Genetics and Phenotype of a Novel Mouse Mutant - Essay Example

BIO-M209 DATA HANDLING LINKED TO MOLECULAR MEDICINE work Using statistical methods for data analysis Tasks in the work: A series of experimental results have been collected and should be analysed by statistical methods. Data sets will be provided and the task will be to: (1) Select adequate statistical methods to analyse the data (2) Describe briefly the rationale why a method is reasonable and useful for the problems (3) Perform the analysis, (4) Answer the question(s) associated with the problems and (5) Draw adequate conclusions on the biological problems given The final results and any conclusions, which can be drawn from the results should be discussed briefly - but conclusively. The report should focus on the clear description, rationale and presentation of the analysis of the data. This is not an essay (!) but it is useful to introduce the question and the main conclusions from the data. The discussions should be short, but clearly define the major conclusions, which can be drawn from the data – or not. Calculations Calculations can be done “by hand” using the adequate formulas as presented in the lectures/seminars. They will also be found in textbooks on statistics. Calculations should be limited to adequate decimals. Statistical analysis packages may be used (like SPSS) or available online calculation programs Raw data are also posted for easier use on Blackboard (Excel format) (>BIO-M209 Coursework > Coursework 1 > File: Coursework 1_Supplementary Data_Raw Data) Tables of critical values for different distributions (t-Test, F-Test, Chi-square test) are found in most textbooks on statistics and freely available programs A selected list of some useful programs and potentially needed tables are also found on Blackboard (>BIO-M209 Coursework > Coursework 1 > Coursework 1_Supplementary Data_Tables.Formulas) Presentation in general: Do not simply present the calculations, but explain the rationale of using a statistical method and draw adequate biological conclusions from the results. It should state the result, its significance and any other conclusion(s), which may be relevant. Part 1: Genetics and phenotype of a novel mouse mutant Introduction: A novel mouse mutant and the corresponding protein, mKIAA058 (in the following ‘KIAA’), was identified and further analysed. In order to understand the functions of this gene/protein, a mouse strain containing a non-functional allele (mKiAA058-; Null-allele) was successfully established. First data indicated an autosomal recessive inheritance as heterozygous (KIAA+/-) as well as homozygous knockout (KIAA-/-) animals were identified. The deficiency of this protein (‘Knockout KIAA’) affects multiple tissues, including skeletal defects (delayed/reduced development of bone and cartilage; growth retardation) as well as a progressive form of vascular degeneration. Later, a corresponding disease in humans was found in a small number of very young patients. The prospects of the patients are not clear at the time and the analysis of the mouse model may provide some hints for the severity of the disease. The analyses of the molecular mechanisms underlying the disease are still ongoing and some problems and experiments linked to these studies are given in the following. Statistical tests may be used to answer some of the questions. Experiments 1 and 2: Genetics In order to define potential effects of the presence/absence of the KIAA protein on the inheritance patterns, a number of breedings were performed. The genotypes of the parents were known and the genotypes of the litters (age: 14-16 days) were analysed by allele-specific PCR reactions. Experiment 1: (Marks: 10) In three parallel experiments (1.1, 1.2, 1.3), crossings of 5 wildtype males (KIAA+/+) with 10 heterozygous females (KIAA+/-) were started and all litters (given as total number of mice) were genotyped and the numbers of all possible genotypes are given in the following Table 1. All tested mice appear normal and show no altered phenotype at the tested age (day 14-16). Breeding: male(KIAA+/+) x female(KIAA+/- ) Table 1: Experiment 1.1 1.2 1.3 Litters Total number 62 90 98 Genotypes KIAA+/+ 34 37 42 KIAA+/- 28 53 56 KIAA-/- 0 0 0 Experiment 2: (Marks: 20) In three parallel experiments (2.1, 2.2, 2.3), crossings of 5 heterozygous males (KIAA+/-) with 10 heterozygous females (KIAA+/-) from experiment 1 were started, all litters were again genotyped (age: day 14-16) and the numbers of mice with all possible genotypes are given in the following Table 2. Breeding: male(KIAA+/-) x female(KIAA+/- ) Table 2: Experiment 2.1 2.2 2.3 Sum (2.1-2.3) Litters Total 71 84 59 Genotypes KIAA+/+ 21 23 18 KIAA+/- 39 46 35 KIAA-/- 11 15 6 Questions to answer: (A) Discuss briefly which type of genetic inheritance would be most adequate to describe the distribution of genotypes? (B) Do the distributions of all genotypes (numbers of mice) in experiments 1 and 2 correspond to the expected numbers according to Mendelian laws? (C) Are there any differences seen in the amount of Wildtype and Heterozygotes? Are the overall numbers of genotypic normal mice (+/+ and +/-) versus Knockout (-/-) significantly different? (D) Are there any further conclusions possible? If yes, what could be possible biological explanations of the observed distribution/s? Which other experiments would be needed? SOLUTION a) A type of genetic inheritance describing the distribution of genotypes in the above experiments is dominant recessive. This is because both the genes of KIAA+/- and KIAA+/- are dominant and therefore the trait is inherited after fertilization by the offspring. b)The genotypes distribution from the two experiments corresponds to the Mendelian law of segregation whereby the two different alleles responsible for the gene KIAA is separate and the recombination which later takes place produced a different type of genotype trait, KIAA-/-, in the offspring. c) The amount of wild type genotypes and heterozygous genotypes in the experiments have no difference as a result of the dominance of the recessive gene in both the alleles. The overall genotypic number of normal mice (+/+ and +/-) and knockout genotypes (-/-) are significantly different in that the fertilization process leads to the emergence of the knockout genotypes. This enables one to identify the exact alleles that carried the mutant gene from the first to the second experiment after which the second fertilization produced the knockouts. d) In conclusion, gene sequence can be altered by many ways. Mutant genes have different effects to the health of an organism, and this majorly depends on the site of occurrence. Alteration of the function of the proteins that are essential also contributes to this. Another experiment that could have been done is isolation of the mutant genes from the genotypes of the parents and running them in polymerase chain reaction (PCR) machine. This will isolate the genotypes of the normal mice (KIAA+/+ and KIAA+/-) from the knockout gene (KIAA-/-) and the overall number in each side identified. Experiment 3: Bodyweight of various genotypes (Marks: 30) The lack of KIAA seems to affect various organ systems (see before) Therefore, a series of experiments were performed to define the growth of mice with various genotypes. Wildtype (KIAA+/+), heterozygous carriers (KIAA+/-) and homozygous knockouts (KIAA-/-) were analysed for their body weight (in grams: g) at different ages, 10-11 days, 20-22 days and 40-42 days. Only female animals were included in this analysis to avoid gender-specific variability. The following tables contain the collected raw data. (Raw data are available: >BIO-M209 >Courseworks >Coursework 1 >2014_Coursework 1_Supplementary Data_Raw data) Day10-11 Day 20-22 Day 40-42 (Bodyweight; gram) (Bodyweight; gram) (Bodyweight; gram) wt (+/-) (-/-) wt (+/-) (-/-) wt (+/-) (-/-) 6.5 6.1 5.7 16 16 14.3 19.8 17.1 14.3 6.9 6.1 6.3 14 15.1 15.7 18.2 17.2 15.9 5.4 6.9 5.8 13.9 14.7 13.9 18.8 17.9 15.1 6.6 5.6 6.5 14.7 14.8 14.7 19.1 17.9 16.1 6.2 6 5.6 16.2 13.7 15.4 17.8 17.8 15.6 5.6 5.6 5.6 15.2 15.5 12.8 18.9 18.4 14.8 6.3 6.3 5.8 14.2 15.7 13.5 16.9 14.4 14.8 6.2 6.4 6.5 15.8 15.2 14.1 17.3 18.7 13.1 6.9 6.1 6.1 16.2 13.4 13.7 17 16.9 5.9 6.3 16.2 15.9 17.2 6 6.5 15.3 13.8 18 5.9 6.5 14.9 17.8 6.1 6.6 16.4 6.5 14.9 6 Questions to answer: (A) Are there differences in the bodyweight of females between genotypes at various ages? Are these differences significant? Which data sets are ‘reasonable’ to be compared? What conclusions could be drawn? In order to know if there are any differences between the bodyweights, a ONE way ANOVA is carried out. Using SPSS the following table was obtained. Ho : µD 10-11 = µD 20-22 = µD 40-42 H1 : µD 10-11 ≠µD 20-22 ≠ µD 40-42  α = .05 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 13 80.5 6.192308 0.205769 Column 2 11 167.7 15.24545 0.836727 Column 3 9 163.8 18.2 1.04 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 892.5459 2 446.273 698.8848 6.87E-26 3.31583 Within Groups 19.1565 30 0.63855 Total 911.7024 32         Where (ss) is sum of square, and MS is the mean square and is given by (SS/df),. f-valye is the fstatic In this example, we reject H0 because both the p-value the 0.00BIO-M209 >Courseworks >Coursework 1 >2014_Coursework 1_Supplementary Data_Raw data) Questions to answer: (A) Which distribution would be most adequate to describe the probability of events? Define the relevant criteria for your choice. Poison distribution. (B) When 1 randomly selected rack is moved to a clean room, what is the probability that no affected cages (with diseased mice) are included in this rack? What is the probability that 2 transferred racks contain no cages with diseased animals, respectively? From the data given the total cages which have been affected is 32 while the total number of cages is (20*40) = 800. Therefore,  λ = (32/800)= 0.04 For P (k=0)= = 0.9608 From the data given the total racks which have been affected is 13 while the total number of cages is 40. Therefore,  λ = (13/40)= 0.325 For P (k=2)= = 0.0382 What is the probability that individual racks contain 1, 2, 3, 4 or 5 cages with affected animals? From the data given the total cages which have been affected is 32 while the total number of cages is (20*40) = 800. Therefore,  λ = (32/800)= 0.04 For P (k=1)= = 0.0384 For P (k=2)= = 0.000769 For P (k=3)= = 0.00001 For P (k=4)= = 0.00000102 For P (k=5)= = 0 Racks 01/2014 02/2014 03/2014 04/2014 05/2014 06/2014 1 0 0 0 0 - - 2 0 0 0 0 0 - 3 0 2 0 - - - 4 0 0 0 0 - - 5 3 0 0 0 - - 6 0 0 0 0 - - 7 0 0 3 0 - - 8 0 0 0 3 - - 9 0 0 0 0 0 - 10 0 3 0 0 0 - 11 0 0 0 0 0 - 12 0 - 0 0 0 - 13 1 - 0 0 0 - 14 0 - - 0 0 - 15 0 - 0 2 0 - 16 0 - - 0 0 - 17 0 - - 0 1 - 18 0 0 - 0 0 - 19 0 0 0 0 0 - 20 0 0 0 0 0 - 21 0 0 0 0 0 - 22 0 1 2 0 0 - 23 0 0 0 0 0 - 24 0 0 - 1 0 - 25 1 0 - 0 0 - 26 0 0 - 1 0 - 27 2 0 0 0 0 - 28 0 0 0 0 0 - 29 0 0 0 0 0 - 30 0 0 0 0 0 - 31 0 1 0 0 0 - 32 0 0 - 0 0 - 33 0 0 - 0 0 - 34 - 0 - 0 0 - 35 - 0 - 0 0 - 36 - 0 - 0 - - 37 - 0 0 2 1 - 38 0 0 0 0 0 - 39 0 0 0 2 0 - 40 0 0 0 0 0 - Coursework 1: Marking scheme Part Experiment Max. Marks Part 1 Experiment 1 Genetics 10 Experiment 2 Genetics 20 Experiment 3 Bodyweight 30 Experiment 4 Bodyweight 10 Experiment 5 Genotypes 10 Part 2 - Sporadic disease 20 Total 100 Read More