Chapter 12 (Part III) Complementation Analysis

Similar documents
Chapter 25 Population Genetics

LAB. POPULATION GENETICS. 1. Explain what is meant by a population being in Hardy-Weinberg equilibrium.

Exercise 8C: Selection

Heredity and DNA Assignment 1

The Making of the Fittest: Natural Selection in Humans

[Presented by: Andrew Howlett, Cruise Slater, Mahmud Hasan, Greg Dale]

! Allele Interactions

Chapter 4 Lecture. Concepts of Genetics. Tenth Edition. Extensions of Mendelian Genetics. Copyright 2009 Pearson Education, Inc.

7.22 ANSWERS to Problem set #1

We can use a Punnett Square to determine how the gametes will recombine in the next, or F2 generation.

Extensions of Mendelism

Evolution. Population Genetics. Targets: Alleles and Genes 3/30/2014

1a. What is the ratio of feathered to unfeathered shanks in the offspring of the above cross?

Exam 1 Answers Biology 210 Sept. 20, 2006

Observing Patterns in Inherited Traits. Chapter 11 Updated Reading Not

Fall 2004 Animal Science 213 Animal Genetics EXAM Points Exam must be completed in INK!

1. You are studying three autosomal recessive mutations in the fruit fly Drosophila

SAMPLE MIDTERM QUESTIONS (Prof. Schoen s lectures) Use the information below to answer the next two questions:

Virtual Lab 2 Hardy-Weinberg

Trasposable elements: Uses of P elements Problem set B at the end

LS50B Problem Set #7

1a. What is the ratio of feathered to unfeathered shanks in the offspring of the above cross?

Chapter 9. Gene Interactions. As we learned in Chapter 3, Mendel reported that the pairs of loci he observed segregated independently

Content Objectives Write these down!

Chapter 14: Mendel and the Gene Idea

Biology 105: Introduction to Genetics Page 1 Midterm Fall 2004

Chapter 11 Reading Guide: Mendel and the Gene Idea

Evolution of Populations (Ch. 17)

Quiz will begin at 10:00 am. Please Sign In

BS 50 Genetics and Genomics Week of Oct 10

Mendel & Inheritance. SC.912.L.16.1 Use Mendel s laws of segregation and independent assortment to analyze patterns of inheritance.

7-1. Read this exercise before you come to the laboratory. Review the lecture notes from October 15 (Hardy-Weinberg Equilibrium)

Genetics Lecture Notes Lectures 6 9

BISC403 Genetic and Evolutionary Biology Spring, Summary of requirements for Exam 2 (to be given on March 24) plus exam 2 from Fall, 2010.

Measuring Evolution of Populations. SLIDE SHOW MODIFIED FROM KIM

Problem Set 2

genetics: epigenetics:

Exploring Mendelian Genetics 11-3

however, you need to purchase the Fly Lab manual from the bookstore in order to obtain a login name and password.

4 th Exam is Thursday, December 9. Review session will be at 5:00 PM Wednesday, December 8

Measuring Evolution of Populations

Chapter 14: Mendel and the Gene Idea

7.03 Problem Set 1 Solutions

AP Biology. Extending Mendelian genetics. Chapter 14. Beyond Mendel s Laws of Inheritance. Incomplete dominance. Incomplete dominance.

5. Alternate versions of the same gene, like purple and white flower color, are termed.

BSCI 410-Liu Homework#1 Key Spring 05

Genetic Equilibrium: Human Diversity Student Version

ch03 Student: If a phenotype is controlled by the genotypes at two different loci the interaction of these genes is called

GENETICS - CLUTCH CH.21 POPULATION GENETICS.

Variation Chapter 9 10/6/2014. Some terms. Variation in phenotype can be due to genes AND environment: Is variation genetic, environmental, or both?

*No in-class activities can be made up for unexcused absences. See syllabus.

Modes of Inheritance Adapted by Ellen G. Dow for QBIC Genetics Lab 2017

Gene Mapping. Biology 20. Principles of Gene Mapping & Practice Problems. See Freeman 2e pp ; or Campbell 7e pp

Fruit Fly Exercise 7 Level 2

Lecture 2-3: Using Mutants to study Biological processes

Genetics IV: Biochemical Genetics

AP BIOLOGY Population Genetics and Evolution Lab

Module 20: Population Genetics, Student Learning Guide

Chapter 14: Mendel and the Gene Idea

Population- group of individuals of the SAME species that live in the same area Species- a group of similar organisms that can breed and produce

The Theory of Evolution

of heritable factor ). 1. The alternative versions of genes are called alleles. Chapter 9 Patterns of Inheritance

LABORATORY 8: POPULATION GENETICS AND EVOLUTION

Chapter 02 Mendel s Principles of Heredity

BIOLOGY 3201 UNIT 4 EVOLUTION CH MECHANISMS OF EVOLUTION

thebiotutor.com A2 Biology Unit 5 Genetics

LAB ACTIVITY ONE POPULATION GENETICS AND EVOLUTION 2017

LINKAGE AND RECOMBINATION. Problem 4 An organism of genotype AaBbCc is test-crossed. The genotypes of the progeny were as follows:

Unit 10: Genetics. Chapter 9: Read P

Name Date Class. In the space at the left, write the letter of the term or phrase that best completes each statement or answers each question.

AP Biology Laboratory 8 Population Genetics Virtual Student Guide

LABORATORY 8. POPULATION GENETICS AND EVOLUTION

CHAPTER 12 MECHANISMS OF EVOLUTION

7.03 Problem Set 2 Due before 5 PM on Friday, September 29 Hand in answers in recitation section or in the box outside of

Chapter 14. Mendel and the Gene Idea

Module 20: Population Genetics, Student Learning Guide

Saccharomyces cerevisiae. haploid =

Unit 6: Genetics & Molecular Genetics Assessment

Exploring Mendelian Genetics

DESIGNER GENES * SOUTHERN POLY REGIONAL 2006

Name: Genetics - Fall 2004 Massachusetts Institute of Technology Professor Chris Kaiser Professor Gerry Fink Professor Leona Samson

Lecture 3A. Population Genetics

Both alleles are expressed / shown (in the phenotype). Accept: both alleles contribute (to the phenotype) Neutral: both alleles are dominant 1

Hardy-Weinberg Principle 4/5/09. Chapter 20. Godfrey H. Hardy: English mathematician Wilhelm Weinberg: German physician

Studies on the sex-specific lethals of Drosophila melanogaster. A third chromosome male-specific lethal mutant

Understandings, Applications and Skills (This is what you maybe assessed on)

What is Genetics? Genetics The study of how heredity information is passed from parents to offspring. The Modern Theory of Evolution =

ANSWERS TO Exam Questions from Exam 1 Basic Genetic Tests, Setting up and Analyzing Crosses, and Genetic Mapping. Number of a alleles

LINKAGE AND CHROMOSOME MAPPING IN EUKARYOTES

Biol 321 April 23, Complementary Gene Action

Linkage & Genetic Mapping in Eukaryotes. Ch. 6

FINDING THE PAIN GENE How do geneticists connect a specific gene with a specific phenotype?

Localized group of individuals belonging to the same species. Group of populations that have the potential to interbreed

The Evolution of Populations

Introduction Chapter 23 - EVOLUTION of

Mendel and The Gene Idea

The Making of the Fittest: Natural Selection in Humans

BIOLOGY 321 SPRING2013 ANSWERS TO ASSIGNMENT SET #4 Answers to Edition 9 text questions:

Biology 163 Laboratory in Genetics Midterm 2, Nov. 14, Honor Pledge: I have neither given nor received any unauthorized help on this exam:

i. allelic frequency c. reproductive isolation j. sexual selection d. allopatric speciation k. founder effect e. sympatric speciation

Transcription:

Biology 234 J. G. Doheny Chapter 12 (Part III) Complementation Analysis You can use a Genetic Dissection analysis to find which genes and which proteins are involved in a biochemical process. (For example, how many proteins are involved in forming the Drosophila Barrier-Insulator complex, that prevents heterochromatin from spreading into euchromatin.) A complementation analysis is something you have to do when you are doing genetic dissection. If you do a random mutagenesis screen you will create several different mutants, but you might have mutated the same gene more than once. A complementation analysis allows you to tell the difference between two mutations to the same gene and two mutations to two different genes. This way you can tell how many genes (and how many proteins) are involved in the biochemical process. (For example, how many genes form the Drosophila Barrier-Insulator.) Why you do a Genetic Dissection Analysis: To find out what genes (and therefore what proteins) are involved in a biochemical process. (For example: how many proteins form the Barrier-Insulator Complex in Drosophila?) Why you do a Complementation Analysis: To find out how many genes (and therefore how many proteins) are involved in the biochemical process. This is necessary because you start a Genetic Dissection Analysis by doing random mutagenesis screen in order to generate mutant strains of Drosophila. But you will probably mutate the same genes several times. Therefore, how do you tell the difference between two different mutations to the same gene, and two different mutations to two different genes? A Complementation analysis allows you to do this. A Random Mutagenesis Screen generates Dominant Negative Mutations: Mutations can be either dominant or recessive. The Wild-Type version of a gene is usually defined as the functional version of the gene, while a mutant version is the non-functional version. A functional version of a gene is usually dominant to a non-functional version (for example, the brown eye gene is dominant to the blue eye gene). However, there is a type of mutation called an antimorphic mutation, where the mutant version of the gene is dominant to the Wild-Type gene. That means that, if you have just one copy of the mutant version of the gene you will get the mutant phenotype, and having a Wild-Type version of the gene will not help. Antimorphic mutations are said to be Dominant Negative, because the mutant version of the gene is dominant over the Wild-Type version of the gene. When you do a random mutagenesis screen, and look for mutant phenotypes, the screen will favor finding Dominant Negative mutations, because the odds of mutating BOTH copies of the gene are extremely low. (The random mutagenesis screen may have generated many recessive mutations, but you d never see them because you would have to mutate BOTH copies of the gene in order to see the mutant phenotype.) Thus, when you do a random mutagenesis screen, and look for mutant phenotypes, the screen will preferentially find Dominant Negative mutations. 1

Dominant Negative Mutations are usually Homozygous Lethal: Antimorphic (dominant negative) mutations are generally very disruptive types of mutations. They have to be in order to give you a mutant phenotype, even when there is only one mutant copy of the gene present. They are usually so disruptive, in fact, that if you have TWO antimorphic mutations present, the organism will not survive long enough to be born. (They are said to be homozygous lethal.) For example, the mutant Fibroblast Growth Factor Receptor that gives rise to Achondroplasia (dwarfism) is a dominant negative mutation which is homozygous lethal. If you have two Wild- Type copies of the gene you will be normal height. If you have one mutant copy of the gene you are a dwarf. If you have two copies of the mutant version of the gene you will not survive to be born, and will be stillborn. The mutant allele of the Huntintin protein gene that gives rise to Huntington s Disease is also a dominant negative mutation that is homozygous lethal. Thus, people who develop Huntington s Disease are heterozygotes. Thus, organisms that carry two copies of an antimorphic, dominant negative mutation usually die before birth. Dominant Negative mutations are usually Homozygous Lethal. Dominant Negative Mutations are (usually) Denoted As A Lower-Case Letter With A d Superscript: The Wild-Type allele of a gene is denoted with an italicized capital letter (for example A) A dominant negative allele would be denoted as a d. The Theory Behind Complementation Analysis: If you do a random mutagenesis screen (in Drosophila, for example) you will generate mutant strains of flies that are carrying ONE dominant negative mutant copy of a gene. If you cross together two heterozygous strains of flies that are both carrying a mutation of the same gene, some of the offspring will get TWO copies of the same mutant allele and die. On the other hand, if you cross together two heterozygous strains of flies that are carrying mutations to two DIFFERENT genes, all of the offspring will be viable (will survive). So the two mutations are said to complement one another. For example, the following cross will complement: PARENT FLIES: Aa d ; BB X AA; Bb d (The parents carry heterozygous mutations to two DIFFERENT genes. One strain is carrying a dominant negative a d mutation and the other is carrying a dominant negative b d mutation) Offspring: AA; BB (viable) Aa d ; BB (viable) AA; Bb d (viable) and Aa d ; Bb d (vable) However, the the following cross will fail to complement, and one quarter of the offspring will be non-viable. (There will be a reduction in the number of offspring.) 2

PARENT FLIES: A ad ; BB X A ad ; BB Offspring: AA; BB (viable) Aa d ; BB (viable) a d a d ; BB (NOT VIABLE!) How You Carry Out A Complementation Analysis: Do a random mutagenesis screen to find mutant phenotypes. Then cross all of the strains to each other (one at a time), and make note of any crosses that give you significantly fewer than the expected offspring. Assume that any crosses that give you a reduced number of offspring are actually mutations to the same gene. Examples of Complementation Analysis Problems: Example 1: You want to find out how many proteins form the Barrier-Insulator complex in Drosophila melanogaster. You use the wm4 inversion as a reporter gene, do a random mutagenesis screen by feeding female flies EMS (a mutagen), and then generating three different strains of E(var) mutations. From the complementation analysis data below, how many genes, and how many proteins form the Drosophila Barrier-Insulator complex? (Answer: two (1,3) and (2)) 3

Example 2: You want to find out how many genes (proteins) are involved in eye formation in the model organism Zebrafish (Danio rerio). You do a random mutagenesis of fish, and generate 5 strains of fish that do not have eyes (five eyeless mutant strains). You cross them together and get the following results. How many genes are involved in Zebrafish eye development? (Answer: three genes (1,3,5) (2) and (4)) 4

Example 3: Vitis vinifera is a species of grapes used to make red wine. These grapes contain three types of sugar: glucose, sucrose, and fructose. You wish to know which genes (proteins) are involved in making fructose (how many genes are involved in fructose anabolism). You do a random mutagenesis screen, and generate 9 strains of grapes that do not make fructose. You do a complementation analysis. How many genes (proteins) are actually involved in making fructose in these grapes? (Answer: four genes (1,2,3) (4,5,6) (7,8) and (9) ) Practice Questions: Answer the following questions with one or two sentences. 1. What do you call a mutant allele of a gene that causes a mutant phenotype to be seen even though a Wild-Type allele of the gene is also present? (Two possible answers.) 2. Do random mutagenesis screens tend to generate dominant or recessive mutant alleles of genes? 3. Do random mutagenesis screens tend to favor the generation of dominant or recessive genetic mutations? Which. 4. Name two human genetic diseases that are caused by dominant negative mutations. 5. Name two human genetic diseases where the mutant allele is homozygous lethal. Be able to answer the following questions in one or two paragraphs: 1. What is an antimorphic mutation? 2. What is a dominant negative mutation? 3. What is a complementation test? 4. What is a complementation analysis used for? 5

5. Explain why random mutagenesis screens tend to generate dominant negative mutations? 6. What does the term homozygous lethal mean? 7. What percentage of the offspring will be non-viable: Aa d ;BB X AA; Bb d 8. What percentage of the offspring will be non-viable: Aa d ;BB X Aa d ; BB 9. What is a reciprocal cross? Be able to explain the following in one hand-written page or less: 1. Random mutagenesis screens create both dominant and recessive mutations, but only the dominant mutations are seen. Explain why. 2. What is a Complementation Analysis, what is it used for, and how do you do it? (You can use a specific example, like finding out how many proteins comprise the Drosophila Barrier-Insulator complex.) Suggested Problems from the Text Book: Chapter 6, Problem 77. 6