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Chromosomal Theory of Inheritance 1. Around 1900, cytologists and geneticists began to notice connections between the behavior of chromosomes and the behavior of genes. a. Evidence for the theory i. Cytologists: 1879—Mitosis worked out; 1890—Meiosis worked out ii. Geneticists: 1860—Mendel proposed laws of segregation, and independent assortment. 1900—Mendel’s work rediscovered iii. Sutton & Boveri: 1902—Identified parallels between Mendel’s factors and behavior of chromosomes. b. These observations led to the development of the chromosome theory of inheritance. i. Genes are located on chromosomes ii. Chromosomes segregate and independently assort during meiosis 2. Thomas Morgan was the first to show that genes are located on chromosomes a. Morgan worked with Drosophila melanogaster, a fruit fly that eats fungi on fruit. b. Morgan spent a year looking for variant individuals among the flies he was breeding. He discovered a single male fly with white eyes instead of the usual red. i. The normal character phenotype is the wild type. ii. Alternative traits are called mutant phenotypes because they are due to alleles that originate as mutations in the wild-type allele. 1. He developed a different notation for genetic symbols. For example, the allele for the white eye mutation is symbolized by w, while the normal allele is symbolized by w+. 2. If the mutation is recessive, a lower case letter is used. Upper case is used for dominant mutant alleles. For example, curly wings is caused by a dominant allele and is symbolized by Cy, while normal wings is Cy+. c. When Morgan crossed his white-eyed male with a red-eyed female, all the F1 offspring had red eyes, suggesting that the red allele was dominant to the white allele. d. Crosses between the F1 offspring produced the classic 3:1 phenotypic ratio in the F2 offspring. i. On closer inspection, however, he noticed that only males had white eyes and, among males, he noticed only half of them had white eyes. ii. Morgan concluded that a fly’s eye color was linked to its sex and that the gene with the white-eyed mutation is on the X chromosome, with no corresponding allele present on the Y chromosome. 1. Females (XX) may have two red-eyed alleles and have red eyes or may be heterozygous and have red eyes. 2. Males (XY) have only a single allele. They will be red-eyed if they have a redeyed allele or white-eyed if they have a white-eyed allele 3. The number of genes in a cell is far greater than the number of chromosomes so it stands to reason that each chromosome must carry many genes. These genes would tend to be inherited together and are called linked genes. a. Linkage—Genes located on the same chromosome are inherited together and DO NOT assort independently. Results of crosses with linked genes are different from those expected according to independent assortment because the genes travel together i. Recombination of Unlinked Genes: Independent assortment of chromosomes ii. When Mendel followed the inheritance of two characters, he observed that some offspring have combination of traits that do not match either parent in the P generation iii. Recombinant Offspring—those that show new combinations of the parental traits; when 50% of all offspring are recombinants, geneticists say that there is a 50% recombination frequency Recombination of UNLINKED Genes YyRr x yyrr (Yellow round x Green wrinkled) ¼ yellow round (PARENTAL TYPE) ¼ green wrinkled (PARENTAL TYPE) ¼ YELLOW WRINKLED (RECOMBINANT) ¼ green round (RECOMBINANT) 50% PARENTAL AND 50% RECOMBINANT b. Linked Genes--- In one cross, Morgan considered body color and wing size i. The wild-type body color is gray (b+), and the mutant is black (b) ii. The wild-type wing size is normal (vg+), and the mutant has vestigial wings (vg) iii. The mutant alleles are recessive to the wild-type alleles iv. Morgan crossed F1 heterozygous females (b+bvg+vg) with homozygous recessive males (bbvgvg). 1. According to independent assortment, this should produce 4 phenotypes in a 1:1:1:1 ratio. 2. He found that most offspring had the same phenotypes as the parents, but other phenotypes were also observed 3. He reasoned that body color and wing shape are usually inherited together because the genes for these characters are on the same chromosome. 4. We would not expect linked genes to recombine into assortments of alleles not found in the parents because they travel on the same chromosome. 5. If the genes are completely linked, we should expect to see a 1:1:0:0 ratio with only parental phenotypes among offspring because no other combinations are possible 6. Although most of the offspring did have the parental phenotype, some of the flies were genetic recombinants, having phenotypes different from the parents. Phenotype Expected if unlinked Black normal 575 Gray normal 575 Black vestigial 575 Gray vestigial 575 Total 2300 Expected if Actual linked 206 1150 965 1150 944 185 2300 2300 #of recombinants /total = recombination frequency 206 + 185 divided by 2300 = .17 = 17% 4. 7. Conclusion: Genes were linked, but not totally. Morgan figured something must have broken the linkage and allowed the genes to be inherited separately. This process is called crossing over. It occurs during meiosis. Geneticists can use recombination data to map the location of genes on a chromosome. a. Crossing over should occur more often between two genes as the distance between those two genes increased. b. The percentage of recombinant offspring, the recombination frequency, is related to the distance between linked genes c. The farther apart two genes are, the higher the probability that a crossover will occur between them, and therefore, the higher the recombination frequency d. This means that the frequency of crossing over could be used to determine the distance between two genes on the same chromosome. e. We can express the distance between genes, the recombination frequency, as map units where “map unit” is defined as 1% crossing over frequency (also called 1 centimorgan) f. Linkage maps are “relative”; they don’t give the exact locations of genes, but now, cytogenetic maps can do just that. g. Example of a linkage map Data b--------17------------vg Loci Recombination Frequency cn-------9-----b b-vg 17% possibility #1: cn----9-----b----------17-------vg cn-b 9% cn-vg 9.5% 9 9.5 possibility #2: b---- -------cn------- --------vg conclusion— has to be possibility #2 because distance between b & vg is 17. (Note: not 18.5 because of double crossovers) 5. Sex Chromosomes a. In humans, anatomical signs of sex appear when embryo is about 2 months old i. XX = female; XY = male ii. Sry gene: sex determination region, only on Y chromosome. Triggers events that lead to testicular formation b. Sex linked disorders: some recessive alleles found on the X chromosome in humans cause certain types of disorders (i.e. colorblindness, duchenne muscular dystrophy, hemophilia) Color blind example: XN = normal; Xn= colorblind allele. Cross: female carrier x normal male XN Xn N N N N n X X X (normal female) X X (carrier female) Y XNY (normal male) XnY (affected male) c. In female mammals, only one X chromosome is active, therefore, males and females have the same number of copies of the genes on the X chromosome. d. During early embryonic development, one X chromosome per somatic cell condenses into a compact object near the nuclear membrane in the nucleus called a Barr body. Cytosine methylation (-CH3) is involved, and the Xist gene is thought to be involved too. The condensed Barr body is re-activated in cells that produce ova. i. Most of the genes on the Barr-body chromosome are not expressed. ii. The selection of which X chromosome will form the Barr body occurs randomly and independently in embryonic cells at the time of X inactivation. iii. As a result, females consist of a mosaic of two types of cells, some with an active paternal X chromosome, others with an active maternal X chromosome. 1. After an X chromosome is inactivated in a particular cell, all the cells that result from that cell by mitosis will have the same inactive X. 2. If a female is heterozygous for a sex-linked trait, approximately half her cells will express one allele, and the other half will express the other allele Modified from original source: http://kvhs.nbed.nb.ca/gallant/biology/chromosomal_theory_notes.html