Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam
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Three hourly exams plus final exam (450 pts), You will have 1.5 hours to complete each exam, PowerPoint PPT Presentation


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…may seem really hard. …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.

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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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…may seem really hard.

…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


Know This…

1

1/2

1/2

1

1/4

1/2

1/4

1/2

1/2

1

Assignment: Correlate this with the observed phenotype.


= 9

= 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 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

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 Method


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

1/4 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 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

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--rr

1/4 YY

1/2 Yy

1/4 yy


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

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

1/16 +

2/16

Genotypes Y--rr

1/4 YY

1/4 rr

1/2 Yy

1/4 rr

= 3/16

yellow/wrinkled


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

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 x 1/2 = 1/8 yyRr

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

2/16 +

1/16

Genotypes yyR--

1/4 yy

1/2 Rr

1/4 yy

1/4 RR

= 3/16

green/round


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

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 yyrr

1/4 YY

1/2 Yy

1/4 yy


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

1/16

Genotypes yyrr

1/4 yy

1/4 rr

green/wrinkled


F2 via Forked Line

  • Y--R--yellow/round9/16

  • Y--rryellow/wrinkled3/16

  • yyR--green/round3/16

  • yyrrgreen/wrinkled1/16


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

YYRR

YYRr

YyRR

YyRr

YR

YYRr

YYrr

YyRr

Yyrr

Yr

YyRR

YyRr

yyRR

yyRr

yR

YyRr

Yyrr

yyRr

yyrr

yr

3/16 p = 0.1875


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

1/4 RR

1/4 yy

1/2 Rr

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

1/4 rr

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

Better

3/16 p = 0.1875


1/4 yy

3/4 R_

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

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

Best (?)


9/16 yellow round

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)


1/4 RrYY

1/2 YY

1/2 Yy

Example

P Rr YY x rrYy

Probability Rr YY in offspring;

1/2 Rr

1/2 rr


1/2 Rr

1/4 RrYy

1/2 RR

1/4 YY

1/4 yy

Example

P Rr Yy x RRYy

Probability of Rr Yy in offspring;

1/2 Yy


Independent Assortment

gametes

Random Segregation

.5

.5

.25

.5

probability

.25

Using ProbabilityLecture 3 Example

YYSs 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.


From Previous Page

Assignment: figure out this pedigree.


Symbols


More Symbols


And more…

2


aa

aa

aa

Where Do you Start?

Aa Aa or AA Aa

Aa

Recessive Trait? or Dominant Trait?


aa

aa

aa

What More Can You Say?

Aa Aa or 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?


Monohybrid

Cross

Conditional Probability

Aa

Aa

?

1/4 AA

1/4 Aa

1/4 Aa

1/4 aa

1 : 1 : 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)

p(even#)

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

1/2

= 1/3


Aa

Aa

?

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

probability without the conditionprobability of the condition

p(probability of being a heterozygote)

= 1/2

p(probability of A_)

= 3/4

= 2/3


A

a

AA

Aa

A

Aa

aa

a

1/4 AA

1/4 Aa

1/4 Aa

1/4 aa

1 : 1 : 1

Conditional Probability

p(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.


p(P1) heterozygous

x p(P2) heterozygous

x p(FF) homozygous recessive

1/2

x

1

x

1/4

Kidney Disease

If 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 Disease

If 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.


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.


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