Genetics Lecture Notes 7 03 2006 Lectures 3 and 4 Lecture 3 Now let s consider diploid organisms The genotype of the zygote will depend on which alleles are carried in the gametes Allele in gamete egg A sperm a A A A A a a a A a a When heterozygotes mate their offspring will have different phenotypes If A is dominant to a the two possible phenotypes will be the phenotype of a a or the phenotype of A A and A a When we do breeding experiments it is important to know the genotypes of the parents But as you can see from the example above individuals with the dominant trait could be either A A or A a A method to control this type of variation is to start with populations that we know to be homozygous One way to do this is to keep inbreeding individuals until all crosses among related individuals always produce identical offspring This is known as a true breeding population and all individuals can be assumed to be homozygous True Breeding homozygous for all genes Say we have a true breeding line of shibire flies these flies are paralyzed and have genotype shi shi First we can test to see whether the shibire allele is dominant or recessive shi shi shi shi wild type x all are shi shi The offspring from a cross of two true breeding lines is known as the F1 or first filial generation The F1 flies appear like wild type therefore shi is recessive not expressed in heterozygote Say we have isolated a new paralyzed mutant that we call par We start with a true breeding par strain that we mate to wild type We find that the mutation is not expressed in the F1 heterozygotes and therefore is recessive To find out whether par is the same as shi we can do a complementation test since both mutations are recessive For this test we cross a true breeding par strain to a true breeding shi strain par par x shi shi F1 these flies must inherit both shi and par Possible outcome Complementation F1 not paralyzed shi and par complement par genotype can supply function missing in shi and vice versa shi and par do not complement par has lost function needed to restore shi F1 paralyzed Explanation Inferred genotype par par shi shi Let s look more carefully at gene segregation in a cross between F1 flies shi shi x shi shi What is the probability of a paralyzed fly in the next F2 generation shi shi Definition p a na N na number of outcomes that satisfy condition a N total number of outcomes of equal probability Probability problems can be solved by accounting for every outcome but usually it is easier to combine probabilities p paralyzed F2 fly p inherit shi from mother and inherit shi from father Product rule p a and b p a x p b note the product rule only applies if a and b are independent which is the case here since the allele from mother does not affect the allele from the father p shi from mother 1 2 p shi from father 1 2 p paralyzed 1 2 x 1 2 1 4 p not paralyzed 1 1 4 3 4 Thus in the F2 generation the phenotypic ratio will be 1 paralyzed 3 not paralyzed A 1 3 phenotypic ratio among the F2 in a breeding experiment shows that alleles of a single gene are segregating This actually constitutes a third definition of a gene Historically this was the first definition of the gene developed by Gregor Mendel in the 1860s Mendel was able to detect single genes segregating in pea plants because he looked at simple traits and started with true breeding strains Let s see how these ideas can be applied to a very interesting problem in the evolution of corn Domestic corn is derived from wild progenitor Teosinte There is no historical record of how the breeding was done to produce Maize but there is a genetic record of the differences between Teosinte and Maize recorded the genomic differences between these two species Maize and Teosinte can be crossed to give viable progeny Teosinte x Maize F1 all same and unlike either parent F2 50 000 plants 1 500 look like Teosinte and 1 500 look like Maize How many genes contribute to the differences between the two kinds of plants Let s designate the genes that differ as A B C D For each gene there are two alleles the allele present in Teosinte and the allele present in Maize For the A gene we will designate these alleles AT and AM respectively For the B gene there will be alleles BT and BM and so on for all the genes that differ Let s follow the A gene through the cross between Maize and Teosinte AT AT x AM AM F1 AT AM Because the F1 don t look like either parent let s assume that the alleles are incompletely dominant Incomplete dominance heterozygote expresses a trait intermediate between the traits of either homozygous parents Alternatively the genes that differ could have a mixture of dominant and recessive alleles F2 AT AT AT AM 1 2 1 4 will look like Teosinte For two genes that differ AT AT 1 4 AM AM 1 BT BT x 1 4 1 16 will look like Teosinte Similarly for three genes the probability will be 1 64 For four genes it will be 1 256 and for five genes it will be 1 1024 Since 1 500 look like Teosinte the conclusion is that 4 5 genes differ between wild corn Teosinte and domestic corn Maize Using modern methods it has been confirmed that there are about five significantly different alleles and several of these have been located using mapping methods Lecture 4 From the last lecture we followed gene segregation in a cross between a true breeding strain with a shibire mutation and flies from a wild type strain shibire mutant x wild type F1 all not paralyzed F2 3 not paralyzed 1 paralyzed This is the segregation pattern expected for a single gene But in an actual experiment how do we know that the phenotypic ratio is really 3 1 There is no logical way to prove that we have exactly a 3 1 ratio Nevertheless we can think of an alternative hypothesis then show that the alternative hypothesis does not fit the data Usually we then adopt the simplest hypothesis that still fits the data A possible alternative hypothesis is that recessive mutations in two different genes are needed to get a paralyzed fly In this case a true breeding paralyzed fly would have genotype a a b b Whereas wild type would have genotype A A B B F1 A a B b not paralyzed F2 p a a and b b 1 4 2 1 16 p a a and B 1 4 x 3 4 3 16 p A and b b 3 16 p A and B the rest 9 16 This is the …
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