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.

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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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Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

…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

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.


Mendel s results f2 dihybrid p generation cross yyrr x yyrr f1 generation cross yyrr x yyrr

= 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


Forked line method

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


Genotypes y r

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


Genotypes y rr

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


Genotypes y rr1

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


Genotypes yyr

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


Genotypes yyr1

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


Genotypes 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

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


Genotypes yyrr1

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

1/16

Genotypes yyrr

1/4 yy

1/4 rr

green/wrinkled


F2 via forked line

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

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?


Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

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


Better

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


Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

1/4 yy

3/4 R_

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

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

Best (?)


Forked line method phenotypes

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)


Example

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


Example1

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


Using probability lecture 3 example

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

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.


Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

Albinism is a recessive trait in humans.

Assignment: figure out this pedigree.


Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

From Previous Page

Assignment: figure out this pedigree.


Symbols

Symbols


More symbols

More Symbols


And more

And more…

2


Where do you start

aa

aa

aa

Where Do you Start?

Aa Aa or AA Aa

Aa

Recessive Trait? or Dominant Trait?


What more can you say

aa

aa

aa

What More Can You Say?

Aa Aa or AA Aa

Aa

Recessive Trait


Predictions

Predictions

What if you were a genetic counselor? What are the odds that this individual carries the trait?


Predictions1

Predictions

What if you were a genetic counselor? What are the odds that this individual carries the trait?


Conditional probability

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 probability1

Conditional Probability

…is the probability of an event occurring given that another event also occurs...

P(event) without the condition

p(condition)


Conditional probability2

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 condition probability of the condition

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


Conditional probability3

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

Kidney Disease

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


A simplification

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 disease1

Kidney Disease

  • non-familial mates: from outside of the family,

  • if k is the recessive trait, then these individuals are KK.


Kidney disease2

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 disease3

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 disease4

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

Practice #1

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


Practice 2

Practice #2


Practice 3

Practice #3


Questions

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.


Three hourly exams plus final exam 450 pts you will have 1 5 hours to complete each exam

a

b

f

c

e

d


Assignments

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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