1. Chapter 15 – The Chromosomal Basis of Inheritance

2. Mendel and Chromosomes

3. Genetic Problem
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 F1 and the F2 generations.
Fruit fly (Drosophila melanogaster) genetics:
Genetic Symbols:
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

4. Morgan’s results
This exact experiment was conducted by Thomas Morgan in the early 1900’s.
Do you think your results match what Morgan found for the F1? The F2?
F1 – yes, F2 – no!
What Morgan found in the F2 generation:
Red-eyed females
Red-eyed males
White-eyed males
Where are the white-eyed females? Can you explain his results?

5. Morgan’s Results
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 sex-linked inheritance
Are there any human traits that could follow a similar pattern?

6. Sex-linked Inheritance
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?

7. X Chromosome Inactivation
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.

8. Genetic Problem Part II
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
Show your results for 2300 offspring; both genotype and phenotype.

9. Morgan’s results Expected results: Morgan’s 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
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

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

11. Genetic Recombination
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
If receive 50% recombinants from a testcross then the genes are on different chromosomes.
Use Morgan’s data as an example.

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

14. 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.
Today, we know that this results from crossing over in meiosis.
Can use this type of information to create a genetic map of a chromosome

15. Challenge problem
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?

16. Linkage Maps
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

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

18. Challenge!
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; tall-downturned snout, 9.
Calculate the recombination frequencies for all three experiments.
Determine the correct sequences of these three genes on their chromosome.

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

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

21. Alterations to Chromosome Structure
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.)

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