c Now let's consider diploid organisms haploid gametes diploid(2n) or (egg) meIosIS diploid zygote c△ somatic cells (sperm) meIOSIs erm cells The genotype of the zygote will depend on which alleles are carried in the gametes Allele sperm gamete A A/a Zygote a/A a/a When heterozygotes mate their offspring will have different phenotypes: If A is dom nant 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 Wa. 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
Lecture 3 Now let’s consider diploid organisms: The genotype of the zygote will depend on which alleles are carried in the gametes. 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. A/A A/a a/A a/a A a A a sperm egg Allele in gamete Zygote
True Breeding: homozygous for all genes Say we have a true breeding line of shibire flies these flies are paralyzed and have geno type shi/shi First, we can test to see whether the shibire allele is dominant or recessive shi/shi x(wild type)shi*/shi+ all are shi/shi+ (The offspring from a cross of two true breeding lines is known as the Fi or first filial generation). The Fi 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 Fi 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/po x shi/shi- F1(these flies must inherit both shi- and par Possible outcome Complementation Explanation Inferred gen par genotype can supply par/pa F1 not paralyzed shi and pe complement function missing in shi and vice versa shi+/shh F1 paralyzed shi and par par has lost function shr/shi do not complement needed to restore shi Let's look more carefully at gene segregation in a cross between Fi flies shi-/shi shi/shi* What is the probability of a paralyzed fly in the next( F2)generation?
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– x (wild type) 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. 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? Explanation Inferred genotype F1 not paralyzed F1 paralyzed 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? par– genotype can supply function missing in shi– and vice versa par– has lost function needed to restore shi– shi– and par– complement shi– and par– do not complement par–/par+ shi+/shi– shi–/shi–
Definition: p()=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(shr from mother)=/2 p(paralyzed)=1/2x112=1/4 p(not paralyzed)=1-114=3, 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 te 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 F1all same and unlike either parent F2 50,000 plants 1/500 look like Teosinte and M1/500 look like Maize
Definition: p(a) = 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(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 na N
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 Br 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 dont look like either parent, let's assume that the alleles are 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) AM/AM 14 will look like Teosinte For two genes that differ: AT/AT 114 x 1/4:1/16 will look like Teosinte Similarly, for three genes the probability will be /64. For four genes it will be/256, and for five genes it will be 1/1024 Since w1/500 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 using mapping methods
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 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) F2: AT/AT AT/AM AM/AM 1 : 2 : 1 1/4 will look like Teosinte. For two genes that differ: AT/AT BT/BT 1/4 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
