DOC PREVIEW
MIT 7 03 - LECTURE NOTES

This preview shows page 1-2 out of 5 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Lecture 2 Lecture 2Lecture 2Lecture 2Lecture 2In this lecture we are going to consider experiments on yeast, a very useful organism for genetic study. Yeast is more properly known as Saccharomyces cerevisiae, which is the single-celled microbe used to make bread and beer. Yeast can exist as haploids of either mating type α (MATα) or mating type aaaaa (MATaaaaa). Haploid cells of different mating type when mixed together will mate to make a diploid cell. Haploids and diploids are isomorphic – meaning that a given mutation will cause essentially the same change in haploid and diploid cells. This allows us to look at the effect of having two different alleles in the same (diploid) cell. All yeast needs to grow are salts, minerals, and glucose (minimal medium). From these compounds, yeast cells can synthesize all of the molecules such as amino acids and nucleotides that are needed to construct a cell. The synthesis of complicated molecules requires many enzymatic steps. When combined, these enzymatic reactions constitute a biochemical pathwaybiochemical pathwaybiochemical pathwaybiochemical pathwaybiochemical pathwayConsider the pathway for the synthesis of the amino acid histidine. AAAAA→ BBBBB → CCCCC → DDDDD → histidine → Protein Enzyme: 11111222223333344444Each intermediate compound in the pathway is converted to the next by an enzyme. For example, if there is a mutation in the gene for enzyme 3 then intermediate C can not be converted to D and the cell can not make histidine. Such a mutant will only grow if histidine is provided in the growth medium.This type of mutation is known as an auxotrophic mutationauxotrophic mutationauxotrophic mutationauxotrophic mutationauxotrophic mutation and is very useful for genetic analysis. growth on minimal growth on minimal + histidine His+ (wild type) + + His– – + Phenotype:Phenotype:Phenotype:Phenotype:Phenotype: All traits of an organism (with an emphasis on trait under investigation) Homozygote:Homozygote:Homozygote:Homozygote:Homozygote: diploid with two like alleles of same gene Heterozygote:Heterozygote:Heterozygote:Heterozygote:Heterozygote: diploid with two different alleles of same gene Recessive Allele:Recessive Allele:Recessive Allele:Recessive Allele:Recessive Allele: trait not expressed in heterozygote genotype phenotype Mate to : diploid genotype diploid phenotype MATaaaaa His3– His– MATα His3– His3–/His3– His– MATaaaaa His3– His– MATα His3+ His3–/His3+ His+ Based on the His– phenotype of the His3–/His3+ heterozygote, we would say that His3– is recessive to wild type. Let’s consider a different kind of mutation giving resistance to copper that occurs in a gene known as CUP1. genotype phenotype Mate to : diploid genotype diploid phenotype MATaaaaa Cup1r copper resistant MATα Cup1+ Cup1r/Cup1+ copper resistant Dominant Allele:Dominant Allele:Dominant Allele:Dominant Allele:Dominant Allele: trait is expressed in heterozygote Cup1r is dominant to wild type (Cup1+). The terms dominant and recessive are simply shorthand expressions for the results of particular experiments. If someone says a particular allele is dominant that means that at some point they constructed a heterozygous diploid and found that the trait was expressed in that diploid.Note: Sometimes an allele will have more than one phenotype and may be recessive for one and dominant for another. In such cases, the phenotype must be specified when one is making statements about whether the allele is dominant or recessive. Consider for example, the allele for sickle cell hemoglobin in humans designated Hbs. Heterozygous individuals (Hbs/Hba) are more resistant to malaria, thus Hbs is dominant for the trait of malaria resistance. On the other hand, Hbs/Hba heterozygotes do not the debilitating sickle cell disease, but Hbs/Hbs homozygous individuals do. Therefore, Hbs is recessive for the trait of sickle cell disease. Once we find out whether an allele is dominant or recessive, we can already infer important information about the nature of the allele. The following conclusions will usually be true. RecessiveRecessiveRecessiveRecessiveRecessive alleles usually cause the loss of something that is made in wild type DominantDominantDominantDominantDominant alleles usually cause increased activity or new activity It turns out that the Cupr allele actually carries more copies of the gene for a copper binding protein and therefore increases the activity of the gene. Last lecture we defined the gene structurally as the DNA needed to encode a protein. We can now define a gene in a new way based on its function. Using the phenotypic difference between wild type and a recessive allele we can use a Complementation testComplementation testComplementation testComplementation testComplementation test to determine whether two different recessive alleles are in the same gene. Say you isolate a new recessive histidine requiring mutation that we will call HisX–. In principle, this mutation could be in His3 or it could be in any of the other genes in the histidine biosynthetic pathway. In order to distinguish these possibilities we need a test to determine whether HisX the same as His3. To carry out a complementation test, one simply constructs a diploid carrying both the His3– and HisX– alleles. An easy way to do this would be to mate a MATα HisX– strain to a MATaaaaa His3– strain. possibility genotype of diploid phenotype of diploid complementa-tion HisX= His3 His3–/His3– His– NoNoNoNoNoHisX≠His3 His3–/His3+, HisX–/HisX+ His+ Yes YesYesYesYesHaving performed this test, if the two mutations don’t complement we conclude that they are in the same gene. Conversely, if they do complement we conclude that they are in different genes. This test only works for recessive mutations. Think about what the outcome would be if HisX– were dominant. The complementation test can be thought of in the following way. If I have an allele with an observable phenotype whose function can be provided by a wild type genotype (i.e., the allele is recessive) — I can ask whether the function that was lost because of the recessive allele can be provided by another mutant genotype. If not, the two alleles must be defective in the same gene. The beauty of this test is that the trait can serve as a read-out of gene function even without knowledge of what the gene is doing at a molecular level.Definitions


View Full Document

MIT 7 03 - LECTURE NOTES

Documents in this Course
Exams

Exams

22 pages

Exams

Exams

64 pages

Exam 1

Exam 1

66 pages

Exam I

Exam I

93 pages

Exam Two

Exam Two

12 pages

Exams

Exams

27 pages

Exam 1

Exam 1

41 pages

Load more
Download LECTURE NOTES
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view LECTURE NOTES and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view LECTURE NOTES 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?