Genetics and Analysis of Data Fall 2023 Lab Guide PDF

Genetics and the Analysis of Data Fall 2023, Lab Guide p.45  Objectives Concepts Outputs Explain how different phenotypes are Complete and answer tables obtained. and...

Genetics and the Analysis of Data Fall 2023, Lab Guide p.45  Objectives Concepts Outputs Explain how different phenotypes are Complete and answer tables obtained. and questions on monohybrid, & dihybrid crosses in your lab Predict results of mono- and dihybrid guide (including chi-square crosses in F1 and F2 generations. tables) Use Chi-Squared test to determine when data supports a hypothesis. Complete table 10 in your lab guide, and all questions on Explain the immunology and genetics blood typing, paternity, and of the human A-B-O blood groups. immunology Roadmap of Today’s Activities 1. Monohybrid Cross (p. 48-55, ~30 minutes) ○ Counting corn kernels! Do they match expectations? 2. Dihybrid Cross (p. 48-55, ~30 minutes) ○ Even more kernel counting 3. Drosophila Salivary Gland (p. 55-56, ~5 minutes) ○ Observing chromosomal banding 4. Blood Typing Exercise (p. 56-61, ~20 minutes) ○ Determining blood type based on clotting/lineage  Genetic Terms A pair of homologous Genes: units of information about specific chromosomes traits, passed from parents to offspring (hereditary). Found at a: Locus: a specific location on the A gene locus chromosome. A pair of alleles Alleles: Different molecular forms of a gene (a gene pair). Usually two for each genetic Three pairs of genes trait. The different forms arise by mutation. Allele Combinations Alleles may be either homozygous or heterozygous to each other: homozygous - have two identical alleles at a locus (AA or aa), heterozygous - two different alleles at a locus (Aa). Dominant alleles mask the expression of the recessive allele, and will be expressed in a heterozygote (a recessive one will not). In meiosis: alleles/genes for characteristics on different chromosomes assort independently of one another during metaphase I , and separate during anaphase I. New alleles/genes are recombined in the offspring. Genotypes and Phenotypes Genotype refers to particular genes an individual carries. Phenotype refers to an individual’s observable traits in respect to a particular characteristic. Cannot always determine genotype by observing phenotype. In a monohybrid and dihybrid cross , you cannot observe the genotype in the first generation of offspring (F1), but can observe the genotype in the second generation of offspring (F2). Mendelian Genetics Assumptions (rules) of Mendelian Genetics 1. Characteristics are discrete (eg: purple vs. white, tall vs. dwarf). 2. Genetic characteristics have alternate forms (Alleles), each inherited from one of two parents. 3. One allele is dominant over the other, and will determine the Phenotype. 4. Heterozygotic individuals produce gametes with an equal frequency of the two alleles (Law of Segregation). 5. Different traits have independent assortment: genes are unlinked (Law of Independent Assortment) Monohybrid Crosses Typically, a monohybrid cross is b/t a homozygous recessive genotype and a homozygous dominant genotype of the parent (P) generation. In the F1 generation, all will be the dominant phenotype. Breeding F1s produce the F2 generation. In the F2, the more abundant phenotype indicates the dominant allele (can be either homozygous dominant or heterozygous). Monohybrid Crosses Typically, a monohybrid cross is b/t a homozygous recessive genotype and a homozygous dominant genotype of the parent (P) generation. In the F1 generation, all will be the dominant phenotype. Breeding F1s produce the F2 generation. In the F2, the more abundant phenotype indicates the dominant allele (can be either homozygous dominant or heterozygous). Monohybrid crosses have an expected phenotypic ratio of 3:1 (dominant: recessive phenotypes). Chi Squared Test A statistical test to determine whether a set of observed values differs significantly from a set of expected values. If we expect Mendelian ratios, we can test whether inheritance of a trait follows a Mendelian pattern by counting phenotypes of individuals. If Mendelian inheritance is hypothesized, and there is significant (p < 0.05) deviation of observed from expected values, you reject the hypothesis. If not, you “fail to reject” the hypothesis (i.e: it is supported by your data). The larger the deviation of observed from expected results, the larger your Chi Squared Value. The larger your sample size, the smaller your Chi Squared Value will be. Chi Squared Example Phenotype 1 Phenotype 2 The larger the deviation of observed Observed from expected Expected results, the larger your Chi Squared Deviation Value. Deviation2 The larger your D2/Expected sample size, the Chi Squared Value smaller your Chi (P1 D2/Expected + Squared Value will P2 D2/Expected) be. Chi Squared Example Phenotype 1 Phenotype 2 To calculate “expected” values : Observed 413 159 1. Sum numbers for Expected observed phenotype 1 + observed phenotype Deviation 2. Deviation2 2. For a monohybrid D2/Expected cross, divide this sum by 4. One fourth will Chi Squared Value be your expected (P1 D2/Expected + number for recessive P2 D2/Expected) phenotypes. 3. Multiply your one fourth value by 3 to find value of three fourths: this will be your expected number of dominant phenotypes. Chi Squared Example Phenotype 1 Phenotype 2 To calculate “expected” values : Observed 413 159 1. Sum numbers for Expected 429 143 observed phenotype 1 + observed phenotype Deviation -16 16 2. Deviation2 256 256 2. For a monohybrid D2/Expected 0.597 1.79 cross, divide this sum by 4. One fourth will Chi Squared Value 2.387 be your expected (P1 D2/Expected + number for recessive P2 D2/Expected) phenotypes. 3. Multiply your one fourth value by 3 to find value of three fourths: this will be your expected number of dominant phenotypes. Chi Squared Table The more data classes you have, the larger the Chi Squared Value (more chances for deviation to occur). The “Degrees of Freedom ” row to use is the number of data classes you have minus 1, (eg: for purple and yellow, it will be 2-1 = 1 df). If your calculated Chi Square Value is smaller than the value in the column with your chosen p-value (usually 0.05), then deviation is not significant. If it is larger, then your deviation is significant. Dihybrid Cross In addition to having different colors, kernels on a corn ear can also have different textures (i.e. they can be smooth or wrinkled). When you are exploring two phenotypes at the same time you are looking at a dihybrid cross. Phenotypes can therefore be: Purple and Smooth Purple and Wrinkled Yellow and Smooth Yellow and Wrinkled Punnett Square- Dihybrid Cross R Su R su r Su r su Purple, Wrinkled Purple, Smooth R Su R/R | Su/Su R/R | Su/su R/r | Su/Su R/r | Su/su Yellow, Wrinkled R su R/R | Su/su R/R | su/su R/r | Su/su R/r | su/su Yellow, Smooth r Su R/r | Su/Su R/r | Su/su r/r | Su/Su r/r | Su/su r su R/r | Su/su R/r | su/su r/r | Su/su r/r | su/su Dihybrid crosses have an expected phenotypic ratio of 9:3:3:1 (dominant/dominant: dominant/recessive: recessive/dominant: recessive/recessive phenotypes). Chi Squared Test with Four Data Classes Class 1 Class 2 Class 3 Class 4 To calculate “expected” values : 1. Sum numbers for observed phenotype 1 + Observed 269 38 96 29 phenotype 2 + phenotype 3 + phenotype 4. 2. For a dihybrid cross, divide this sum by 16. Expected One sixteenth will be your expected number for recessive/recessive phenotypes. 3. Multiply your one sixteenth value by 3 to Deviation find your expected number of dominant/recessive and recessive/dominant Deviation2 phenotypes. 4. Multiply your one sixteenth value by 9 to D2/Expected find your expected number of dominant/dominant phenotypes. Chi Squared Chi Squared Test with Four Data Classes Class 1 Class 2 Class 3 Class 4 To calculate “expected” values : 1. Sum numbers for observed phenotype 1 + Observed 269 38 96 29 phenotype 2 + phenotype 3 + phenotype 4. 2. For a dihybrid cross, divide this sum by 16. Expected 243 81 81 27 One sixteenth will be your expected number for recessive/recessive phenotypes. 3. Multiply your one sixteenth value by 3 to Deviation 26 -43 15 2 find your expected number of dominant/recessive and recessive/dominant Deviation2 676 1849 225 4 phenotypes. 4. Multiply your one sixteenth value by 9 to D2/Expected 2.78 22.83 2.78 0.15 find your expected number of dominant/dominant phenotypes. Chi Squared 28.54  Observing the Giant Salivary Gland Chromosomes of Drosophila Many flies exhibit very large salivary gland chromosomes During the larval stage of metamorphosis the chromosomes in the salivary gland cells undergo many rounds of DNA replication, but fail to divide Before pupation these salivary gland cells are enormous and each will have giant chromosomes containing 1000’s of copies of DNA, easily viewed under a regular compound microscope (note that each of these chromosomes has a banded appearance) Due to their large size it is possible to correlate changes in chromosomal band patterns with genes that get expressed and are important during Drosophila larval development  Drosophila Giant Salivary Gland Chromosomes, 40x and 100x Dr. Iwantsch, Fordham University  Drosophila Giant Salivary Gland Chromosomes, 400x Dr. Iwantsch, Fordham University Human Blood Groups (A, B, O) The four phenotypes of the ABO blood groups in humans are determined by three alleles for the enzyme (I) that attaches A or B carbohydrates to red blood cells: IA, IB, and i. ○ The enzyme encoded by the I A allele adds the A carbohydrate, ○ The enzyme encoded by the I B allele adds the B carbohydrate, ○ And the enzyme encoded by the i allele adds neither. A and B are dominant over O. A and B are coexpressed if both genes are present (Hence AB blood), and thus exhibit codominance. An individual with blood type O is the recessive condition and has to have two copies of ii allele. Blood groups are an example of multiple alleles (genes exist in populations in more than two allelic forms). Immunology and Human Blood Groups (A, B, O) Antigens: substances/molecules recognized as being different/foreign from those that we have in our bodies naturally. When the immune system recognizes an antigen it produces antibodies to attack each specific antigen. Antibodies bind with their specific antigens, leading to their inactivation, and targeting them for destruction: Type A people have the A carbohydrate on RBCs and anti-B antibodies in their plasma (recognizing B carbohydrate as foreign). Type B people have the B carbohydrate on RBCs and anti-A antibodies in their plasma (recognizing A carbohydrate as foreign). Type AB people have both A and B carbohydrates on their RBCs have neither antibodies in their plasma. Type O people have neither A or B carbohydrates on their RBCs and have both antibodies in their plasma. Blood Types and Genotypes (multiple alleles) Human Blood Types and Antibodies Group Blood Contains Antibody Reaction Carbohydrate/ Antibody Addition of Addition of antigen Anti-A Anti-B A A Anti-B Yes No B B Anti-A No Yes AB A and B None Yes Yes O None Anti-A No No Anti-B Blood Typing Anti-A antibodies will attack and clot (agglutinate ) type A blood. Anti-B antibodies will attack and clot type B blood. Anti-A and Anti-B antibodies will attack and clot type AB blood. Anti-A and Anti-B antibodies will not attack and clot type O blood. Blood Transfusions In order to have a successful transfusion: Blood type A can be transfused with type A or type O. Blood type B can be transfused with type B or type O. Blood type AB can be transfused with any blood type. Blood type O can only be transfused with type O Blood Typing Exercise Following the instructions on page 53 of your lab guides, we will use commercially prepared sera to determine the blood types four fictional people, based on the results of an agglutination (clotting) test. Complete the following table and answer the questions on pages 54 and 55. Name Anti-A Serum clotting? Anti-B Serum clotting? Blood Type? ??? Mr. Smith Yes No A Mr. Jones No Yes B Mr. Green Yes Yes AB Ms. Brown No No as part of your activity!!)O (Determine these Transfusion of type B blood into a recipient with Type A blood The blood of the donor will clot in the recipient’s blood: The many anti B antibodies in the recipient’s blood will bind to and coat all the donor’s RBCs, making them stick and clump together. Not as concerned about donor antibodies sticking to recipient’s antigens- too few antibodies in the blood of the donor to trigger large amount of clotting by recipient. Transfusion of AB Type Blood into a recipient with Type A blood The blood of the donor will clot in the recipient’s blood: The many anti B antibodies in the recipient’s blood will bind to and coat all the donor’s RBCs, making them stick and clump together. Not concerned about donor antibodies sticking to recipient’s antigens because AB does not have serum antibodies. Transfusion of O Type Blood into a recipient with Type A blood The blood of the donor will not clot in the recipient’s blood (type O RBCs have no surface antigens): The recipient’s anti B antibodies do not have antigens to bind to in the type O donor’s blood. Not as concerned about donor antibodies sticking to recipient’s antigens- too few antibodies in the blood of the donor to trigger large amount of clotting by recipient. Transfusion of Type B Blood into Type AB Recipient The blood of the donor will not clot in the recipient’s blood: The recipient does not have surface antigen antibodies to attack the antigens on the donor’s RBCs. Not as concerned about donor antibodies sticking to recipient’s antigens- too few antibodies in the blood of the donor to trigger large amount of clotting by recipient. Blood Typing and Paternity Exercise We will look at paternity cases: blood (pheno) type of mother and child are stated. Specify the mother’s possible genotypes, then the child’s, to help you deduce the father’s. ○ Eg: If the mother is blood type B she could either be IBIB (homozygous) or IBi (heterozygous), so she should be listed as: IB__. Repeat above step for the child. Sometimes you can find the mother’s genotype from the information given: if the mother is type B and her child is type O (ii), the mother’s genotype has to be IBi (child has one allele from each parent). The father can have the following phenotype/genotype combinations: A/I Ai, B/I Bi, O/ii. Which blood (pheno) types would rule out an alleged father? Add a column to the table for whether the mother could successfully give a blood transfusion to the child. Outputs Notebook check: Complete tables and Chi square tests with two and four data classes (tables 3, 5 and 6) Complete anonymous blood typing exercise (table 7) Complete all questions about immunology and blood typing (pages 54 and 55)