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,

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,No Make-up Quizzes, absolutely no exceptions,

can drop two (2) lowest quiz scores (except 1&2).

Total course points - 550

Grades

…may seem really hard.

…should be relatively easy.

Know This…

11/2 1/2

11/4 1/2 1/4

1/2 1/21

Assignment: Correlate this with the observed phenotype.

Mendel’s Results, F2 DihybridP generation cross: YYRR x yyrr F1 generation cross: YyRr x YyRr

• Y_ R_ = 315

• yyR_= 108

• Y_rr = 101

• yyrr = 32

= 9

= 3

= 3

= 1

Forked-Line Method

1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

1/4 rr

1/4 RR

1/2 Rr

1/4 rr

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

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 x 1/2 = 1/8 yyRr

1/4 x 1/4 = 1/16 yyrr

Genotypes Y--R--

1/4 YY

1/2 Yy

1/4 RR

1/2 Rr

1/4 RR

1/2 Rr

yellow/round

1/4 x 1/4 = 1/16 YYRR

1/16 +

1/4 x 1/2 = 1/8 YYRr

2/16 +

1/2 x 1/4 = 1/8 YyRR

2/16 +

1/2 x 1/2 = 1/4 YyRr

4/16 = 9/16

Genotypes Y--rr

1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

1/4 rr

1/4 RR

1/2 Rr

1/4 rr

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

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 x 1/2 = 1/8 yyRr

1/4 x 1/4 = 1/16 yyrr

Genotypes Y--rr

1/4 YY

1/2 Yy

1/4 rr

1/4 rr

yellow/wrinkled

1/4 x 1/4 = 1/16 YYrr

1/16 +

1/2 x 1/4 = 1/8 Yyrr

2/16 = 3/16

Genotypes yyR--

1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

1/4 rr

1/4 RR

1/2 Rr

1/4 rr

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

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 x 1/2 = 1/8 yyRr

1/4 x 1/4 = 1/16 yyrr

Genotypes yyR--

1/4 yy

1/4 yy

1/2 Rr

1/4 RR

green/round

1/4 x 1/2 = 1/8 yyRr

2/16 +

1/4 x 1/4 = 1/16 yyRR

1/16 = 3/16

Genotypes yyrr

1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

1/4 rr

1/4 RR

1/2 Rr

1/4 rr

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

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 x 1/2 = 1/8 yyRr

1/4 x 1/4 = 1/16 yyrr

Genotypes yyrr

1/4 yy 1/4 rr

green/wrinkled

1/4 x 1/4 = 1/16 yyrr

1/16

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

yr

yr

YYrr Yyrr

yyRR yyRr

YYRR YYRr

YYRr

YyRR YyRr

YyRr

YyRR YyRr

YyRr Yyrr yyRr yyrr

3/16 p = 0.1875

BetterYyRr x YyRr (F1)

1/4 yy

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

3/16 p = 0.1875

Best (?)YyRr x YyRr (F1)

1/4 yy 3/4 R_ 1/4 x 3/4 = 3/16 yyR_

Sum Law: 1/4 RR + 1/2 Rr

Forked-Line Method(phenotypes)

3/4

yellow

1/4 green

3/4 round

1/4 wrinkled

3/4 round

1/4 wrinkled

9/16 yellow round

3/16 yellow wrinkled

3/16 green round

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 Yy1/2 Rr

1/2 RR

1/4 RrYy

1/4 YY

1/4 yy

Using ProbabilityLecture 3 Example

YY x Yy Ss x Ss

YYSs x YySs

Independent Assortment

YY or Yy SS Ss ss

gametes

.5 .5 .5.25 .25 probability

(p) Y_ = 1 (p) S_ = .75 Product Rule: (p) Y_S_ = .75

(p) ss = .25 Product Rule: (p) Y_ss = .25

Random Segregation

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.

Albinism is a recessive trait in humans.Assignment: figure out this

pedigree.

Assignment: figure out this pedigree.

From Previous Page

Symbols

More Symbols

And more…

2

Where Do you Start?

Aa

Aa Aa or AA Aa

Recessive Trait? or Dominant Trait?

aa

aa aa

What More Can You Say?

Aa

Aa Aa or AA Aa

Recessive Trait

aa

aa aa

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

?

1/4 AA

Aa Aa

1/4 Aa 1/4 aa1/4 Aa

1 : 1 : 1 (p)Aa = 2/3 = .66

Monohybrid Cross

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) p(even#)

p(2 roll | even #) = 1/6 1/2

= 1/3

probability without the conditionprobability of the condition

?

Aa Aa

p(probability of A_)

p(probability of being a heterozygote) = 1/2

= 3/4

= 2/3p( heterozygous | A_ ) = 1/2 3/4

Conditional Probability

• Use the formula,

• Or use a Punnett Square,

• Or...

A a

A

a

AA Aa

Aa aa

1/4 AA 1/4 Aa 1/4 aa1/4 Aa

1 : 1 : 1

p(event) without the condition p(condition)

p(A|B) =

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

p(P1) heterozygous

1/2 = 1/8 = .125

x p(P2) heterozygous

1x

x p(FF) homozygous recessive

1/4x

If 1 and 2 had an offspring, what is the probability that their first kid would show the phenotype?

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

p(P1) heterozygous

1/2 = 1/16

x p(P2) heterozygous

1x

x p(FF) homozygous recessive

1/4x x 1/2

p (boy)

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?

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

c

de

f

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.