
of
rh
e
X
ch
ro
mo
so
me there
.f<Jre
can
expre
ss
th
e
ms
e
lv
es even
if
th
ey are
recessive .
CJ
Example:
c
ol
o
ur
blindn
ess
and
ha
emophilia
a.
Red-green colour blindness
=
Pers
on
s suffe
rin
g.
from
red
-g
r
ee
n
co
lour
blindn
ess
ca
nnot
di
s
tingui
sh be
t,
veen
re
d and gr
ee
n colo
ur
s.
Basic Genetic
e
A
higher
!)
lT
l'
Cllt a μ c
01
·
l
lll
' I
..
, '
.
. .
.
.
~
· 1 dl
t.:
et
il()u1hl111d
111
a
pr,pula
11
on.
c:,
Explanation:
Th
C
g
C
11
('
ro
r
C ()
I
l)
ll
r
h
1
· I
.
.
,
,.
lll l
ll
C...,...,
I ',
a r
ct.:
c-,...,
, vc
;
1l
le lc
l,
1r
r1
cd
r;n th e X
Lh1omosnn
1
e
Th
ey
.
1
•
1...
1
rnmn
-,
om
c
()I"
111..il
c c ,rric
.,
ncil ht r
th
e
c
()
l
<>U
rhli
ncJ
gene
nor
it
s
normal
al lclc .
C)
Exampl
e:
A normal
\.\
'
.
.
.
L
om
...in
1s
married to a colourblinrJ man
Let:
X
rep
·e
x
1
se
nt
s
chromosome w
ith
th
e normal a
ll
ele
X'
represents
X
chromo
so
me w
ith
colourblind a
ll
ele
y
represe
nt
s
y
chromosome
Parental phenotype:
Normal female
X
Parental genotype:
XX
Gametes:
X
F
1
genotype:
xx
c
F
I
phenotype:
normal female
(carrier)
F1:
Normal female
X
(carrier)
xx
c
/~
Gametes:
X
Xe
F
.
2·
XY
CoJourblind male
xcy
xc
XY
normal male
Normal male
XY
y
/~
X
y
xx
normal
female
normal
normal
colour-blind
mal
e
female (carrier)
male
e
Cris
s
-cross
pattern of inheritance
-
a
colourblind man transmits his
colo
ur
blind gene through
hi
s daughter who
is
phenotypically normal but carries
the colourblin
cl
g
en
e to half of her sons, i.e.,
the
se
x-linked recessive character
of the g
ra
nd
fat
he
r reappears in
the
grandson.
109
-:::J?J!
d
E
xe
rcises
,vc
Note
s
an
-----:-
N~S
;c:;
S n
9,
i
o/ogy
-
/ntens
oTJ
hilia
. d
on
th
e
X
ch
romosom
e.
B
l
oo
u
ol
th
I,.
Haem
" ,
.
rre
ne
ca
rn
e
t
.
,
..
a r
ecess
i
ve
o
.
. .
a
nd
ma
y
ble
ed
lo
death
.
~
T
hi
s ,s
du
e
10
·, .
.
o c
lot
att
cr
inJur
y
. . . pe
rson
fa
1
Is
t
hac111
op
l11li
ac
,
.
. .
,o
n
in
man
.
0
;\~:
till
ii
,s
(11L}I
C
c
ot11n
c:,
Exp
lanation:
y
allele
for
th
e
trail,
a man w
ith
a
e
do
es
not
ca
rr
y
an
Since
th
ey
c
hr
omoso
m
.
haemophili
ac.
h
.. X c
hromo
s
ome
ts
recess
i
ve
ge
ne
on
15
20.
5
Variati
on
----
-------------
-~
11
0
A.
Types
of
variation
a.
Discontinuous
variation
C)
It
is
the
variation
where
the
character
can
be
grouped into
two
or a few
distin
ct
classes
with
no
intermediate
phenotypes.
c:>
Characteristics
showing
di
sc
ontinuous variation are usually controlled by one
pair
(o
r
two
pair
s)
of alleles
and
are
not
affected
by
environmental factors
easily.
c:>
Human
examples:
tongue-rolling,
eye
colour, shape of ear lobe, blood gro
up
,
etc.
b.
Continuous
variation
c:>
It is the variation of a character where there is a co
t·
t·
n rnuou s range o
intermediate
phenotypes
between
two
extremes
of the character.
c:>
A
normal
distribution
curve
is
produced.
c:>
Ch
arac
teri
s
tic
s s
howing
continuous variation
are
produ
d b
.
.
.
ce
Y
the
combined
effec
ts
of
many genes
(
many
paus of alleles)
and
are
affect
db
.
.
.
e
Y
environmental
factors
more
eas
il
y.
c:>
Human
exa
mples: hei.g
ht
,
we
ig
ht
, IQ, etc.
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1
200
1
100
1 000
900
800
C
C'il
E
700
::i
.c.
0
(/)
600
u
C
C'il
t/)
::i
5
00
0
.c
I-
400
300
200
100
I
14
~0
~~
1 ~
45
~-:
1-:=-5
:::---
0
-:
1~
~-=-5
-
1
J_
6_0 _
1
J_
6_5 _
1
j_
;-
o _
117_5
_1180
::::::~1
:i:::
85=
__,
=1
:i..__
90
Height (cm)
B.
Causes
of
genetic variation
C)
/
ndependent
assortment
of
chromosomes
and
crossing over
at
meiosis
e
to give gametes with
different combinations
of
alleles.
~
Random
fertilisation
C,
Mutation
e
a sudden, inheritable
change in the genetic material (DNA) .
e
Cause:
exposure
to
ionising radiation
and
chemicals
C.
Effects
of
environment
on
gene expression
Ex
amples
:
a.
Effect
of
light
on
chlorophyll production
c:,
The
all
ele
for c
hl
orophyll production (dominant) can express
it
self o
nl
y when
light
is
present.
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fect
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0
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lt
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ters
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ot
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ch
ar
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sho
win
g
continuous
variation
ar
e
aff
ected
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di
ff
crc
nt
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body
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ight
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significantly
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wh
il
e hei
oht
aw'
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arc
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112
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