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Lab. 7 Mendelian Genetics “Law of Independent assortment” Main topics: Mendel’s Laws of Inheritance: Law of Independent assortment Polygenic Trait. Law of Independent Assortment The Law of Independent Assortment states that different pairs of alleles are passed onto the offspring independently of each other. Therefore, inheritance of genes at one location in a genome does not influence the inheritance of genes at another location. Mendelian’s Second Rule (Law of Independent Assortment) Mendel observed that, when peas with more than one trait were crossed, the progeny did not always match the parents. This is because different traits are inherited independently – this is the principle of independent assortment. For example, his cross-bred pea plants with round, yellow seeds and plants with wrinkled, green seeds. Only the dominant traits (yellow and round) appeared in the F1 progeny, but all combinations of trait were seen in the self-pollinated F2 progeny. The traits were present in a 9:3:3:1 ratio (round, yellow: round, green: wrinkled, yellow: wrinkled, green). Explanation: The cross was made between plants having yellow and round cotyledons and plants having green and wrinkled cotyledons. The F1hybrids all had yellow and round seeds. When these F1 plants were self-fertilized they produced four types of plants in the following proportion: Yellow and round Principles of Genetics Yellow and wrinkled Green and round 1 Green and wrinkled =9 =3 =3 =1 The above results indicate that yellow and green seeds appear in the ratio of (3:1) Similarly, the round and wrinkled seeds appear in the ratio of (3:1). This indicates that each of the two pairs of alternative characters viz. (3:1) X (3:1) = 9:3:3:1 Yellow - green cotyledon color is inherited independent of the round - wrinkled character of the cotyledons. It means at the time of gamete formation the factor for yellow color enters the gametes independent of R or r, i.e., gene Y can be passed on to the gametes either with gene R or r. Cytological explanation of the results: In the above experiment yellow and round characters are dominant over green and wrinkled characters which can be represented as follows: Phenotype Alleles of genes (i) yellow color of cotyledons Y (ii) green color of cotyledons y (iii) round character of cotyledons R (iv) wrinkled character of cotyledons r Therefore, plants with yellow and round cotyledons will have their genotype YYRR and those with green and wrinkled cotyledons will have a genotype yyrr. These plants will produce gametes with gene YR and yr respectively. When these plants are cross pollinated, the union of these gametes will produce F1 hybrids with YyRr genes. When these produce gametes all the four genes have full freedom to assort independently and, therefore, there are possibilities of four combinations in both male and female gametes. (i) YR (ii) yR (iii) Yr (iv) yr This shows an excellent example of independent assortment. These gametes can unite at random producing in all 16 different combinations of genes, but presenting four phenotypes in the ratio of 9: 3: 3: 1. Principles of Genetics 2 Figure: Dihybrid ratio 2. The Chi-Square Analysis Evaluates the Influence of Chance on Genetics Data “∑χ2 = (O-E) 2/ E” Chi Square Example Let's Test the following data to determine if it fits a 9:3:3:1 ratio. Observed Values Expected Values 315 Yellow,Round Seed (9/16)(556) = 312.75 Round, Yellow Seed 108 Green,Round Seed (3/16)(556) = 104.25 Round, Green Seed 101 Yellow, WrinkledSeed (3/16)(556) = 104.25 Wrinkled, Yellow 32 Green, Wrinkled (1/16)(556) = 34.75 Wrinkled, Green 556 Total Seeds Principles of Genetics 556.00 Total Seeds 3 Probability Degrees of Freedom 0.9 0.5 0.1 0.05 0.01 1 0.02 0.46 2.71 3.84 6.64 2 0.21 1.39 4.61 5.99 9.21 3 0.58 2.37 6.25 7.82 11.35 4 1.06 3.36 7.78 9.49 13.28 5 1.61 4.35 9.24 11.07 15.09 Grain Expected Observed Expected [Obs No. - [Obs No. - Exp Phenotype Ratio Number Number Exp No.] (Column 1) (Column (Column (Column 5) 4) 2) 9/16 315 yellow& round green& 3/16 108 & 3/16 101 556 x 9/16 = 2.25 0.0161 556 x 3/16 = 3.75 0.134 556 x 3/16 = -3.25 0.101 104.25 1/16 32 wrinkled Total No. (Column 6) 104.25 wrinkled green& ÷ Expected No. 312.75 round Yellow No.]2 381 x 1/16 = -2.75 0.217 34.75 ---------- 556 0 Number of classes (n) = 4 df = n-1 =(4-1) = 3 Chi-square value tab. = 7.82 Principles of Genetics 4 X2= 0.4681 Enter the Chi-Square Table at df = 3 and we see the probability of our chi-square valuetabulated=7.82. By statistical convention, we use the 0.05 probability level as our Critical Value. If the Calculated chi-square value is less than the 0.05 value, we accept the hypothesis. If the value is greater than the value, we reject the hypothesis. Therefore, because the calculated chi-square value is greater than we accept the hypothesis that the data fits a 9:3:3:1 ratio. Polygenic Trait: Examples: If two AaBbCc people would get married and have children. What are they get from the following: Number of gametes Genotypes resulting Number of individuals Answer: AaBbCc X AaBbCc N= The resulting number of gametes = 2N = heterozygous The number of genotypes resulting = 3N The resulting number of individuals = 4N = References: http://www.uic.edu/classes/bios/bios101/x202_assortment.pdf http://www.houstonisd.org/cms/lib2/tx01001591/centricity/domain/5363/s_bi_mendeliangenetics09_t.pdf http://www.ndsu.edu/pubweb/~mcclean/plsc431/mendel/mendel3.htm Principles of Genetics 5 23= 33= 43= Exercise 7 1- How many different kinds of gametes could an individual with the following genotypes produce? HhMm HHMmoo AaBbCC VvGgSsEe 2- When you cross the following a true-breeding strain of corn with purpled, smooth kernels to a true-breeding strain of corn with yellow, wrinkled kernels you find that the F1 progeny all have purple, smooth kernels. - You are unsure of the genotype of an individual F2 corn plant that has purple, smooth kernels and would love to know its genotype. A. What cross would allow you to determine the genotype of the purple, smooth corn plant? B. Describe the four different possible scenarios from your cross above. Indicate phenotypes and genotypes. 4- Use Punnett squares to show the genotype and phenotype of the offspring that would result from the following crosses: GgRr x GgRr GgRr x GgRR 5- From a cross between two plants, one completely heterogyzous tall green (Ssyy) and the other heterozygous tall yellow (SsYy), the following results were observed: Principles of Genetics 6 52 tall green 55 tall yellow 26 short green 27 short yellow A.Show by Punnett square (or other labeled diagrams) the cross between the two original plants and the expected phenotypic ratio of offspring. B.Use chi square calculations to determine whether or not the variations in the observed results could be due to chance. ( at level of frequency 5%) 6- Cross between two pea plants one of them is yellow seed with round shape (heterozygous traits) and other is green seed with wrinkled shape (homozygous traits). What is genotype and phenotype of first generation? 7- For the following crosses, determine the probability of obtaining the indicated genotype in an offspring Cross Offspring AAbb x AaBb AAbb AaBB x AaBb aaBB AABbcc x AaBbCc aabbCC AaBbCc x aabbcc AaBbcc Principles of Genetics 7 Probability