a. After cutting the plasmid with EcoR1 restriction enzymes, the gel electrophoresis results would look like diagram b. The plasmid has three EcoR1 sites, meaning the cut would result in three seperate pieces. The lengths of those pieces would be 500, 900, and 1,000 bps. Option b displays three bands at those locations.
b. Diagram c. Sal1 cuts in two places and creates two pieces of plasmid DNA, one 800 bps long and one 1,600 bps. The bands shown in option c correspond with the sizes of those two pieces.
c. Cutting the plasmid with EcoR1, Sal1, and Bam1 all at once would result in 6 fragments, one 200 bp, two 300 bp, two 500 bp, and one 600 bp long. This would create four bands. Diagram d most nearly matches those four bands.
2.) HindIII SmaI
a. ___________________|_________________________|________________
|--------------2.5-----------------------3.0----------------------------2.0---------|
b. -------1.5----EcoR1----1.5-------
___________________|____________|_____________|________________
|--------------2.5-----------------------3.0----------------------------2.0---------|
3.) Sophie's coat color is a unique combination of brown, black, and white patches. Many factors contribute to this.
The brown and black portions of her fur can be explained by X-inactivation in females. Sophie is heterozygous for fur color, a sex-linked trait, so we'll let XF represent brown hair and Xf represent black hair. Sophie's genotype would, according to our example, be XFXf. For females, cells during embryonic development undergo X-inactivation, which is the repression of one of two X chromosomes by methylation. The methylated chromosome, either XF OR Xf, supercoils into a structure known as a barr body. Transcription of the barr body is blocked, and so only the allele on the uncoiled chromosme is expressed. Since X-inactivation occurs randomly in each cell, some of Sophie's fur is black and some of it is brown. Size of the patches of brown or black fur is determined by when during embryonic development X-inactivation occurs.
. The white spots on Sophie's coat are the result of a white spotting gene. This gene has an epistatic relationship with the other pigmentation genes, the ones that make her fur brown or black. The spotting gene causes the absence of color, therefore "covering up" portions of brown or black.
4.)
a. sex-linked (recessive)
b. Let XA= normal, dominant and Xa= disease, recessive
I (1: XAY) (2: XAXa)
II (1: XAXA) (2: XaY) (3: XAXa / XAXA) (4: XAY) (5: XAXa) (6: XAY)
III (1: XAXA) (2: XAY) (3: XAXa) (4: XAXa) (5: XAY) (6: XAXA) (7: XaY) (8: XAXA / XAXa)
(9: XAY)
c.In the third generation, couples 1&2 and 6&7 would have no chance of having an affected child. However, couple 6&7 have a 100% chance of having daughters that carry the disease. Couple 4&5 have a 25% chance of having an affected child (male) and a 25% chance of having a carrier (female).