Fly Lab Report - Genetics

CROSS 1: ______Purple (PR)______ (female’s phenotype) x _____Shaven (SV)_______(male’s phenotype) Female’s genotype______pr/pr/sv+/sv+_______ x male’s genotype___pr+/pr+/sv/sv____ Female’s gametes:______pr_______ Male’s gametes:________sv_______ Describe the F1 generation produced by FlyLab JS: how many flies are there and what are their phenotypes? There are 993 flies. There are 503 observed Female +, resulting in a proportion of 0.5065 and 490 observed male +, resulting in a proportion of 0.4935. Are the mutations dominant or recessive? How do you know? The mutations are recessive because neither purple eyes nor shaven bristles are present in the offspring. F1 genotype:___________Wild type (+) x prsv__________ F1 phenotype:________Wild type (+)_____ F1 gametes:_____________________WT_____________________________ Draw a Punnett square which represents the production of the F2 generation: + SV + ++ +,SV PR +,PR SV,PR What is the expected phenotypic ratio? Be sure to include the phenotypes. Expected phenotypic ratio = 9:3:3:1 9 = Wild type (+) 3 = SV 3 = PR 1 = SV,PR Copy the data table representing the F2 generation from “Analyze” or “Lab Notes” into the space below. Add three columns to it. Title the first column “Expected”, and calculate the expected number of flies for each phenotype using the expected ratios and the total number of flies. Title the second column “Difference”, and fill it in by calculating the difference between the observed number of flies and the expected number of flies (observed - expected). Finally, title the third column “Difference /Expected”, and calculate the values by squaring the difference in the previous column and dividing it by the expected number of flies. The 2 sum of all the entries in the third column is 𝛘 . Copy your F2 data table here: 2 Expected Difference Difference^2 / Expected Phenotype 571.5 5.5 0.05 + 190.5 4.5 0.1 SV 190.5 5.5 0.16 PR 63.5 6.5 0.67 SV,PR What is 𝛘 ?______0.98_______ 2 How many degrees of freedom are there in these data?_____7______ What probability does this 𝛘 correspond to?___Difference between experimental and expected ratio_____ 2 Are the data from your cross consistent with your hypothesis about what they should have been? Explain. To a degree, they are consistent. The experiment demonstrates a 10:4:4:1 ratio with +, SV, PR, SV,PR respectively. Although the expected ratio is 9:3:3:1. It is not in 100% correlation but it is not far off either. TEST CROSS OF F1 FLY: Select the F1 female and then mate her to a double mutant male fly. Female’s genotype/phenotype:__________Wild type (+)________________________________ Female’s gametes:_____________________+,+____________________________________ Male’s genotype/phenotype:_______________SV,PR____________________________ Male’s gametes:________________________SV,pr_____________________________________ Draw a Punnett “square” which represents this test cross (NB: it does not have to be a square!): + SV + +,+ SV,+ PR +,PR SV,PR What is the expected phenotypic ratio? Include the phenotypes. Expected phenotypic ratio = 1:1:1:1 Copy the results of the test cross onto the next page and do 𝛘 analysis on them. What is 𝛘 ?__0.91____ 2 2 How many degrees of freedom are there in these data?_____3______ What probability does this 𝛘 correspond to?____Difference between experimental and expected ratio______ 2 Are the data from this cross consistent with your hypothesis about what they should have been? Explain. The experiment resulted in a 1:1:1:1 proportion of observed phenotypes. The data from this cross is consistent with the hypothesis that the result should have theoretically been. RECESSIVE EPISTASIS When two genes with two alleles each (one completely dominant over the other) affect different structures, then 4 different phenotypes are possible for the offspring. But what if both genes affect the same structure? Then it is possible to have anywhere between 2 and 4 phenotypes generated by the action of the two genes. We will consider the case in which two genes contribute to the formation of the same structure, but the second gene is reliant upon the first gene having successfully carried out its function before it can act. In this case, the ratio characteristic of recessive epistasis is seen. Although Mendel described the behavior of genes as they passed from one generation to the next, he did not have the slightest idea of how they influenced phenotypes or what made one allele dominant over another. But over 150 years have gone by since he published his work, and you are already aware that most genes encode proteins, which carry out most of the functions necessary to keep organisms alive. Generally speaking, a dominant allele encodes a functional protein which is able to do its job/carry out its function, while a recessive allele encodes a protein which is partly or wholly non-functional: in other words, unable to do its job properly. So for example, the gene that Mendel studied which determined whether a pea plant had purple or white flowers might have encoded an enzyme which synthesized purple pigment; a dominant allele would encode a functional form of that enzyme, resulting in purple pigment being made and deposited in flower petals, while the recessive allele would encode a non-functional form of the enzyme which could not synthesize purple pigment. If a plant had 2 recessive alleles, it could not make purple pigment at all and would consequently have white flowers (resulting from the absence of any pigment in the petals), while the one dominant allele in a heterozygous plant could produce enough purple pigment for purple flowers. (In actuality, Mendel’s purple flower gene does not encode the enzyme which makes purple pigment itself, but rather a transcription factor which determines whether the pigment-making enzyme is expressed. However, the principle is exactly the same - the dominant allele encodes a functional transcription factor which allows the enzyme to be made, while the recessive allele encodes a non-functional transcription factor which does not cause the enzyme to be made). The example of recessive epistasis given in the textbook is that of coat color in Labrador retrievers: the first gene (MC1R) encodes a protein called melanocyte-stimulating hormone receptor which determines whether eumelanin (a black or brown pigment) is made by melanocytes (cells which produce the various forms of the melanin pigments). If the dog has at least one dominant allele of this gene, the resulting functional receptor will activate the melanocytes to make eumelanin, resulting in a brown or black dog, but if the dog has only recessive alleles of this gene, the absence of functional receptor means the melanocytes will not make eumelanin. (However, they will make the yellow/orange pigment pheomelanin). The second gene encodes a protein which determines whether the eumelanin is black or brown: a dominant allele encoding a functional protein results in black eumelanin, while a recessive allele encodes a non-functional protein which cannot convert brown eumelanin to black eumelanin. However, this gene will never get a chance to act unless eumelanin is being made in the first place through the action of MC1R. R E Everts, J Rothuizen, & B A van Oost : “Identification of a premature stop codon in the melanocyte‐stimulating hormone receptor gene (MC1R) in Labrador and Golden retrievers with yellow coat colour” (2002); https://onlinelibrary.wiley.com/doi/abs/10.1046/j.1365-2052.2000.00639.x Michael D. Morgan, Erola Pairo-Castineira, Konrad Rawlik, Oriol Canela-Xandri, Jonathan Rees, David Sims, Albert Tenesa & Ian J. Jackson : “Genome-wide study of hair colour in UK Biobank explains most of the SNP heritability” (2018); https://www.nature.com/articles/s41467-018-07691-z CROSS 2: ________EY_________(female’s phenotype) x ___________PR________(male’s phenotype) Female’s genotype___________Wild type (+)_________ x male’s genotype________Wild type (+)_________ Female’s gametes:_____ey_______ Male’s gametes:___pr_______ Describe the F1 generation produced by FlyLab JS: how many flies are there and what are their phenotypes? The F1 generation produced a total of 1042 flies. 526 of them were female and 516 of them were male. They were all of the Wild type (+) phenotype. Are the mutations dominant or recessive? How do you know? The mutations are recessive as they are not present in any individuals of the F1 generation that was observed. F1 genotype:_____Wild type (+) x ey,pr__________ F1 phenotype:_____Wild type (+)________ F1 gametes:____________Wild type (+)___________________ Draw a Punnett square which represents the production of the F2 generation: + EY + +,+ EY,+ PR +,PR EY,PR What is the expected phenotypic ratio? Include the phenotypes. The expected phenotypic ratio is 9:3:4 9 = Wild type (+) 3 = PR 4 = EY For this cross, you may check the box that says “Ignore sex of flies” above the F2 data table. After you have done so, copy the data table representing the F2 generation from “Analyze” or “Lab Notes” onto the next page and calculate 𝛘 as before.Copy your 2 F2 data table here and calculate 𝛘 : 2 What is 𝛘 ?______1.81________ How many degrees of freedom are there in these data?_____2________ 2 What probability does this 𝛘 correspond to?____Difference between experimental and expected ratio______ 2 Are the data from your cross consistent with your hypothesis about what they should have been? Explain. The data from the cross is consistence with the hypothesis about what they should have been. The data shows a 9:3:4 ratio between +, PR, and EY respectively. The hypothesis was a 9:3:4 ratio. For each gene, what phenotype results from having at least one dominant allele and what phenotype results from having only recessive alleles? The phenotype that results from having at least one dominant allele is a wild type (+) offspring. The phenotype that results from having only recessive alleles results in either an EY or PR offspring. Which gene relies upon the upon the successful action of the other gene in order for it to be able to express itself? What must the other gene do for it? The recessive gene relies on the successful action of the other gene in order for it to be able to express itself. The other gene must not be expressed. CHI-SQUARE TEST The chi-square test is a way of determining whether a set of data fall inside or outside normal statistical variation for a model that predicts what is expected. For the theory behind it, see your textbook! Here is a set of instructions on how to perform the chi-square test: You should have (1) a set of experimental observations and (2) a model/hypothesis which predicts what your experimental results should be. If you have a number of offspring or individuals with different phenotypes, your hypothesis should predict what fraction of the offspring/individuals should be of each phenotype, e.g., if Mendel’s Laws of Segregation and Independent Assortment apply to the pea traits of shape and color, 9/16 of Mendel’s peas from a dihybrid cross should be round and yellow, 3/16 should be round and green, 3/16 should be wrinkled and yellow, and 1/16 should be wrinkled and green. 1. Add up the number of offspring/individuals in each category (i.e., of each phenotype) to determine the total number of offspring/individuals. 2. Using the total number of offspring/individuals and the expected fractions, calculate the expected number of each category (phenotype) of offspring/individual. 3. Calculate 𝛘 (chi-square) using the following formula: 2 𝛘 =𝛴(observed - expected) /expected 2 2 Be sure to include all categories/phenotypes of offspring/individuals when you calculate this. It is a sum, so you should have a single value of 𝛘 for a given cross/experiment, not a separate value for each category/phenotype! A low value for 2 𝛘 2 is good. 4. Determine the degrees of freedom, which is the number of categories/phenotypes minus one. This is because if the total number of offspring/individuals is regarded as fixed, there could be any number of offspring/individuals in all of the categories (phenotypes) except for the last one, which will have to assume a particular value so that the total number of offspring/individuals can be correct. 5. Use the table or the website below to ascertain the probability which corresponds to 𝛘 with the right degrees of freedom. A high probability is good, though we accept anything over 5% (0.05). If the probability is over 5%, your experimental results fall within the normal range of statistical variation of your predicted results. If the probability is under 5% but not by much, you should probably redo the experiment. If the probability is significantly under 5%, you should re-examine your experimental protocol and/or your hypothesis! 2