CHROMOSOMAL BASIS OF INHERITANCE

Transcription

1 CHROMOSOMAL BASIS OF INHERITANCE The fruit fly (Drosophila melanogaster) has been widely used for the study of eukaryote genetics because of several features: short generation time (two weeks) prolific breeder (one female = 200 eggs) easy to maintain in large numbers numerous mutants are available Morgan was one of the first to use fruit flies to study inheritance in animals. He needed a mutant of the wild-type fly and eventually discovered one spontaneous mutant with white eyes. (wild-type = w + or +; white eye = w) P 1 wild-type X white eye F 1 ALL wild-type F 2 3/4 wild-type; 1/4 white eye This looks like simple Mendelian inheritance, however, when the F 2 were examined closely the following was discovered: 2/4 wild-type females 1/4 wild-type males 1/4 white eye males!!! Morgan concluded that the mutant gene for white eye must be SEX-LINKED and therefore on the sex chromosome.

2 To be sex-linked, the gene was either on the X chromosome or the Y chromosome (one of the two kinds of sex chromsomes). Based on further genetic crosses, the gene was found to be on the X chromsome. Therefore, we can symbolize the original crosses as follows: P 1 X + X + x X w Y F 1 X + X w X + Y F 1 x F 1 X + X w x X + Y F 2 X + X + : X + X w : X + Y: X w Y LINKED GENES - genes for different characters that are located on same chromosome. Inheritance of linked genes IF NO CROSSING-OVER: In fruit fly- b + = gray body (wild-type) b = black body (mutant) vg + = normal wing (wild-type) vg = vestigial wing (mutant) P 1 (b + b + vg + vg + ) X (bbvgvg) chromosomes b + b + b b vg + vg + vg vg

3 F 1 (b + b vg + vg) chromosomes b + b vg + vg TESTCROSS of F 1 (expected if no crossing over): P 1 b + b vg + vg X bb vgvg gametes b + vg + : b vg X b vg F 1 1/2 b + b vg + vg (gray body, normal wing) 1/2 bb vgvg (black body, vestigial wing) Morgan s group actually observed the following results from the testcross: 965 gray body, normal wing (b + _vg + _) 944 black body, vestigial wing (bb vgvg) 206 black body, normal wing (bb vg + _) 185 gray body, vestigial wing (b + _vgvg) The last two types of individuals represent a new combination of the two traits NOT found in either of the parents (i.e. recombinants)!!! The results show the inheritance of linked genes WITH CROSSING OVER. recombination frequency = total # recombinants X 100 total # offspring

4 recombination frequency = 391 X 100 = 17% 2,300 17% of the time a crossover will occur between gene for body color (b) and gene for wing shape (vg). The frequency is related to distance between the two linked genes, i.e. two linked genes close to each other on a chromosome will show less frequent crossing over than two different linked genes that are farther apart on the chromosome. Two genes extremely far apart will have a cross over frequency of 50% and thus behave as unlinked genes, i.e. genes on different homologous chromosomes, like Mendel observed. Two of Mendel s traits were in fact linked but showed 50% crossing over! Would his results with these two genes have supported the Law of Independent Assortment if they were very close together on the chromosome? CHROMOSOME MAPPING (based upon crossover data) - Since the crossover frequency is related to distance between two linked genes we can use frequency to determine relative position of several genes along a single chromosome and speak of percent frequency as mapping units of distance, e.g. 1% cross-over frequency would be the same as 1 map unit. Example of locating the relative positions of linked genes based upon map units (cross-over frequency): Genes Map units A-B 8 A-C 28 A-D 25 B-C 20 B-D 33

5 C B A D CHROMOSOMAL BASIS OF SEX DETERMINATION - Several sex determination systems occur in nature. 1. XY/XO system Humans (Y determines maleness ) XXY = male XXXY = male XXX = female Fruit flies (Y has no activity) XY = male XXY = female XXXY = female

6 2. Z-W system (birds, some inseclts, some fishes) ZW = female ZZ = male (Why are parthenogenetically produced turkeys always male?) 3. Haplo-diploidy (most bees and ants) No sex chromosomes diploids = females (result from fertilized egg) haploids = males (develop parthenogenetically without chromosome duplication to form a diploid cell at beginning of development) SEX-LINKED DISORDERS IN HUMANS - Some inheritable diseases of humans are due to mutated genes found on the sex chromosome (especially the X chromosome). A recessive sex-linked gene will be expressed more in the male population than the female population because of the lack of an allelic counterpart on the Y chromsome that could mask the recessive gene on the X chromosome. That is to say, for the female to express the mutation she would need to have two recessive alleles (one on each X chromosome) while the male only needs one (on the single X chromosome). Some examples are: color blindness, Duchenne s muscular dystrophy, hemophilia GENE DOSAGE COMPENSATION - How does an organism compensate for one sex having a double dose of sex-linked genes while the other sex only has one dose?

7 In human females, one of the X chromosomes condenses into a compact, inactive chromosome called a Barr body. It can be seen in stained cells under the light microscope as a dark staining spot just inside the nuclear membrane. At the time of Barr body formation (sometime in embryonic development), each cell of the embryo will randomly decide which of the two X chromosomes to condense. Once formed, all cells derived via mitosis from each of these cells will have the same X chromosome as a Barr body. Bar bodies and the Calico cats. Calico cats allow us to visually see the results of this mosaic pattern of X chromosome inactivation. Male calico cats are very rare. How could you get a calico male? CHROMOSOMAL ALTERATIONS - 1. Alterations of chromosome numbers: a. aneuploidy - an extra chromosome or missing chromosome due to nondisjunction during meiosis. (2n + 1 or 2n - 1 chromosome number) b. polyploidy - one or more extra sets of chromosomes. (triploidy = 3n, tetraploidy = 4n, etc.) 2. Alterations of chromosome structure: a. deletions - missing a piece of the chromosome b. duplications - chromosome fragment joins other homologue c. inversions - rejoined fragment may be inverted from original orientation

8 d. translocations - chromosome fragment joins to a nonhomologue EXTRANUCLEAR INHERITANCE ( cytoplasmic inheritance or maternal inheritance )- Both mitochondria and chloroplasts contain genes (DNA) In most plants, chloroplasts are donated exclusively by the egg and thus all chloroplasts in the offspring have genes from the egg (not the sperm). In plants and animals, the mitochondria are donated exclusively by the egg and thus all mitochondria in the offspring have genes form the egg.