Grades
 Three hourly exams plus final exam (450 pts),
 You will have 1.5 hours to complete each exam,
 You will be allowed one (1) 11” x 8.5” crib sheet, both
sides, for each exam,
…may seem
really hard.
 Exams - 150 points each, Final Exam cumulative.
 Quizzes will be given every Wednesday (total 100 pts),
 will cover the basics of the assigned reading
(including that day's assignment),
 quizzes 12.5 points each, ~15 minutes,
 No Make-up Quizzes, absolutely no exceptions,
 can drop two (2) lowest quiz scores (except 1&2).
 Total course points - 550
…should be
relatively easy.
Know This…
1
1/2
1/4
1/2
1
1/2
1/2
1/4
1/2
1
Assignment: Correlate this with the observed phenotype.
Mendel’s Results, F2 Dihybrid
P generation cross: YYRR x yyrr
F1 generation cross: YyRr x YyRr
• Y_ R_
= 315
=9
• yyR_
= 108
=3
• Y_rr
= 101
=3
• yyrr
= 32
=1
Forked-Line Method
1/4 x 1/4 = 1/16 YYRR
1/4 YY
1/4 RR
1/2 Rr
1/4 rr
1/2 Yy
1/4 RR
1/2 Rr
1/4 rr
1/2 x 1/4 = 1/8 YyRR
1/2 x 1/2 = 1/4 YyRr
1/2 x 1/4 = 1/8 Yyrr
1/4 x 1/4 = 1/16 yyRR
1/4 yy
1/4 RR
1/2 Rr
1/4 rr
1/4 x 1/2 = 1/8 YYRr
1/4 x 1/4 = 1/16 YYrr
1/4 x 1/2 = 1/8 yyRr
1/4 x 1/4 = 1/16 yyrr
Genotypes Y--R-1/4 YY
1/2 Yy
1/16 +
1/4 RR
1/4 x 1/4 = 1/16 YYRR
1/2 Rr
1/4 x 1/2 = 1/8 YYRr
1/4 RR
1/2 x 1/4 = 1/8 YyRR
1/2 Rr
1/2 x 1/2 = 1/4 YyRr
2/16 +
2/16 +
yellow/round
4/16
= 9/16
Genotypes Y--rr
1/4 YY
1/2 Yy
1/4 yy
1/4 RR
1/4 x 1/4 = 1/16 YYRR
1/2 Rr
1/4 rr
1/4 x 1/2 = 1/8 YYRr
1/4 x 1/4 = 1/16 YYrr
1/4 RR
1/2 x 1/4 = 1/8 YyRR
1/2 Rr
1/4 rr
1/2 x 1/2 = 1/4 YyRr
1/2 x 1/4 = 1/8 Yyrr
1/4 RR
1/2 Rr
1/4 rr
1/4 x 1/4 = 1/16 yyRR
1/4 x 1/2 = 1/8 yyRr
1/4 x 1/4 = 1/16 yyrr
Genotypes Y--rr
1/4 YY
1/4 rr
1/2 Yy
1/4 rr
1/16 + 2/16
1/4 x 1/4 = 1/16 YYrr
1/2 x 1/4 = 1/8 Yyrr
= 3/16
yellow/wrinkled
Genotypes yyR-1/4 YY
1/2 Yy
1/4 yy
1/4 RR
1/4 x 1/4 = 1/16 YYRR
1/2 Rr
1/4 rr
1/4 x 1/2 = 1/8 YYRr
1/4 x 1/4 = 1/16 YYrr
1/4 RR
1/2 x 1/4 = 1/8 YyRR
1/2 Rr
1/4 rr
1/2 x 1/2 = 1/4 YyRr
1/2 x 1/4 = 1/8 Yyrr
1/4 RR
1/2 Rr
1/4 rr
1/4 x 1/4 = 1/16 yyRR
1/4 x 1/2 = 1/8 yyRr
1/4 x 1/4 = 1/16 yyrr
Genotypes yyR-1/4 yy
1/2 Rr
1/4 x 1/2 = 1/8 yyRr
1/4 yy
1/4 RR
1/4 x 1/4 = 1/16 yyRR
2/16 +
1/16
= 3/16
green/round
Genotypes yyrr
1/4 YY
1/2 Yy
1/4 yy
1/4 RR
1/4 x 1/4 = 1/16 YYRR
1/2 Rr
1/4 rr
1/4 x 1/2 = 1/8 YYRr
1/4 x 1/4 = 1/16 YYrr
1/4 RR
1/2 x 1/4 = 1/8 YyRR
1/2 Rr
1/4 rr
1/2 x 1/2 = 1/4 YyRr
1/2 x 1/4 = 1/8 Yyrr
1/4 RR
1/2 Rr
1/4 rr
1/4 x 1/4 = 1/16 yyRR
1/4 x 1/2 = 1/8 yyRr
1/4 x 1/4 = 1/16 yyrr
Genotypes yyrr
1/4 yy
1/4 rr
1/4 x 1/4 = 1/16 yyrr
1/16
green/wrinkled
F2 via Forked Line
• Y--R--
yellow/round
9/16
• Y--rr
yellow/wrinkled
3/16
• yyR--
green/round
3/16
• yyrr
green/wrinkled
1/16
Why use Forked-Line Method?
• Based on a classic dihybrid cross (YyRr x YyRr),
what is the probability that an organism in the F2
generation will have round seeds and breed true
for green cotyledons?
OK?
YR
Yr
YR
Yr
yR
yr
YYRR
YYRr
YyRR
YyRr
YYRr
YYrr
YyRr
Yyrr
YyRR
YyRr
yyRR
yyRr
YyRr
Yyrr
yyRr
yyrr
yR
yr
3/16
p = 0.1875
Better
1/4 yy
1/4 RR
1/2 Rr
1/4 x 1/4 = 1/16 yyRR
1/4 rr
1/4 x 1/4 = 1/16 yyrr
3/16
1/4 x 1/2 = 1/8 yyRr
p = 0.1875
Best (?)
1/4 yy
3/4 R_
Sum Law: 1/4 RR + 1/2 Rr
1/4 x 3/4 = 3/16 yyR_
Forked-Line Method
(phenotypes)
3/4
yellow
1/4
green
3/4 round
9/16 yellow round
1/4 wrinkled
3/16 yellow wrinkled
3/4 round
3/16 green round
1/4 wrinkled
1/16 green wrinkled
Example
P
Rr YY x
rrYy
Probability Rr YY in offspring;
1/2 Rr
1/2 rr
1/2 YY
1/2 Yy
1/4 RrYY
Example
P
Rr Yy
x
RRYy
Probability of Rr Yy in offspring;
1/2 Yy
1/4 YY
1/4 yy
1/2 Rr
1/2 RR
1/4 RrYy
Using Probability
Lecture 3 Example
YYSs x YySs
YY x Yy
Independent
Assortment
Ss x Ss
gametes
YY or Yy
.5
.5
(p) Y_ = 1
SS
Ss
ss
.25 .5
.25
Random Segregation
probability
(p) S_ = .75
Product Rule: (p) Y_S_ = .75
(p) ss = .25
Product Rule: (p) Y_ss = .25
Humans ?
• Is it possible to ascertain the mode of inheritance
of genes in organisms where designed crosses and
the production of large numbers of offspring are
not practical?
Pedegree: an orderly diagram of a
families relevant genetic features.
Assignment: figure out this pedigree.
Albinism is a recessive trait
in humans.
From Previous Page
Assignment: figure out this pedigree.
Symbols
More Symbols
And more…
2
Where Do you Start?
Aa Aa or AA Aa
Aa
aa
aa
aa
Recessive Trait? or Dominant Trait?
What More Can You Say?
Aa Aa or AA Aa
Aa
aa
aa
aa
Recessive Trait
Predictions
What if you were a genetic
counselor? What are the odds
that this individual carries the
trait?
Predictions
What if you were a genetic
counselor? What are the odds
that this individual carries the
trait?
Conditional Probability
Aa
Aa
?
1/4 AA
Monohybrid
Cross
1
1/4 Aa
:
1
1/4 Aa
:
1/4 aa
1
(p)Aa = 2/3 = .66
Conditional Probability
…is the probability of an event occurring
given that another event also occurs...
P(event) without the condition
p(condition)
Conditional Probability
Example: With a 6-sided die, what is the probability of
rolling a 2, given that an even number is rolled on the die:
p(2 roll | even #) =
p(2 roll | even #) =
=
p(2 roll)
p(even#)
1/6
1/2
1/3
probability without the condition
probability of the condition
Aa
Aa
?
p(probability of being a heterozygote)
= 1/2
p(probability of A_)
= 3/4
p( heterozygous | A_ ) =
1/2
3/4
= 2/3
Conditional Probability
p(A|B) = p(event) without the condition
• Use the formula,
p(condition)
• Or use a Punnett Square,
• Or...
1/4 AA
1
1/4 Aa
:
1
1/4 Aa
:
1
1/4 aa
A
a
A
AA
Aa
a
Aa
aa
Kidney Disease
If 1 and 2 had an offspring, what is the probability that
their first kid would show the phenotype?
A Simplification
• Unless otherwise specified (or the pedigree
suggests otherwise), the traits that we will
track will be rare,
• We will assume a p = 0 that a non-familial
mate carries the trait.
Kidney Disease
• non-familial mates: from outside of the family,
• if k is the recessive trait, then these individuals are KK.
Kidney Disease
If 1 and 2 had an
offspring, what is
the probability
that their first kid
would show the
phenotype?
p(P1) heterozygous
x p(P2) heterozygous
x p(FF) homozygous recessive
1/2
x
1
x
1/4
= 1/8 = .125
Kidney Disease
If 1 and 2 had an offspring, what is the probability that the first kid
would be a boy, and show the phenotype?
Kidney Disease
And, what is
the probability
that this boy is
a carrier?
If 1 and 2 had an offspring, what is
the probability that the first kid
would be a boy, and show the
phenotype?
p(P1) heterozygous
p (boy)
x p(P2) heterozygous
x p(FF) homozygous recessive
1/2
x
1
x
1/4
x
1/2 = 1/16
Practice #1
Round (R) and Yellow (Y) are dominant.
Practice #2
Practice #3
Questions
• Don’t rely on the answers in the back of the
book to solve your problems…
• Don’t just solve them, but understand the
principles needed to solve them.
a
b
f
c
e
d
Assignments
• Read from Chapter 3, 3.6 (pp. 100-105),
• Master Problems…3.12, 3.15, 3.20,
• Chapter 4, Problems 1, 2,
• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 - 4.14,
4.16, 4.19 - 4.20.