LAB #6:
Mendelian Genetics
Lab Manual Exercise 13
G
Genotype
& Phenotype
Ph
• Genotype refers to particular genes
an individual carries
• Phenotype
yp refers to an individual’s
observable traits
• Cannot always determine genotype by
g phenotype
p
yp
observing
Alleles
• Different forms of a gene
– distinct in DNA sequences
• Originally arose by mutation
– which changes DNA sequence
F
f
• Alleles come in dominant and recessive
genotypes
– Dominant allele masks a recessive allele of a gene
We denote dominant
m
alleles with CAPITAL
L letters
– W
Homologous
H
l
Chromosomes
When talking about genotypes:
• Homozygous
– two identical alleles at a locus
- AA: homozygous dominant
- aa: homozygous recessive
• Heterozygous
yg
– two different alleles at a locus - Aa
Mendel’ss Theory of Segregation
Mendel
• An individual inherits a particulate unit of
i f
information
ti (allele)
( ll l ) about
b t a trait
t it f
from each
h
parent
formation the two alleles of
• During gamete formation,
a gene in an organism segregate from each
other. (MEIOSIS!)
• Genes are inherited as discrete “particles.”
• Blending does not occur, so that an allele
th t is
that
i presentt iin th
the parental
t l generation
ti
and is not evident in the F1 generation can
reappear in the F2 generation.
Law of Independent Assortment
• Mendel concluded that the two “units”
(alleles) for the first trait were to be
assorted
d into gametes independently
d
d
l
of the two “units” (alleles) for the
other trait
• The two chromosome in each
homologous pair are sorted into
gametes during meiosis
Nucleus of a
diploid (2n)
Reproductive cell
with two pairs of
homologous
chromosomes
OR
Possible alignments
of the two homologous
chromosomes during
metaphase I of meiosis
The resulting alignments
at metaphase II:
allelic combinations
possible in gametes:
A
A a
a
A
A a
a
B
B b
b
b
b B
B
A
A
a
a
A
A
a
a
B
B
b
b
b
b
B
B
B
A
B
A
1/4 AB
b
a
b
a
1/4 ab
b
A
b
A
1/4 Ab
B
a
B
a
1/4 aB
Fig. 10.8, p. 158
Th Garden
The
G d
Pea
P
Plant
Pl
•
•
•
•
Used to study genetics
Self pollinating
Self-pollinating
True breeding
C be
Can
b experimentally
i
t ll crosspollinated
Mendel’s Experiments
p
Monohybrid Cross: A cross between two
individuals in the same species in which ONE
genetic trait is documented.
purple flowers
AA
white flowers
aa
TRUEBREEDING
PARENTS:
purple
flowers
white
Flowers
x
aa
AA
GAMETES:
A
A
a
a
Aa
F1 HYBRID
OFFSPRING:
In-text, p. 158
Monohybrid Punnett Square
True-breeding
homozygous recessive
parent plant
F1
PHENOTYPE
S
aa
True-breeding
homozygous dominant
a
parent plant
Aa
Aa
Aa
Aa
a
A
Aa Aa
A
Aa Aa
AA
F1 PHENOTYPES
Aa
An F1 plant
self-fertilizes
and produces
gametes:
F2
PHENOTYPES
Aa
A
AA
A
Aa
Aa
aa
a
A AA Aa
a
Aa aa
Dihybrid
y
Cross
Dihybrid Cross: A cross between two
indi id als in the same species in which
individuals
hich
TWO genetic traits are documented.
purple flowers AA
t ll
tall
BB
white flowers aa
dwarf
What are parental genotypes?
bb
TRUEBREEDING
PARENTS:
purple
Flowers,
Tall
x
white
Flowers,
dwarf
aabb
AABB
GAMETES:
AB
AB
ab
ab
AaBb
F1 HYBRID
OFFSPRING:
What is the F1 generation
phenotype?
In-text, p. 158
F1 Generation:
flowered, tall
AaBb
AaBb
PurplePurple
Trait Studied
SEED
SHAPE
H PE
SEED
COLOR
POD SHAPE
POD
COLOR
FLOWER COLOR
Dominant
Form
Recessive
Form
F2 Dominant-toRecessive Ratio
5,474
round
1,850
wrinkled
6,022
yellow
2,001
green
882
inflated
428
green
705
purple
FLOWER
POSITION
651 long
stem
STEM
LENGTH
787
tall
299 wrinkled
2.96:
1
3.01:
1
2.95:
2
95:
1
152 yellow
2.82:
1
224
white
3.15:
1
207 at
tip
3.14:
1
277
dwarf
2.84:
1
Fig. 10.5, p. 156
Monohybrid Cross
Exercise 13.1 B pg. 183
Monohybrid Cross
Lab handout exercise
•
•
•
–
–
–
Tobacco Seedlings
•
Make individual counts of white and green
phenotypes
DO NOT PULL PLANTS OUT OF PETRI DISH
Pool class data
Data analysis
y
NOTE: Difference between corn and
seedling counts = one piece of corn is
THE population,
l ti
while
hil each
hP
Petri
t i dish
di h of
f
seedlings is a population.
TAKE HOME: each seed is an individual
(whether its part of an ear of corn or
laying in a Petri dish).
Dihybrid Cross
Exercise 13.2 B pg.
189
Observable human traits
Exercise 13.3, pg. 190
E
• There are many traits that follow the
M d li pattern
Mendelian
tt
of
f inheritance.
i h it
• Ex: Earlobe Anatomy depends on a single
gene
n with
ith ttwo alleles
ll l s
– Dominant allele specifies detached earlobes – E
– Recessive allele specifies attached lobes - e
In today
today’ss lab,
lab we will examine the
genotypes for several human traits.
We’ll Use Statistics Today
y
–
Scientists use statistical tests to examine the likelihood
that their results are due to random events (vs. “real”
results)
–
For example, if you flip a coin 100 times, you probably won’t
get a clean 50 heads, 50 tails. The same problem arises
for genetic ratios (e.g.,
(e.g. you might get 2.81:1.03 ratio,
ratio
instead of 3:1 ratio)
–
Chi-square (X2) is one such statistical procedure that
m
the difference
ff
between y
your actual data versus
examines
some prediction
–
If you calculate a probability less than or equal to 5% (p ≤
) then y
your results significantly
g
y differ from the
0.05),
predicted
d
d genetic ratio
–
Don’t get bogged down in the math behind chi-squares (e.g.,
there are many electrical tools we use in our day-to-day
lif without
life
ith t knowing
k
i the
th actual
t l electrical
l t i l circuitry…but
i
it
b t
they still get the job done for us!)
Chi Squared Analysis
Lab and handout exercise
• Scientists use statistical tests to examine
the likelihood that their results are due to
random events (vs. “real”
“
results)
• Otherwise stated: Chi-square (χ2) tests if an
observed
b
d frequency
f
distribution
di ib i fits
fi an expectedd
distribution
(O − E )
Χ =∑
E
2
2
Monohybrid Chi-square
Chi square (χ2)
Phenotype
Geno- Observed
type
(O)
Expected (O-E)2
(E)
(O-E)2
E
Purple
P_
64
60.75
10.56
0.174
White
pp
17
20.25
10.56
0.522
81
81
Total
Expected
p
ratio in a monohybrid
y
cross?
How do we calculate E?
((3/4)) x 81 = 60.75
(1/4) x 81 = 20.25
0 696
0.696
3:1
χ2
Chi-square
2
(χ )
• Once you have calculated your chi-squared
compare
p it to the table on page
p g 186.
• If your chi-squared is GREATER than the
.05
05 cutoff = chance alone cannot explain the
deviation
• Your number is LESS than the .05
05 cutoff =
variation is due to chance.
Monohybrid Chi-square
2
(χ )
• What is the expected ratio?
– 9:3:3:1
• How do we calculate expected values?
–
–
–
–
(9/16) x Total = E
(3/16) x Total = E
(3/16) x Total
T t l=E
(1/16) x Total = E
TODAY’S
TODAY
S PLAN
• Work in pairs, but complete the post-lab BY YOURSELF!
• Follow the lab book and do all of the exercises in order,
except SKIP section 13.1.C1, 2, & 3 (all Fern exercises).
• Look at the tobacco seedlings & count number of green vs.
albino. As soon as you have your tobacco data, give it to
your TA
TA. Do the X2 test for these data…
data Show all work!
• Answer all the questions in your lab book.
• Post-lab
P t l b
– Have TA check off your lab book work
– Turn in post-lab sheet with completed tables and problems
• When you’re all finished turn in the post-lab assignment.
FOR POST-LAB QUESTIONS, RECALL:
Inheritance of
X-Linked
Genes
• X and Y
chromosomes
function as a
homologus pair
•X
X is
i chromosome
h
on which diseasecausing
g allele is
found.
•No comparable
allele is found on
the Y chromosome.