n Chapter 15 The Chromosomal Basis of Inheritance

n Chapter 15 – The Chromosomal Basis of Inheritance

Mendel and Chromosomes

Genetic Problem n n Cross a white-eyed male fruit fly with a wild-type red-eyed female fruit fly. Red is dominant to white. Show your results for both the F 1 and the F 2 generations. Fruit fly (Drosophila melanogaster) genetics: Genetic Symbols: ¨ ¨ ¨ n n If mutant is recessive, first letter is lower case ( w = white eye) If mutant is dominant, first letter is capital (Cy = curly wings) Wild-type is designated by a superscript (Cy+ = straight, normal wings) Wild-type = normal or most frequently observed phenotype Mutant phenotypes = alternatives to wild-type due to mutations in wild-type gene

Morgan’s results n n This exact experiment was conducted by Thomas Morgan in the early 1900’s. Do you think your results match what Morgan found for the F 1? The F 2? F 1 – yes, F 2 – no! What Morgan found in the F 2 generation: ¨ Red-eyed females ¨ Red-eyed males ¨ White-eyed males n Where are the white-eyed females? Can you explain his results?

Morgan’s Results n n n Deduced that the gene for eye color is linked to sex, and is therefore on the X chromosome Possible female genotypes: w+w+, w+w, ww Possible male genotypes: w+ and w Led to the discovery of sexlinked inheritance Are there any human traits that could follow a similar pattern?

Sex-linked Inheritance n n A gene located on either sex chromosome is called a sex-linked gene Females can be carriers, while males are hemizygous (one allele) Examples in humans include: color blindness, Duchenne muscular dystrophy, and hemophilia If only need one X to survive, what do females do with the other X?

X Chromosome Inactivation n n One of the 2 X chromosomes in every female cell will become inactive and condense into a barr body The genes on this barr body are not expressed This process is random and independent of other embryonic cells (all subsequent mitotic divisions will include that active X chromosome) Discovered a gene called XIST (X-inactive specific transcript) that is active only in barr bodies. The RNA from this gene covers the X.

Genetic Problem Part II n n Cross a dihybrid wild-type fly (gray body, normal wings) with a double mutant (black body, vestigial wings). Use the following letter combinations: ¨ Gray body = b+ ¨ Black body = b ¨ Normal wings = vg+ ¨ Vestigial wings = vg n Show your results for 2300 offspring; both genotype and phenotype.

Morgan’s results n Expected results: ¨ Wild-type gray normal (b+bvg+vg) = 575 ¨ Black body, vestigial wing (bbvgvg) = 575 ¨ Gray body, vestigial wing (b+bvgvg) = 575 ¨ Black body, normal wing (bbvg+vg) = 575 n Morgan’s results: ¨ Wild-type gray normal (b+bvg+vg) = 965 ¨ Black body, vestigial wing (bbvgvg) = 944 ¨ Gray body, vestigial wing (b+bvgvg) = 206 ¨ Black body, normal wing (bbvg+vg) = 185

Morgan’s Results cont. n n n How could these numbers be so different? Discovered the concept of linked genes Genes on the same chromosome tend to be inherited together

Genetic Recombination n the production of offspring with new combinations of traits different from those combinations found in the parents; results from the events of meiosis and random fertilization ¨ Parental types = progeny same as parents ¨ Recombinants= progeny different from parents n n If receive 50% recombinants from a testcross then the genes are on different chromosomes. Use Morgan’s data as an example.

Morgan’s Example Phenotype Genotype Expected if unlinked Expected if completely linked Actual results Black, normal bbvg+vg 575 Gray, normal b+bvg+vg 575 1150 965 Black, vestigial bbvgvg 575 1150 944 Gray vestigial b+bvgvg 575 206 185 Recombination Frequency: 391 recombinants / 2300 offspring * 100 = 17%

Morgan’s Results Since 17% of the progeny were recombinants, the linkage must be incomplete. Meaning they are on the same chromosome, but at different locations on that chromosome. n Today, we know that this results from crossing over in meiosis. n Can use this type of information to create a genetic map of a chromosome n

Challenge problem n A wild-type fruit fly (heterozygous for gray body and red eyes) is mated with a black fruit fly with purple eyes. The offspring are as follows: wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45. What is the recombination frequency between these genes for body color and eye color?

Linkage Maps n Create a gene or linkage map for the following Drosophila traits: ¨ Frequency between b and vg loci is 17% ¨ Recombination frequency is only 9% between b and cn (cn a third gene locus on the same chromosome for cinnabar eyes) ¨ The recombination frequency is 9. 5 between cn and vg

Linkage Map n n The recombination frequency is slightly less for bvg because of double crossovers. A second cross over would cancel out the first thus reducing the frequency. Only an approximation

Challenge! n A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T=tall, t=dwarf), head appendages (A=antennae, a=no antennae), and nose morphology (N=upturned snout, n=downturned snout). Since the creatures are not “intelligent”, Earth scientists are able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For tall with antennae, the offspring are tall-antennae, 46; dwarf-antennae, 7; dwarf-no antennae, 42; tall-no antennae, 5. For antennae and an upturned snout, the offspring are: antennae-upturned snout, 47; antennae-downturned snout, 2; no antennae-downturned snout, 48; no antennae-upturned snout, 3. A third testcross using a heterozygote for height and nose morphology was also conducted. The offspring are: tall-upturned snout, 40; dwarf-upturned snout, 9; dwarf-downturned snout, 42; talldownturned snout, 9. n Calculate the recombination frequencies for all three experiments. n Determine the correct sequences of these three genes on their chromosome.

Answer! A and T = 12% n A and N = 5% n T and N = 18% n Sequence = T-A-N n

Alterations in Chromosome Number n n n Usually caused by nondisjunction Aneuploidy – having an abnormal number of a particular chromosome Trisomy – having one extra chromosome (2 n + 1) Monosomy – missing one chromsome (2 n – 1) Polyploidy – having an extra set of chromosomes (triploidy – 3 n, tetraploidy – 4 n, common in plants, rare in mammals)

Alterations to Chromosome Structure n n Deletion – removes a chromosomal segment Duplication – repeats a segment Inversion – reverses a segment within a chromosome Translocation – segment of one chromosome is moved to another chromosome (reciprocal translocation – exchange, nonreciprocal translocation – only one chrom. )

Human Examples Down syndrome – Trisomy 21 n Klienfelters syndrome – XXY n Turners syndrome – X n Cri du chat – specific deletion of chromosome 5 n Chronic myelogenous leukemia – translocation of chromosomes 9 and 22 n