GENETICS F’13

Biology 2030

GENETICS F’13

Biology 2030

Renée Dawson, Ph.D. dawson@biology.utah.edu JTB335 T H 12:30-1:50 pm by appointment or after class

Tuesday 8:35 - 9:25 am LS107 2030-004 John McCormick

Tuesday 9:40-10:30 am LS107 2030-002 John McCormick

Tuesday 10:45-11:35 am LS107 2030-003 Miles Christensen *

Tuesday 11:50am-12:40pm LS107 2030-005 Ani Schjelderup *

 

Wednesday 8:35 - 9:25 am LS111 2030-006 Michael Guernsey

Wednesday 9:40-10:30 am LS111 2030-008 Michael Guernsey

Wednesday 10:45-11:35 am LS111 2030-009 Daniel Ence

Wednesday 11:50am-12:40pm LS111 2030-007 Daniel Ence

* Contact Miles and Ani to arrange office hours

 

Biol 2030 F’132-3:20pm T&H

Genetic Analysis:An integrated Approach

$130 ($275)

Turning Point clicker optional but recommended:$60 new $45 usedRegister Clicker on Canvasbefore the next class

-see Canvas for syllabus, slides Exam Practice, etc.

Device ID

1 Mastering Genetics (required)2 End of chapter questions (even # answers in back)3 Additional Problems Sets on Canvas

Biol 2030 is a problem-based class and the concepts build. The key to your success is doing many, many problems.

Be aware of your thought patterns. Make of list of places you get stuck

Blooms Hierarchy of Thoughtlisted from the lowest to highest levels of thought!

1. Knowledge - memorize, define, recall2. Comprehension - summarize, describe, discuss3. Application - use ideas in different situations, solve problems4. Analysis - compare and contrast, ask questions, separate

concepts into parts, understand relationships5. Synthesis - combine known facts with new idea, creative

solutions, hypothesize6. Evaluation - relevance, significance, identify new frontiers,

personal judgments.

For THIS lecture

• TEXT Basic Mendel (Chapter 2: 25-43)

For NEXT lecture

• TEXT Probabilities(Chapter 2: 44-50)

• First discussion review problems• APS #1 (Mendel)• Mastering due this Sunday

– Introduction to MG– Ch 2 Mendel pt 1

GENETICSBiology 2030

Lecture 1

Chapter 2

MENDEL’s

LAWS

GENETICSBiology 2030

Lecture 1

Chapter 2

MENDEL’s

LAWS

Alternative forms of a gene

lead to different traits• We are diploid meaning each person inherits 2 homologs

– 1 from mom and 1 from dad • Chromosome contain a DNA molecule• A gene can be thought of as:

the portion of DNA that encodes a protein (or functional RNA)

• Homologous chromosomes have genes at same loci (place)

• Can have alternative forms of a gene (alleles)

• Phenotype the appearance of an individual ABd• Genotype the genetic composition of an individual AABbdd–Homozygous carry the same alleles AA or dd–Heterozygous carry different alleles Bb

•Dominant trait is the one that shows in the heterozygote•Recessive trait does not show in heterozygote

–Bb is a carrier of recessive t–bb is homozygous recessive–not called a carrier even though it carries two copies of b

• Modern genetics began with Gregor Mendel’s quantitative experiments with pea plants1854-1863

• Without knowledge of DNA, chromosomes or meiosis described the units of inheritance and how they pass from generation to generation

In the garden of Mendel…

1 Removed stamensfrom purple flower

White

Stamens

Carpel

Purple

PARENTS(P)

OFF-SPRING

(F1)

2 Transferred pollen from stamens of white flower to carpel of purple flower

3 Pollinated carpel matured into pod

4 Planted seeds from pod

PEAS good choice of model system

-lots of variation, easy to detect

-easy to grow

-short generation time

-can do out crosses

-self fertilizing, keel closed

Mendel followed the Scientific Method

- Make initial observations about a phenomenon or process

- Formulate a testable hypothesis

- Design a controlled experiment to test the hypothesis

- Collect data from the experiment

- Interpret the experimental results, comparing them to those expected under the hypothesis

- Draw a conclusion and reformulate the hypothesis if necessary

see Figure 2.3

see Figure 2.4

• Mendel started with true breeding plants,

self-fertilization produces same traits generations after generation (homozygous)

• F1 showed only one of two parental traits

• Trait not seen in F1

reappeared in 25% of F2 (recessive)

• One gene with two alleles

– Purple (A) is dominant to white (a)

Monohybrid CrossAA aa

Aa

Aa x Aa

A_ aa

parental

First Filial

Second Filial

offspring of F1 X F1

Phenotypic ratio 3:1 3 A : 1 aGenotypic ratio 1:2:1 1 AA : 2 Aa : 1 aaThis tells you that one gene is involved!

Figure 2.4

see Figure 2.6

MENDEL’S FIRST LAW:SEGREGATION

From his experimental data, Mendel concluded that:

• An organism has two distinct “elements” or units (alleles) for each inherited characteristic (diploid).

• Two alleles of a trait will separate or segregate from each other into (haploid) gametes

• Gametes combine randomly to form offspring

Mendel's Inferences (Abstractions):

• Idea that a discrete particle of information (gene) transmits information governing each trait.

• Two copies of this information are present in the adult (diploid).

• One copy is transmitted to the progeny from each parent via the gametes

• There are different forms or versions of each bit of information (alleles).

• Dominant and recessive types of alternative versions.

• Phenotype vs genotype. Because of dominance the effect of recessive alleles are masked in the heterozygote (skip a generation)

Mendel started with 34 different varieties and found 7 traits that behaved the same way.

We call these “Mendelian” traits. They all have a trait that is encoded by a single gene with 2 alleles, and simple dominant/recessive

Figure 2.1

Quantification

Replication

Mendelian Traits

All crosses were the same; it did not matter if pollen came from purple or white plant (autosomal, not on sex chromosome)

Reciprocal Crosses

Figure 2.5

If RR, then all are round

r rR Rr Rr

R Rr Rr

If Rr, then only half are round

r rR Rr Rr

r rr rr

Figure 2.5

recessivedominantPhenotype

• Mendel was not an isolated genius– lots of botanist were working on plant hybridizations– not as lucky with choice of model systems, or traits (height)

– cultural difference observed didn't use math

• mendelweb.org original paper and discussion– publish 1866 and went virtually unnoticed– not an obscure journal widely distributed in England, US and

Russia – Darwin saw, took notes on next article in journal, no marks on

Mendel, didn't read?, wasn't wondering about source of variability

• Seed shape gene with two alleles: Round and wrinkled

• Seed color gene with two alleles:

yellow and green

• By looking at two characteristics

at once (dihybrid),

Mendel found 9:3:3:1 ratio,

each trait 3:1

so not affecting each other

3:1 Round:wrinkled

3:1 Yellow: green Round and yellow are independent

Dihybrid Crosses

See Figure 2.11

Independent Assortment of Two Traits

9

3

3

1

3:1 Round:wrinkled

3:1 Yellow: green

Round and yellow are independent

Figure 3.4

During gamete production, two (unlinked) genes segregate their alleles independently. The inheritance of one does not influence the chance of inheriting the other.

MENDEL’S SECOND LAW: INDEPENDENT ASSORTMENT

Do ONE 16 square dihybrid cross so that

you can SEE the results. After that do two independent monohybrid crosses and multiply.

P AABB x aabb

F1 AaBb

F1 self cross

AaBb x AaBb

9/16 A_B_

3/16 A_bb

3/16 aaB_

1/16 aabb

A 9:3:3:1 ratio tells you that two independent(non-interacting, unlinked)genes are involved!

A a

A AA Aa

a Aa aa

B BB Bb

b Bb bb

B b

The forked-line diagram is used to determine gamete genotypes and frequencies

See Figure 2.10

GENETICSBiology 2030

Lecture 2

Chapter 2

PROBABILITIES

GENETICSBiology 2030

Lecture 2

Chapter 2

PROBABILITIES

• The likelihood that an event will occur • The probability that a coin will land

heads up is ½

• The probability that a heterozygous individual (Bb) will produce a gamete with the B allele is ½

• Inheritance follows the rules of probability– multiplication and addition can be

used to determine the probability of certain events occurring

Probability

1/2 x 1/2 = 1/4 BB1/2 x 1/2 = 1/4 bb1/4 Bb + 1/4 bB = 2/4 = 1/2 Bb

READ 2.4 and “When to add and when to multiple 3 methods”document on Canvas

The probability of independent events = the product of the probability of each event

• If both parents are heterozygous (RrYy) what is the probability that they will produce a rryy child?

• Probability of a sperm with r allele = ½• Probability of a ova with r allele = ½• Probability of a rr child is ½ X ½ = ¼

• rr 1/4 x yy 1/4 = rryy 1/16

Product Rule

rr yy

rryy

see page 44

Sum RuleThe probability of dependent events

= the sum of probability of each event

•Parents are heterozygous for a trait, R. •What is the chance that their child carries at least one dominant R allele?

•Probability of child carrying RR = ¼•Probability of child carrying Rr = ½

•Probability of child carrying R_ = ¼ + ½ = ¾

see page 45

p probability of a boy ½; q probability of a girl ½

Binomial expansion: (p + q)3 p3 3p2q 3pq2 q3

p3 1/8 (3 boys); 3p2q 3/8 (2 boys, 1 girl); 3pq2 3/8 (1 boy, 2 girls); q3 1/8 (3 girls)

Binomial Probabilitypages 45-47

Pascal’s TriangleA shortcut to the binomial expansion.

see Figure 2.15

In a self-fertilized Gg pea plant, give the proportion of yellow and green peas in pods with six peas each

p probability of yellow peas 3/4; q probability of green peas ¼

Chi-Square (2) Analysis Tests the Fit Between Observed and Expected Outcomes

2 = (O E)2/E; 0 = observed valuesE expected values

Probability (P) value

is the probability that the results of an experiment will deviate from the expected results by chance

degrees of freedom (df)is the number of independent variables= number of outcome classes, n, minus 1

A statistically significant result from 2 analysis has a P value less than 0.05

P<0.05 means that the experimental result has less than 5% probability (1/20) of happening by random chance, thus the null hypothesis (random chance) is rejected

P values above 5% indicate a nonsignificant deviation between observed and expected results. failure to reject the null hypothesis (random chance)

Statistical Significance

The Chi Squared method allows you to determine if the observed and expected numbers are significantly different. GgLl x ggll (if unlinked)

Observed Expect (if unlinked) O-E

Phenotype Genotype O E d d2(dxd) d2/E

GL G_L_ 965 1/4 (575) 390 152,100 264.5

gl ggll 944 1/4 (575) 369 136,161 236.8

Gl G_ll 206 1/4 (575) -369 136,161 236.8

gL ggL_ 185 1/4 (575) -390 152,100 264.5

2= ∑ = 1002.6

The formula for chi squared is 2 = ∑ d2/E ∑ means the sum of E is expectedd(difference)=O(observed)-E(expected) d=O-E

df (degrees of freedom)=# phenotypes -1= (1 in this case) fix thisfor 1 df a 2 >110.83 gives you a P value of < 0.001 which is extremely significant.

• The "null hypothesis" is always that the "Observed values are NOT different than the Expected values".