Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5,...

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Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25? CH3 problem set Today’s lecture: Review material from 9/18 Human Pedigrees Begin CH4 Today’s Lab: Maize segregating ears: Dihybrid cross and chi- square analysis *Exam I is one week from today! Reading assignment: Omit sections 4.8 and 4.9

Transcript of Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5,...

Page 1: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Lecture for Tuesday September 23, 2003What’s due?

CH2 problem setAssignments:

CH4 problems:

1-5, 8, 10, 11, 14, 16, 17, 21, 22

What’s due Thursday 9/25?

CH3 problem set

Today’s lecture:

Review material from 9/18

Human Pedigrees

Begin CH4

Today’s Lab:

Maize segregating ears: Dihybrid cross and chi-square analysis

*Exam I is one week from today!

Reading assignment:

Omit sections 4.8 and 4.9

Page 2: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Review:Trihybrid cross- A genetic cross between two individuals involving three characters (also referred to as a three-factor cross)The Forked-Line Method (branch diagram):

Recall:

*The F1 that result from a monohybrid cross (AA x aa) all have the genotype Aa and the phenotype represented by A*The F2 that result from a cross between 2 individuals from the F1, have a phenotypic ratio of 3:1*Assume independent assortment of the 3 gene pairs

KEY: We are

examining the

resulting phenotypes

!

Page 3: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Review Chi-Square Analysis:Mendel’s monohybrid and dihybrid ratios are predictions based on the following assumptions:

1. Each allele is dominant or recessive

2. Random segregation of alleles

3. Independent assortment

4. Fertilization is random

NOTE:

*The outcomes of 2-4 are “chance events” and are subject to random fluctuation

*As sample size increases, the average deviation from expected results decreasesEstablishing a null hypothesis (H0):

States that there is no difference between the observed and expected data

An Example (for a monohybrid cross):

The observed phenotypic ratio is 3:1

Page 4: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Review Chi-Square Analysis:The null hypothesis is analyzed statistically:

*It may be rejected or

*It may fail to be rejected

Chi-Square (X2) Analysis:

*Examines deviation between observed and expected numbers

*Accounts for sample size(o-e)2

X2 = e

Interpretation:

*determine df (n-1)

*typically use p value of 0.05 or greater (i.e. 0.01, 0.001)

*reject or fail to reject null hypothesis

Page 5: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Review Chi-Square Analysis:p value (probability): consider as a percentage (i.e. 0.05 = 5%)

*A level of error that is acceptable to the researcher in analysis of data

*5% of the time your result (or outcome) is due to chance

*95% of the time your results are not due to chance*If your calculated X2 is GREATER than that shown at p = 0.05, then you reject your null hypothesis

*Therefore, we CAN NOT reject our null hypothesis!

Example from Table 3.1:

Calculated X2 = 0.53

Page 6: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Human PedigreesPedigree- a family tree that shows the phenotype of a particular trait for each family member

= Female

= Male

=Unknown

*Shaded symbol=expressed phenotype*Individuals KNOWN to be heterozygous are half shaded

*Horizontal lines connect parents, vertical lines lead to offspring

*Proband (p)= individual in whom a genetically determined trait of interest is first determined

Page 7: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Chapter 4: Modification of Mendelian RatiosAllele- (short for allelomorph) alternative forms of the same gene

*Wild-type allele- allele that occurs most frequently in a population (arbitrarily designated as “normal”); usually dominant

*Mutant allele- allele that contains modified genetic information and often specifies an altered gene product

Conventional symbols for alleles:

recessive allele- initial letter of the name of the recessive trait, lowercased and italicized

dominant allele- same letter in uppercase

Tall = D Dwarf = d

Example:

BRCA1 or BRCA2- (humans) Breast Cancer susceptibility

SUPERMAN- (Arabidopsis) regulates genes involved in floral development

Genetic nomenclature is extremely diverse!

Page 8: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Incomplete or Partial Dominance

Incomplete dominance- expression of a heterozygous phenotype which is distinct from, and often intermediate to, that of either parent

Cross between parents with contrasting traits: Red flowers or white

flowers

Offspring with an intermediate

phenotype: pink flowers

Page 9: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Incomplete or Partial Dominance con’tCRCR x CWCW

CRCW

CRCW x CRCW

¼ CRCR ½ CRCW ¼ CRCW

Page 10: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Codominance:Codominance- Condition in which the phenotypic effects of a gene’s alleles are fully and simultaneously expressed in the heterozygote

Example:

MN Blood group- red blood cells contain a transmembrane glycoprotein (glycophorin); two different forms of this protein exist, M and N

Genotype LMLM

LMLN LNLN

Phenotype M MN

NLMLM X LMLN

¼ LMLM ½ LMLN ¼ LMLN

We can predict genotypic and

phenotypic ratios

Page 11: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Multiple Alleles- three or more alleles of the same gene

Examples:

*Table 4.1: over 100 alleles at a given locus in Drosophila

*ABO Blood group in humans

Multiple Alleles:

*Characterized by the presence of glycoprotein antigens on the surface of red blood cells

*Distinct from the M and N antigens

*Also exhibits codomiance

Genotype IAIA IAIO IBIB IBIO IAIB

IOIO

Antigen A A B B A,B

Neither

Phenotype A A

B B AB

O

Page 12: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Lethal Alleles:Lethal Allele- recessive allele in which a homozygous genotype results in death

Example: Coat color in mice

*A = agouti = wild-type allele

*AY = yellow = mutant allele

Dominant Lethal: Huntington’s disease (H); heterozygous individuals (Hh) have late onset

Page 13: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Combining modified modes of inheritance:

Page 14: Lecture for Tuesday September 23, 2003 What’s due? CH2 problem set Assignments: CH4 problems: 1-5, 8, 10, 11, 14, 16, 17, 21, 22 What’s due Thursday 9/25?

Gene interaction:Individual characteristics (discrete phenotypes) are often under the control of more than one gene

Epistasis- from the greek “stoppage”, interaction between genes such that one gene interferes with or prevents the expresion of another gene

Example:

In Drosophila, the recessive gene eyeless (when homozygous) prevents the expression of eye color genes present in genomeNovel phenotypes due to gene interaction

Example:

disc-shaped fruit (AABB) X long fruit (aabb)

F1 are all AaBb and disc-shaped

F2 Ratio 9/16 3/16 3/16 1/16

GenotypeA-B- A-bb aaB- aabb

Phenotype disc sphere sphere

long

Final phenotypic ratio 9/16 disc

6/16 sphere

1/16 long