Genetics Lecture

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 sperm A a A A/A A/a egg Zygote a a/A a/a When heterozygotes mate their offspring will have different phenotypes: If A is domia or the phenotype of nant to a, the two possible phenotypes will be the phenotype of a/a A/A A and A/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/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 genoshi–. type shi–/shi First, we can test to see whether the shibire allele is dominant or recessive. shi– shi–/shi x shi+ (wild type) shi+/shi ↓ shi+ 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 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 par– x shi–/shi shi– ↓ F1 (these flies must inherit both shi– and par–) Possible outcome F1 not paralyzed F1 paralyzed Complementation? Explanation Inferred genotype par– genotype can supply function missing in shi– and vice versa shi– and par– complement shi– and par– do not complement par– has lost function needed to restore shi– Let’s look more carefully at gene segregation in a cross between F1 flies. shi+ shi–/shi x shi+ shi–/shi What is the probability of a paralyzed fly in the next (F2) generation? par–/par par+ shi– shi+/shi shi–/shi shi– n Definition: p(a a) = a 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 a and b) = p(a a) x p(b 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 shi– from mother) = 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/A AT x AM/A AM AM F1 : AT/A Because the F1 don’t look like either parent, let’s assume that the alleles are codominant. Codominant Codominant: heterozygote different than either homozygote. Incomplete dominance: heterozygote expresses the traits of both homozygous parents. (Alternatively, the genes that differ could have a mixture of dominant and recessive alleles) F 2: AT/A AM AT/A AT 1 : AM/A AM 2 : 1 1/4 will look like Teosinte. For two genes that differ: AT/A AT 1/4 BT/B 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.