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Three hourly exams plus final exam (450 pts), You will have 1.5 hours to complete each exam,

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…should be relatively easy.

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,
- Exams - 150 points each, Final Exam cumulative.
- Quizzes will be givenevery 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

= 3

= 3

= 1

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

1/4 RR

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

1/4 RR

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

1/2 Rr

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

1/2 Yy

1/4 rr

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

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

1/4 RR

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

Forked-Line Method1/4 x 1/2 = 1/8 YYRr

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

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

1/16 +

2/16 +

2/16 +

4/16

Genotypes Y--R--1/4 RR

1/4 YY

1/2 Rr

1/4 RR

1/2 Yy

1/2 Rr

= 9/16

yellow/round

1/4 RR

1/2 Rr

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

1/4 rr

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

1/4 RR

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

1/2 Rr

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

1/4 rr

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

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

1/4 RR

1/2 Rr

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

1/4 rr

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

Genotypes Y--rr1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

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

1/4 rr

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

1/4 RR

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

1/2 Rr

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

1/4 rr

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

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

1/4 RR

1/2 Rr

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

1/4 rr

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

Genotypes yyR--1/4 YY

1/2 Yy

1/4 yy

1/4 RR

1/2 Rr

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

1/4 rr

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

1/4 RR

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

1/2 Rr

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

1/4 rr

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

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

1/4 RR

1/2 Rr

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

1/4 rr

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

Genotypes yyrr1/4 YY

1/2 Yy

1/4 yy

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?

3/4 round

3/4 yellow

1/4 wrinkled

3/16 yellow wrinkled

3/16 green round

3/4 round

1/4

green

1/16 green wrinkled

1/4 wrinkled

Forked-Line Method(phenotypes)gametes

Random Segregation

.5

.5

.25

.5

probability

.25

Using ProbabilityLecture 3 ExampleYYSs x YySs

YY x Yy Ss x Ss

YY or Yy SS Ss ss

(p) Y_ = 1

(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.

Albinism is a recessive trait in humans.

Assignment: figure out this pedigree.

Assignment: figure out this pedigree.

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

…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

Aa

?

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

probability without the conditionprobability of the conditionp(probability of being a heterozygote)

= 1/2

p(probability of A_)

= 3/4

= 2/3

a

AA

Aa

A

Aa

aa

a

1/4 AA

1/4 Aa

1/4 Aa

1/4 aa

1 : 1 : 1

Conditional Probabilityp(event) without the condition

p(condition)

- Use the formula,
- Or use a Punnett Square,
- Or...

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.

x p(P2) heterozygous

x p(FF) homozygous recessive

1/2

x

1

x

1/4

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

= 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?

And, what is the probability that this boy is a carrier?

p(P1) heterozygous

p (boy)

x p(P2) heterozygous

x p(FF) homozygous recessive

1/2

x

1

x

1/4

Kidney DiseaseIf 1 and 2 had an offspring, what is the probability that the first kid would be a boy, and show the phenotype?

x 1/2

= 1/16

Practice #1

Round (R) and Yellow (Y) are dominant.

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.

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.

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