Transcript Genetics - Lincoln Park High School

OR
Genetics
Junior IB Biochemical Biology
I. Mendel’s Law of Segregation
IB Review Book p23
R r
G1 of Interphase
RR rr
Locus:
the
position
of a gene
on a
chromosome
S of Interphase
(chromosomes replicated)
• These chromosomes are homologous
• One came from the mother and one from the
father
R r
RR rr
• The R and the r are “alleles”
• Alleles are alternate forms of the same gene
• Genes are pieces of DNA that code for proteins
(one gene = one protein)
• This individual is HETEROZYGOUS for “R”
• The diploid (2n) number is 2
• The haploid (n) number is 1
•Mendel’s Law of Segregation – An individual’s
two homologous chromosomes separate
during anaphase I.
This assures that…
1. Each gamete will carry ONE version of
every single gene
2. When two gametes join, they will create a
new individual with exactly TWO copies of
every single gene
• r
R
r
R
• Alleles separate from each other during
gamete formation
Fertilization:
P1
Rr x
Rr
R
r
R
R R
R r
r
r R
r r
These are the two
possible gametes that
the father could
produce – there are
only 2 possible
sperm: R and r, and
there is an equal
chance of each…
F1 (1st Filial) = offspring of P1
Thus, a Punnett square gives every possible offspring
that could be produced between two parents
•
•
•
•
R
r
R
RR
Rr
r
Rr
rr
If R is a protein that
makes a seed round
and r is a protein that
makes a seed
wrinkled…
Phenotypic Ratio (looks):
3 round : 1 wrinkled
Genotypic Ratio (homo- or hetero-)
1 homozygous dominant : 2 heterozygous :
1 homozygous recessive
II. Mendel’s Law of Independent
Assortment
• Alleles segregate in a random fashion (no
In other words:
pattern, no linkage)
T
because of the way
chromosomes
randomly line up
during metaphase I, “T”
has the same chance of
ending up with “R” as it
does with “r”
t
R
r
4 possible combinations in gametes
TR
Tr
tR
tr
A. Dihybrid Cross…
… of unlinked, autosomal genes
TtRr x TtRr
T R
T R TTRR
T r TTRr
t R TtRR
t r TtRr
T r
TTRr
TTrr
TtRr
Ttrr
t R
TtRR
TtRr
ttRR
ttRr
t r
TtRr
Ttrr
ttRr
ttrr
Which are recombinants (new combinations)?
Phenotypic ratio: 9 Tall, Round : 3 Tall, wrinkled
9:3:3:1
3 short, Round : 1 short, wrinkled
B. Dihybrid x pure recessive
TtRr x ttrr
t r
TR
Tr
tR
TtRr Ttrr
ttRr
1 : 1 : 1
Tall
Round
Tall
wrinkled
short
Round
tr
ttrr
: 1
short
wrinkled
*Recombinant (does not have parental combinations)
C. Parents: AABbccDdEe x aabbCcDdEe
• Chance of having a child: AabbccDDEe
• Aa
bb
cc
DD
Ee
1/1 x ½ x ½ x ¼ x 1/2
A
A
a Aa Aa
a Aa Aa
B b
b Bb bb
b Bb bb
= 1/32
C. Parents: AABbccDdEe x aabbCcDdEe
• Chance of having a child: AaBbCcddee
• Aa
Bb
Cc
dd
ee
1/1 x ½ x ½ x ¼ x 1/4
= 1/64
• Aa
bb
cc
Dd
Ee
• 1/1 x ½ x ½ x ½ x ½
= 1/16
Pedigree
• Sickle Cell Anemia
– HbS – Normal Dominant
– Hbs – Sickle Cell
HbSHbs
carrier
male
female
HbSHbs
HbSHbs
HbsHbs
HbSHbS
sickle cell
normal
Dominant or Recessive?
III. Multiple Alleles and
Codominance (as illustrated by blood
groups…)
• 3 alleles: IA, IB, and i
• Phenotype
O
A
B
AB
Genotype
ii
IAIA or IAi
IBIB or IBi
IAIB
TYPE
O
A
CELLS
A-antigen that can
be recognized and
destroyed by
A
foreign bodies
A
that will produce
antibodies against
A
it
A
A
+ or -
A
B
B
Rh
+ or -
B
B
B
+ or -
B
A
A
B
A
AB
+ or B
A
Rh factor… ‘+’ have it or ‘-’ don’t
• Rh factor works the same way – if you are Rh+ it
means you have Rh antigens on the surface of your
blood cells
• If you are Rh- then you have no antigens, and will
produce antibodies in the presence of Rh+ blood
• Universal Donor:
Type O• Universal Recipient:
Type AB+
CHINA
U.S.
Type AB
Type O
Problem #1:
Mother is type A and father is type A.
What are all possible children?
IA
i
A
A
IA IA IA IA i
A
i
IAi
O
ii
Problem #2:
Mrs. Brown AB
Mr. Brown B
Mrs. Smith A
Mr. Smith B
Babies: Baby 1 is type O… Baby 2 is type A
IV. Testcross
• Sheep:
White is dominant
Black is recessive
• White male
• $1,000.00 per vial of sperm from a pure white male.
How do we know if a male is homozygous dominant or
not?
• Cross with a homozygous recessive
W? x ww
…if it is pure, all offspring will
W ?
be white (you need to have a lot
w Ww ww
of offspring to be sure)
w Ww ww
2 white : 2 black
(not a pure homozygous dominant)
*What are the chances that a heterozygous male crossed
with a black female would give rise to 4 white offspring
in a row?
*A: ½ x ½ x ½ x ½ = 1/16th chance
V. Sex-linked traits (ex: hemophilia &
colorblindness)
• Found on X chromosome but not on Y
B
B
NORMAL
B
b
CARRIER
b
b
COLORBLIND
B
NORMAL
b
COLORBLIND
Ex 1: Cross colorblind male with carrier
female
• XbY x XBXb
XB
Xb
–
–
–
–
Xb
Y
XBXb XBY
XbXb XbY
Of the offspring, what are the chances of having a normal son?
1/4
What are the chances of having a colorblind son?
1/4
Ex 2: Cross a hemophilia-carrier female
with a normal male…
XHXh x XHY
XH
Y
XH
Xh
XHXH XHXh
XHY XhY
• Chances of a child being a hemophiliac?
• ¼
Ex 3: Colorblindness (sex-linked)
P1
(parents)
F1
Parents must be…
DAD: XBY
MOM: XBXb
Note: If a normal mother has a son who is
colorblind, then she MUST be heterozygous (a
carrier)
Genetics review…
•This is a karyotype (all of the
chromosomes in the human
body)
•Each chromosome has genes on
it (to code for proteins)
•There are multiple versions of
each gene (example: T = tall, t =
short)
B
b
•These multiple versions are
R
R
called “alleles”
q
q
•You always have two alleles,
because you have two versions
T
t
of each chromosome (one from
Example: This individual’s
mom, one from dad)
genotype would be BbRRqqTt
VI. Modifying Mendel with
linkage and crossing over
• TtBb x ttbb
A. If no linkage, you would expect to get 1TtBb : 1Ttbb :
1ttBb : 1ttbb ratio
B. Suppose there is the following linkage:
T B
t b
t b
t b
• With this linkage,
you would expect:
tb
tb
TB
TB
tb
TB
tb
tb
tb
tb
tb
tb
a 1 : 1 ratio!
C. But… what if the actual ratio of the cross is 7:1:1:7?
• There must be linkage (we are close to 7:7) with some
crossing-over occurring (a source of recombination)
T
B
t
b
t
b
t
b
Tb
tb Tb
tb
tb Tb
tb
tB
tB
tb
tB
tb
1 : 1
This
happened 2
out of 16
times!
7 : 1 : 1 : 7
TB Tb tB tb
tb
tb
tb
tb
#’s like this imply crossing over AND linkage
2/16
14/16
With a cross of a double heterozygote (TtRr) and a
double homozygous recessive (ttrr), REMEMBER:
WITHOUT linkage we expect 1:1:1:1, and WITH linkage
we expect 1:1
• You observe a ratio of 84 : 79 : 88 : 81. Do you think
there is linkage?
• NO
• You observe a ratio of 118 : 121. Do you think there is
linkage?
• YES
• You observe a ratio of 131 : 12 : 9 : 137. What is going
on?
• Linkage with crossing over! This is basically a 1:1 ratio,
with 21 (12 + 9 = 21) “out of place” results mixed in.
VII. Analysis of COV data to construct
gene maps
• If ratio is 7:1:1:7, then 2/16 were recombinants
• 2 / 16 = .125
COV = 12.5%
= 12.5 cM (centi-Morgans apart)
• 1% COV = 1 centi-Morgan
Problem #1:
pure white (r), smooth leaves (p) x pure red (R), pointed (P) leaves
• F1 are all red, pointed
• A test-cross is done… RrPp x rrpp (to detect linkage)
40 white, smooth
36 red, pointed
10 white, pointed
14 red, smooth
100
• Q1: Is there linkage?
• Q2: How far apart are these 2 genes (R & P)?
R
P
?
A1: The RrPp x rrpp cross will determine if
there is linkage…
• If no linkage…
RP
Rp
rp RrPp Rrpp
1 : 1 :
• If linkage…
RP
rp
rp RrPp rrpp
1 : 1
rP
rrPp
1 :
rp
rrpp
1
• If both linkage & crossing over…
something like 7 : 1 : 1 : 7
• (40:36:10:14 implies that there IS linkage and crossing
over)
A2…distance apart
• 24/100 = 0.24
COV = 24%
= 24 cM
R
P
24 cM
Problem 2: Assume Gene T and B are 12.5 cM
apart and another gene A is linked with T & B.
A has a COV with T of 5 cM and a COV with B of
7.5 cM. Draw a map of TBA.
T
A
B
12.5 cM
5 cM
7.5 cM
Problem #3: COV of R & T = 6.0
COV of S & T = 13.5
COV of R & S = 19.5
R
T
S
6.0
13.5
19.5
5
R /r
17
T/t
2
B/b E/e
VIII. A. Modifying Mendel with
Polygenic Inheritance
• Many genes control one trait
• Producing continuous variation
• NOT producing a ratio
• Ex1: Human skin color
• Ex 2: Human height
• Ex 3: Hair color
B. Codominance (and Incomplete
Dominance)
• Codominance: both alleles are dominant,
and so both are represented (ex: blood
type… IAIB is “type AB”
• Incomplete Dominance: neither is
completely dominance, and so there is a
blend (if RED snapdragons crossed with
WHITE snapdragons give PINK flowers)
C. Mendelian ratios can also be modified
by “gene interaction”
• Studied by Bateson
• Ex: Epistasis (when genes act on one another)
CcPp
x
CcPp
purple flowers purple flowers
• Mendelian genetics expects 9:3:3:1
• Actual ratio is 9 purple : 7 white. WHY?
Purple
Dom; Dom
C_P_
:
White
one or no dominant genes
Ccpp, ccPp, ccpp
IX. Nondisjunction – errors in meiosis
• Causes extra chromosomes or too few chromosomes
• Chance occurrence
• Failure of chromosomes to separate
• Tested for by amniocentesis or CVS (chorionic villi sampling)
A. In Sex chromosomes…
spermatogenesis
XY
Interphase
X-X
Anaphase I
X-X
Prophase II
Anaphase II
oogenesis
X-X
XX
Y-Y
X-X
Y-Y
Y
X-X
Y
X-X
X-X
X-X
X
X
• XX = normal female
• XY = normal male
• X = sterile female (called “Turner syndrome”)…
chromosome # is 45
• XXY = sterile male (called “Klinefelter’s syndrome”)…
chromosome # is 47
B. Nondisjunction in autosomes ex: Down’s syndrome
• 95% due to maternal age
egg
sperm
#21 #21
Trisomy-21
is Down’s
#21
#21 #21 #21
ZYGOTE
Terms:
• Karyotype –
photograph of an
individual’s
chromosomes:
• Autosome – non-sex
chromosome
• Linkage group – group
of genes who have their
loci on the same
chromosome
a
B
A
b
C
d
D
c
• As far as these chromosomes are concerned, what is the
individual’s genotype?
• AaBbCcDd
• Do genes B and C assort independently (randomly)?
• Yes
• Which genes are in the same linkage group?
• A, B, and D
• Set up a Punnett square for the cross between this individual and
an individual with the genotype bc (we are only concerned with the
B and C genes).
BC
Bc
bC
bc
bc
• Set up a Punnett square for the cross between this individual and
an individual with the genotype abd (we are only concerned with the
A, B, and D genes).
aBd
abd
AbD
T = tall
t = short
B = brown
b = white
R = round
r = wrinkled
M = loud
m = quiet
Z = hairy
z = bald
• TtBb x ttbb
45 tall,brown:3 short,brown:5 tall,white:47 short, white
• BbRr x bbrr
27 brown,round:26 brown,wrinkled:22 white,round:25 white, wrinkled
• MmTt x mmtt
41 loud,tall:9 loud,short:5 quiet,tall:45 quiet, short
• TtZz x ttzz
21 tall,hairy:28 tall,bald:30 short,hairy:21 short,bald
• BbMm x BbMm
47 brown,loud:4 brown,quiet:1 white,loud:48 white,quiet
Determine which are linked, and map the chromosome.
14
T
B
M
8
5