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Premium member Presentation Transcript CHI-SQUARE TEST: CHI-SQUARE TESTPowerPoint Presentation: A garden pea plant, heterozygous for the gene pair Ss, produced 31 smooth & 19 wrinkled Effect of the smooth allele (S) is dominant over that for wrinkled (s), a 3:1 ratio would have been expressed if the plant have been self fertilized or exact numerical ratio of 37.5 smooth to 12.5 wrinkled. Deviation from the expected sufficient reason to discard self-fertilization hypothesis and look for another explanation. Substitute the hypothesis that Ss plant was fertilized by a ss plant, which would have ideally yielded 25 smooth: 25 wrinkled seedPowerPoint Presentation: To decide whether to accept or reject hypothesis, we must therefore evaluate the size of the discrepancy between the observed and expected ratios Chi-square Chi-square is to find out the actual measurement of the size of the discrepancy between observed and expected result, so that we may accept or reject hypothesis. 2 = (Observed – Expected number) 2 Expected NumberPowerPoint Presentation: Smooth Wrinkled Total Observed (O) 31 19 50 Expected (E) ¾ x 50 = 37.5 ¼ x 50 = 12.5 50 O - E - 6.5 6.5 ( O - E ) 2 42.25 42.25 ( O - E ) 2 E 1.13 3.38 4.51PowerPoint Presentation: Level of Significance Discrepancies can be large/small Assign values of these two kinds of discrepancies we might say that large discrepancies are the largest 5% and small discrepancies are the remaining 95% On the basis of this we can propose to reject the hypothesis if the discrepancies fails into large class This particular 5% frequency value, that enables us to reject a hypothesis is called the 5% level of significance.PowerPoint Presentation: Goodness of Fit When the calculated value is less than table value. Then the fit is said to be good or the assumed ratio is correct. Degrees of Freedom Independent comparison that can be made in a set of data is called as degrees of freedom Degrees of freedom = Total number of class - 1 Limitation of Applied to frequencies and not to ratio or percent. Proper application the number of frequency in any of the class should not be below 5.PowerPoint Presentation: Multiple factor It is quite natural that small differences exist among individuals of similar genotype due to the effect of environment on genotype. On the other hand, there are some heritable differences also exist with continuous variation. Most of the economical traits show continuous variation and they are measurable or quantifiable.PowerPoint Presentation: Quantitative characters Quantitative characters are traits which show continuous variation and governed by a large number of genes called multiple genes or multiple factors or polymeric genes or polygenes . Their inheritance follows same mendelian principles.PowerPoint Presentation: Qualitative characters Qualitative characters show discontinuous variation and are governed by one or two major genes or oligognes .PowerPoint Presentation: Multiple factor Hypothesis (Nilson – Ehle)PowerPoint Presentation: Nilson-Ehle studied Kernel colour in wheat concluded that is a quantitative character He crossed true breeding red kernel whet (RR) with true breeding white (rr) and the F 1 was red (Rr) and the F 2 segregated for red and white in 3:1 ratio indicating the dominance of red over white. However, careful examination indicated the variation in red color among the red color progeniesPowerPoint Presentation: F 1 red was not as intense as one of the parents In F 2 he could observe two grades of red ie., one was red as that of one of its parent, two were higher red as that of F 1 individuals. In some crosses, a ratio of 15 red : 1 white was found in F 2 indicating that there are two pairs of genes for red colour that either or both of these can produce red kernels.PowerPoint Presentation: Finally he observed different shades of red in F 2 for red kernel types. The F 2 showed red shades and white as follows; Dark red : 1 Medium dark red : 4 Medium red : 6 15 Light red : 4 White : 1 Total : 16 It was concluded two duplicate dominant alleles R 1 and R 2 cumulatively decide the intensity of red colour and both R 1 and R 2 are in completely dominant over white. The high intensity of red colour depends on the numberPowerPoint Presentation: If four genes are determining the kernel color then , R 1 R 1 R 2 R 2 x r 1 r 1 r 2 r 2 dark red White R 1 r 1 R 2 r 2 dark redPowerPoint Presentation: ♂ ♀ R 1 R 2 R 1 r 2 r 1 R 2 r 1 r 2 R 1 R 2 R 1 R 2 R 2 R 2 (dark red) R 1 R 1 R 2 r 2 (Medium dark red) R 1 r 1 R 2 R 2 medium dark red R 1 r 1 R 2 r 2 light red R 1 r 2 R 1 R 1 R 2 r 2 medium dark red R 1 R 1 r 2 r 2 light red R 1 r 1 R 2 r 2 light red R 1 r 1 r 2 r 2 very light red r 1 R 2 R 1 r 1 R 2 R 2 medium dark red R 1 r 1 R 2 r 2 light red r 1 r 1 R 2 R 2 light red r 1 r 1 R 2 r 2 very light red r 1 r 2 R 1 r 1 R 2 r 2 light red R 1 r 1 r 2 r 2 very light red r 1 r 1 R 2 r 2 very light red r 1 r 1 r 2 r 2 whitePowerPoint Presentation: Genotype Genotypic ratio Phenotype R 1 R 1 R 2 R 2 1 Dark red R 1 R 1 R 2 r 2 2 Medium dark red R 1 r 1 R 2 R 2 2 Medium dark red R 1 r 1 R 2 r 2 4 Medium red R 1 R 1 r 2 r 2 1 Medium red r 1 r 1 R 2 R 2 1 Medium red R 1 r 1 r 2 r 2 2 light red r 1 r 1 R 2 r 2 2 light red r 1 r 2 r 2 r 2 1 white The F2 ratio in wheatPowerPoint Presentation: Hence, if two parents differ for the two genes the segregation was 1:4:6:4:1 provided both R 1 and R 2 contribute equally to the colour. If three genes are involved in F 2 segregation showed 1:6:15:20:15:6:1 for red shades and 1 for white.PowerPoint Presentation: Thus, Nilson-Ehle ’ s multiple factor states that i) for a given quantitative trait there could be several genes , which were independent in their segregation, but had cumulative effect on phenotype ii) Dominance is usually incomplete iii) Each gene contributes something to the strength of expression of character whereas its recessive allele does not of genes present dominance gene.LINKAGE: LINKAGE The tendency of genes to be passed on to next generation in group is known as linkagePowerPoint Presentation: SYMBOLS FOR LINKED GENES: Linked genes in an organism are indicated by two horizontal lines with the genes on one chromosome above the line and the genes on other homologous chromosome below the line e.g. CS/cs. Some times a single line or an oblique line is used. e.g. CS or CS/cs csPhases:: Phases: 1. Coupling phase: When the dominant allele at one locus with dominant allele at the others are situated on the same chromosome in one parent and two recessive alleles lie on the same chromosome in the other parent the condition is known as coupling phase.Phases:: Phases: 2. Repulsion phase: When the one dominant allele at one locus and recessive allele at the other locus (Ab) are situated on the same chromosome in one parent this condition is known as coupling phase. In this case one dominant and one recessive gene are linked and the genotype of hybrid will be AB/ab.PowerPoint Presentation: Chromosome theory of Linkage The chromosome theory of linkage of Morgan and Castle states that: i) The genes which show linkage, are situated in the same pair of chromosomes. ii) The linked genes remain arranged in a linear fashion on the chromosome. Each linked gene has a definite and constant order in its arrangement. iii) The distance between the linked genes determines the degree of strength of linkage. The closely located genes show strong linkage then the widely located genes which show weak linkage. iv) The linked genes remain in their original combination during the course of inheritance.PowerPoint Presentation: A linkage group is a group of genes situated on the same chromosome. The number of linkage groups will be equal to the haploid number of chromosomes which the species possesses. Thus, maize which has 10 pairs of chromosomes has 10 linkage groups.PowerPoint Presentation: Difference in linkage and independent assortment. Mendel’s law of independent assortment is applied only to those genes which are located on separate chromosomes, because, the linked genes of a linkage group (chromosome) inherit together. A dihybrid contains either linked genes or independently assorted genes, can be determined by test crossing it with a double recessive parent. The independently assorted genes give the test cross ratio of 1 : 1 : 1 : 1 and linked genes give the test cross ratio of 1 : 1PowerPoint Presentation: Example I. If genes occur on different chromosomes, they assort indecently and give a test cross ratio of 1 : 1 : 1 : 1 as follows: P 1 AABB x aabb P 1 Gametes : ( AB ) ( ab ) F 1 : A a B b Test cross : A a B b x aa bb Gametes : ( AB ) ( A b ) ( a B ) ( ab ) ( ab ) F 2 : ¼ A a B b : ¼ A a bb : ¼ aa B b : ¼ aa bb Or 1 : 1 : 1 : 1 (Test cross ratio).PowerPoint Presentation: The linked genes do not assort independently, but tend to stay together in the same combinations as they were in the parents. The linked genes give the test cross ratio of 1 : 1 as follows : Parents : AB/ab x ab/ab Gametes : (AB) (ab) (ab) (ab) F 1 : ½ AB/ab : ½ abab or 1 : 1 (test cross ratio)PowerPoint Presentation: The degree of intensity with which two independent genes are linked together is called the linkage value. It can be calculated from F2 frequency and is expressed as percentage. Intensity of linkage is measured in inverse sense as recombination fraction. Smaller this fraction, more intense or close is the linkage.PowerPoint Presentation: CALCULATION OF LINKAGE FROM F 2 DATA Following methods are used to calculate linkage from F2 data 1) Emmersons additive method 2) Immers product ratio method 3) Square root method 4) Maximum likelihood methodPowerPoint Presentation: In Emmersons additive method linkage value is calculated by following formula p 2 = E - M n p = linkage value E = Sum of parental classes (End classes) M = Sum of recombinant classes (middle ) n = Total population. Cross over value = 1 - P If linkage value or c.o. value is to be expressed in percentage it is multiplied by 100.PowerPoint Presentation: LINKAGE AMONG TWO LINKED GENES FROM TEST CROSS DATA From test cross data the linkage value in percentage can be determined by following formula. Linkage (%) = Parental combinations x 100 Total combinations You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.