1. health and fitness
2. disease

V\
Sym oiss_ b lack - B
Chestnut -b
Phenotype
s
trotting - T
pacing -f
Mate
x
black, trotter
x
Female
chestnuts_pac^
b
Genotype
Gametes
bt 1 BtTt
BbTt
8 bTr
(G enotype)
BbTt I BbTt
bt
-
black. trotter
(Phenotype)
Female
P2
Genctype
Gametes
P2 Punnett square
F Phenotypes
black, trotter
chestnut, trotter
(from F1)
t
i man, brown eye color is due to a dornin^
(B), and blue eye color is due to its
allele (b), Right-handedness is due to a dominant gene
(t?), and Left-handedness is
A man homozygous for both brown eyes and right-handedness
blue-eyed, left-handed woman. What is the genoty
pe
arried a girl With the same genotype, what kinds
m
of offspring could they produc
proportions?
In garden peas, the gene for a smooth speed
coat is due to its recessive allele (s). Also, the
and a dwarf plant is due to its recessive allele M. Cross
heterozygous one for both tallness and smoo th
seeds. What kinds of off sprin
and in what proportions?
inkled seed
ntgene(r),
rt
a
they produce
lashi
_ erk hair . They ha
l:. iv
short eyelashes and dark hai
an w ith short Avolashes
nd red hair and one with
and children.
ypes of parents
inant to blue. A
nt to straight and brown eyes is dom
ina
dom
is
r
hai
ly
cur
n
ma
In
2..
marries awoman with blue eyes and
He
r.
hai
t
igh
stra
and
s
eye
wn
man has bro
l, too have
i_-our have blue eyes and curly hai
.
outly flair, They have six cniidren.
pes of the parents and the offspring
oty
gen
the
e
Giv
r.
hai
t
igh
stra
brown eyes and
u,porn
inant allele (B), and chestnut
dom
a
n
upo
ent
end
dep
is
ck
bla
,
in horses
inant over its
to an allele (T) which i s dom
ecessive ("b). T he trotting gait is due
ies.
Ni w an,) on, different chtomoso
ssive pacing allele (t). T hese. g
tt er.
mated to a homozv ous chestnut tro
a) If a homozygous black pacer is
?
what will the F 1 caeneration be like
Draw a purane
o l e i&^divid ills are Gro ssed?
it
ult
res
h\ that will he the
o.
uare: and derive the genotypic rati
ive allele
dominant gene (E), red to its recess
. `rn poultry black color is due to a
allele
e (C), plain head to its recessive
gen
nt
ina
dom
a
to
due
is
d
hea
d
(e). Creste
duce many
to a black, plain female. They. pro
the
(c). A red crested mate bird is mated
sted. What were the genotypes of
cre
ed
i
f
hal
and
,
sted
cre
ck
bla
f
offspring, hal
parents?
finds some
-haired dachshunds. but he always
5. A man owns a pair of short, red
m repeated
puppies which are produced. Fro
the
in
r
hai
ck
bla
o
yom
and
air
long-h
black. Could
; 8 long red; 4 short black; 4 long
ratings he obtained: 16 short red
breeders
1 ratio, thus indicating that both
:3:
9:3
ofa
le
mp
exa
an
be
s
ult
res
these
es?
are heterozygous for the two gen
b_ What phenotypes would you exp ect from the
ddVVw
,so
In pine trees, long ne edles are dominant to short and rough
bark is do rtin
to smooth. Use a punnett square to show the possible geno
types of offspring
odu d from 2 entirely heterozygous pine trees.
.
3. A pea plant heterozygous for height and for yellow seeds
is crossed with a
dwarf plant having green se eds- Using a p unnett squar
e, dete rmine the
expected genotypic and phenotypic ratios of the offsprin
g.
4. Using the `fraction multiplication" method, dete rm ine the
phenotypic ratios of
offspring born to a mother with blood type 0 and Rh - and
a father with blood
type AB and Rh+. ( Rh - is recessive, use rr for it and
R for Rh+. The father
is heterozygous for Rh type))
5. Long hair and yellow eyes are dominant in
cats, Dete rmine whether or not
short -haired, green-eyed kittens could be born from the
following parents:
mother long hair, yellow eyes
father long hair, green eyes
The following question mixes one trait that exibits complete
dominance and one
trait that exibits incomplete dominance.
6. Roan is a coat color produced when red cattle are bred
with white. That is,
Red is incompletely dominant to white. Hornless cattle
are dominant .
at
and
phenotype
as woul
d be
r
expec
o
ted from
a
a roan
n
horno
le ss med
h
02.049
CONTINUITY
Karyotype p reparation
LIF E SCIENCES ON FILE
02°04
H
0
0
A Blood sample removed from donor
13 Sample suspended in saline; red cells settle out
C Catchicine added - stops cell division at metaphase
0 Water added - cells swell and burst
E Cells spread onto slide
F-G Cellsobserved under micr
H Chromosomes photographed
€ Individual chromosomes cut out
J Chromosomes arranged in order of diminishing size
0 2a 050
Bx N XA B^rin6 n
6
7
ofi^A h
A
katyotype
3 d karyotype
11
8
zx
AAA )%
12
x xxx
--SDL-UNKAG
Sex ChromosomesX chromosomeY chromosome:
In humans there are 23 pairs of chromosomes (46). Of these, 22 pairs are
autosomes and 1 pair, the 23 rd pair, are the sex chromosomes.
organ. Morgan worked
iat sex chromosomes exist was discovered by
fruit flies known as [rophila . H e discovered that when 2 red eyed
ents were mated, some individuals in the F 1 generation had white eyes and
that the white eyed individuals were always males.
To explain this phenomenon, Morgan suggested that the gene for eye color was
carried on the sex chromosomes. When genes are carried on the sex
genes. Most sex-linked
`
chromosomes, they are said to be aex-lbked or x-JmkW
genes are carried only on the X chromosome and not the Y.
To illustrate:
Parental Cross:
Genotypes:
Red eyed male x Red eyed female (red is dominant)
F1 offspring:
In humans, there are many traits carried on the X chromosome. When a defect
or mutation occurs in one of the genes on an X chromosome, a female still has a
chance of having a normal gene on her other X chromosome, whereas a male
who inherits an X that carries a gene defect, will almost always express the
s
phenotype governed by the mutation. This is
humans are also referred to as Male-Disorders.
married a blue
e father was blue eyed
os
wh
an
m
d
in
r-b
lo
e () is
co
4. A brown eyed,
ess). Assume brown ey
dn
in
bl
r
lo
co
r
fo
us
go
normal vision.
eyed carrier ( heterozy
blindness recessive to
r
lo
co
d
an
b)
e(
blu
d color blind?
er
dominant ov
ren will be blue eyed an r blind?
ild
ch
eir
th
of
ge
ta
en
rc
a) What pe
ed and colo
e girls will be brown ey
ind?
b) What percentage of th
own eyed and color bl
br
be
ll
wi
ys
bo
e
th
of
c) What percentage
eyed gene
d dominant to the white
an
d
ke
lin
xse
is
)
(A
e
5. In the fruit fly, red ey
notypes of:
(r). Give the possible ge
a) white eyed males
b) red eyed males
c) white eyed females
d) red eyed females
wing crosses:
e offspring of the follo
th
of
s
pe
ty
no
ge
e
th
e
Giv
red eyed male
a) white eyed female X
ale
ed female X white eyed m
b) heterozygo us red ey
e
al
m
ed
ed female X red ey
c) heterozygous red ey
problgms
inked)
its which follow a sex-linked (or X-l
To solve problems involving tra
must be combined with the symbols
les
alle
the
for
s
bol
sym
the
n,
inheritance patter
omosome(s).
ressed as superscripts on the X chr
for sex. The allele symbols are exp
les and in
omosome since it carries so few alle
chr
Y
a
on
put
is
bol
sym
le
alle
No
me w not
h sex (:X) linkage, the Y chromoso
wit
ling
dea
ns
stio
que
s
etic
gen
st
mo
l" for a'trait
assume that individuals are "norma
ays
Alw
it.
tra
the
for
le
alle
an
carry
ise.
unless the question tells you otherw
ce pattern. The 3
ich follow a sex-linked inheritan
't'here are many human traits wh
most common traits are:
i
ui` o (oniina #^
• eessive)
-^
Red/green colorblind ess
-
eWGreen color vision
Ability to blood clot
Normal visual acuity
Hemophilia
Myopia (.nearsightedness)
-linked recessive trait. Show the
Kxample.l: Colorblindness is a sex
s:
genotypes of the following individual
Normal male
Colorblind male
Carrier male
Normal female
Colorblind female
Carrier female
inant x-linked
feather characteristic follows a dom
tain
cer
a
,
ltry
pou
In
2:
e
pk
Eu
allele (B) produces barred feathers
inheritance pattern. The dominant
the
es non-barred feathers. What is
and the recessive allele (b) produc
notypes from a cross between a
ratio of offspring genotypes and phe
non-barred male?
heterozygous barred female and a
Parent Phenotypes:
Parent Genotypes:
Puimett Square:
F1 Genot )es:
h rre_d_f a
XX
x
non-barred ttm
XY
F1 PhCJIO tVpes:
HemoohWa
- a sex-linked disorder.
- a blood clotting disorder.
- one factor is missing in the "blood clotting cascade" which prevents the
proper clotting of blood when a wound occurs.
- people with this disorder face the possibility of bleeding to death with the
smallest of injuries - even a paper cut!
- there are several different kinds of hemophilia - the kind depends on which
factor is missing.
- historically, most forms of this disorder were found only in males - as
females would bleed to death at puberty with the onset of menstruation.
- also, ? "mutant" alleles are required in order for a female to inherit the
disorder, while males require only one. (Xh`Y = male, h h = female) -this
explains why more males have the disorder in any case.
- today, many females live with hemophilia past puberty, but they need to
routinely have blood transfusions containing the clotting factor that they are
missing so that they will not bleed to death (either upon injury or
menstruation).
today, males with hemophilia also receive transfusions of their missing
blood clotting factors in order to reduce the possibility of them bleeding to
death upon receiving injuries.
- so there are treatments for hemophilia, but there is no cure
Pnhlemll
A man with hemophiliav-ishes to have a child with a woman who is normal.
but whose father also had hemophilia.
a) Show the parent genotypes:
b) Show the punnett square:
c) What are the chances that a. of their Ch
children will be nnormal?
Q
(1) \\ dt are tl e chances that their bD v ill I)e normal'?
a
e) What are the chances that their girls will be normal?
°%,
M
ybrid Crosses
-
Summit ry and Short Cuts
B = Black
b= white
(2 phenottypes)
Cron rte:
BB x bb
100%Bb
all Ft = Black
Fl Cross.
BWxBW
1 BB
2BW
1WW
Bb x Bb
I BB
2 Bb
I- bb
F2 phenolyxc ratio:
F2 phenotypic ratio:
1 Black: 2 B& W: 1 White
3 black: lwh ite
Dihybrid Cross Short Cut
s a peter tzct<s for both it
IND both saes fallow
ring wwiU be
M . e1"s Law o f Dominance, tae Perot is ratio of their offsp
the outcome will b
9:3:3:1. For example, iii a cross between Aal3.b x A iEb,
9 Trait (A) with Trait (B)
3 Trait (A) with trait (b)
3 trait (a) with Trait (B)
I trait (a) with trait (b)