The Analysis of the Mitosis and Meiosis - Lab Report Example

Number: Lab Day: Monday Tuesday AM Tuesday PM Lab Group: A B C D E F (delete the other options to leave your lab day and group) Introduction (500 word limit) Q1 Describe the differences between meiosis and mitosis (5%)? Mitosis and meiosis are two types of cell division process. They have the same phases of division, but these processes and their results are significantly different1. During mitotic cell division, each chromosome is divided into two identical chromosomes, which are distributed over the two new cells and formed two diploid somatic cells. During this division in mitosis, the hereditary factors do not change. At meiosis, the division process has two general stages. At the first stage, the number of chromosomes is decreased twice. As a result of the division of a single diploid cell, two haploid cells emerge. Therefore, during the meiosis, four meiotic sexual haploid cells with altered heredity are produced and then genetic information is mixed. The second meiotic division is the same as mitosis one. As in mitosis, in meiotic anaphase II the single sister chromosomes (called chromatids) separate and move to opposite poles of the cell1. Meiosis is the basis of sexual reproduction because it occurs in maturing germ cells and leads to the appearing of new hereditary characteristics in the chromosomes. Q2 Explain the difference between a gene and an allele using specific examples (5%) Gene is the coding DNA sequence that placed in the specific locus on the chromosome. Such coded chains displayed in different traits, such as color of skin or length of the stem. Allele is a variant of gene and can be dominant or recessive. For example, in diploid organism one gene contains two alleles; therefore, traits can be different. If both alleles are identical, the gene locus is homogenous. When alleles are different, such gene locus called heterogenic. A good example is the Huntington disease. Huntingtons disease has the autosomal dominant inheritance. Consequently, an affected person has at least one mutant allele of Huntingtin gene (HTT)2. The child, whose one parent is homogenous by disease, carries two different alleles of the HTT gene and is heterogenic by this gene. Therefore, this child inherited a mutant allele of HTT gene and the Huntingtons disease will progress. Another example is blood groups. Gene locus that encoded blood groups has three alleles — IA, IB, and IO3. These alleles determine compatibility of blood transfusions. The person carries one of the six possible genotypes – AA, AB, BB, AO, BO and OO. Each of genotypes produce one of the four possible phenotypes: "AB" heterozygotes, and "O" homozygotes, "A" AA homozygous and AO heterozygous and "B" BB homozygous and BO heterozygous genotypes4. Q3 What is the purpose of crossing-over during meiosis? (10%) Crossing-over is the process that occurs in prophase I of meiosis while homologous chromosomes (bivalents) are close together because of conjugation. During the crossing-over, the exchange between the corresponding sites of the intertwined chromatids of homologous chromosomes is occurred5. This process provides a recombination of paternal and maternal alleles in each linkage group. In different progenitors of gametes, crossing-over occurs on different sites on the chromosomes, thereby forming a large variety of combinations of alleles in the parental chromosomes. It is clear that the crossing-over as recombination mechanism is effective only when the corresponding genes of paternal and maternal chromosomes are represented by different alleles. Identical linkage groups after crossing-over do not obtain new combinations of alleles. Therefore, due to crossing-over new combinations of genes and more genetic variation are formed5. Such emergence of new features may be very useful during natural selection. Hypothesis and Analysis (25%) We have predicted that the underlined alleles are dominant: S Sweet: nice rounded corns T Starchy: slightly shrivelled corns & Y yellow corn P purple corn Complete the punnet square for each experiment based on the prediction for your monohybrid cross: Parents S T S SS ST Genotype sweet sweet Phenotype T ST TT Genotype sweet starchy Phenotype What is the expected ratio of phenotypes: 3 sweet : 1 starchy Complete the punnet square for each experiment based on the prediction for the dihybrid cross: Parents SY SP TY TP SY SSYY SSPY STYY STPY Genotype sweet yellow sweet purple sweet yellow sweet purple Phenotype SP SSPY SSPP STPY STPP Genotype sweet purple sweet purple sweet purple sweet purple Phenotype TY STYY SSPY TTYY TTPY Genotype sweet yellow sweet purple starchy yellow starchy purple Phenotype TP STPY STPP TTPY TTPP Genotype sweet purple sweet purple starchy purple starchy purple Phenotype 3 sweet yellow : 9 sweet purple : 1 starchy yellow : 3 starchy purple Describe in your own words how to perform a chi-squared analysis Firstly, the expected number of samples in a certain phenotype class should be counted. For this purpose, the total amount of observed samples should be divided by the number of phenotypic classes. Then, the degrees of freedom should be counted in such a way: the number of phenotypic classes minus 1 and further using formula for the X2 count. The last step is to compare obtained X2 and probability P (tabulated values for certain degrees of freedom) to verify null hypothesis. Results (20%) Present and analyse your data to determine if the predictions were correct using chi-squared analysis. Remember tables must include a title and a legend. Dihybrid Cross. The amount of dihybrid crossing samples was: Yellow sweet 150 Yellow starchy 48 Purple sweet 441 Purple starchy 142 The total amount is 781 samples. The total number of phenotypes classes is 16 (9+3+3+1). 781/16= 48.8125. Therefore, the expected number of samples in certain phenotype classes will be: 3 sweet yellow – 348.8125=146.4375 9 sweet purple – 948.8125=439.3125 1 starchy yellow – 148.8125= 48.8125 3 starchy purple – 348.8125=146.4375 The chi-squared analysis should be made. Therefore, X2 is 0.2409. The Degrees of freedom is 3 (4-1=3). For P 0.05 and 3 degree of freedom, the value of X2 is 7.8115. Obtained X2 is 0.2409, which is lower than 7.8115, so the null hypothesis (segregation in 9:3:3:1 ratio, traits are controlled by two genes) is accepted. Monohybrid Cross. The amount of monohybrid crossing samples was: Sweet 562 Starchy 209 The total amount is 771 samples. The total number of phenotypes classes is 4 (3+1). 771/4= 192.75. Therefore, the expected number of samples in certain phenotype classes will be: 3 sweet – 3192.75=578.25 1 starchy – 1192.75=192.75 The chi-squared analysis should be made. Therefore, X2 is . The Degrees of freedom is 1 (2-1=1). For P 0.05 and 1 degree of freedom, the value of X2 is 3.841. Obtained X2 is 1.8265, which is lower than 3.841, so the null hypothesis (segregation in 3:1 ratio, traits are controlled by one gene) is accepted. Discussion (500 word limit) Q4 Does your own data or class data provide the more robust statistical information? Explain the importance of sample size in statistical analysis? (10%) Our own data provide relevant statistical information because we obtain X2 value within the limits of statistical significance. Nevertheless, there can be a mistake, because in our research we relied on a small sample. The sample is a small group of objects of a particular class. For precise data, it is necessary to investigate all the objects of this class, however, because for practical reasons, only part of the population can be studied. This part of the population is called the sample. Further, statistical analysis allows the researcher to disseminate the patterns on the entire population with a certain degree of accuracy6. Creating a sample regulated by a number of mandatory requirements, the violation of which can lead to erroneous conclusions from the results of the study. Firstly, the sample size is important. The accuracy of evaluation of investigated parameters depends on the sample size. Here we should pay attention to the word "accuracy". Large sizes studied groups provide more accurate (but not necessarily correct) results (6). If the results of sample studies are necessary to interpret on the entire population, the sample should be representative. Representativeness of the sample suggests that it reflects all the essential properties of the population. In other words, in the studied classes of corn with different traits (yellow and purple, starchy and sweet) occur with the same frequency as in the entire population. Going back to sample size, it should be noted that it is closely related to the probability of occurrence of first and second types statistical errors. Such statistical errors can be caused by the fact that there studied not whole population but only a part. The first type error is the erroneous rejection of the null hypothesis. An error of the second type is an incorrect deviation of an alternative hypothesis, which means that the differences or relationships between classes are caused not by coincidence, but because of the influence of the studied factors6. Q5 In your Sordaria experiment we measured the frequency of crossing over. Explain how this can be used to determine the gene to centromere distance (10%). Crossing over is a process that conducts the exchange of genetic material between homologous chromosomes. The frequency (proportion) of crossover-produced ascospores can be used as a measure of the relative distance between the gene locus and the centromere. In Sordaria, the recombinant frequency would be equal to the number of recombinant asci per 100 total asci (both recombinant and non-recombinant). It can be shown in the following formula: Recombinant asci actually contains four non-recombinant as well as four recombinant versions of the chromosome (where the gene of interest resides) that is why the multiplication of the frequency by 0.5 is necessary7. Cited works 1. Ohkura H. Meiosis: An Overview of Key Differences from Mitosis. Cold Spring Harb Perspect Biol [Internet]. 2015 Jan 20 [cited 2015 Feb 2]; Available from: http://www.ncbi.nlm.nih.gov/pubmed/25605710 2. Sturrock A, Leavitt BR. The clinical and genetic features of Huntington disease. J Geriatr Psychiatry Neurol [Internet]. 2010 Dec [cited 2015 Apr 23];23(4):243–59. Available from: http://www.ncbi.nlm.nih.gov/pubmed/20923757 3. Reid ME, Mohandas N. Red blood cell blood group antigens: structure and function. Semin Hematol [Internet]. 2004 Apr [cited 2015 Apr 23];41(2):93–117. Available from: http://www.ncbi.nlm.nih.gov/pubmed/15071789 4. Cartron JP, Colin Y. Structural and functional diversity of blood group antigens. Transfus Clin Biol [Internet]. 2001 Jun [cited 2015 Apr 23];8(3):163–99. Available from: http://www.ncbi.nlm.nih.gov/pubmed/11499957 5. Handel MA, Schimenti JC. Genetics of mammalian meiosis: regulation, dynamics and impact on fertility. Nat Rev Genet [Internet]. 2010 Feb [cited 2015 Apr 23];11(2):124–36. Available from: http://www.ncbi.nlm.nih.gov/pubmed/20051984 6. Devane D, Begley CM, Clarke M. How many do I need? Basic principles of sample size estimation. J Adv Nurs [Internet]. 2004 Aug [cited 2015 Apr 3];47(3):297–302. Available from: http://www.ncbi.nlm.nih.gov/pubmed/15238124 7. Glase J. A study of gene linkage and mapping using tetrad analysis in the fungus Sordaria fimicola. Proc 16th Work … [Internet]. 1995 [cited 2015 Apr 23];1–24. 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