homework 09

Version Beta digest it with a restriction enzyme, then either the DNA is cut or the DNA is not cut. Further if the DNA is not cut, is it because of a failure to cut OR that the DNA is not supposed to be cut? This question is related to "important practical issues" #2 and #3 described in the "Designing a Protocol to Determine Alleles Present at the 'Type O' SNP Locus" section of this lab. Use the fragment lengths of the actual PCR product we are making in lab as determined in the previous question using the actual primers 6-1 and 6-2. NOT the virtual primers answer. The first row is done for you. On the ‘Digest condition’ column, those indicate if the enzyme is functioning or not. In the row filled out for you, both restriction enzymes are functional and working, and thus there is no known defect of Kpn I or BstE II. It does not mean that both are being used at the same time in the same sample of DNA, just that they are functional and should work if the DNA has the recognition sequence within it. Under ‘O,O homozygote’, you see that cell has two sub-columns underneath it. This is to imply that the exact same sample of DNA is used in both separate digests. Thus, there are 6 digests being performed on three distinct DNA samples. Recall that humans are diploid also…but we are not magically able to separate the chromosomes from each other. For the row where ‘Kpn I fails’, Kpn I is simply non-functional wherever Kpn I occurs in that row . A. In the table below, write ALL of the fragment sizes (in basepairs) that you would predict under the conditions described in the first column. Refer to Fig. 8 and consider the circumstances in actual human cells that are diploid. [13 pts] Genotype Digest condition O,O homozygote O,A/B heterozygote A/B,A/B homozygote Digest w/Kpn I (only) Digest w/BstE II (only) Digest w/Kpn I (only) Digest w/BstE II (only) Digest w/Kpn I (only) Digest w/BstE II (only) Both restriction enzymes are active 220;476* 696** 697**;220;476 696**;220;476 697 ** 220;476* ONLY Kpn I fails 696 696 697 220,476 697 220;476 ONLY BstE II fails 220;476 696 697; 220,476 696,697 697 697 Both fail 696 696 696,697 696,697 697 697 * Approximate, because the overhangs formed in the restriction digest do not contain complete basepairs.

Version Beta Answer: The O/O homozygous genotype will result in a person having O blood. This is because it is a homozygous recessive trait. There is a guanine in both chromosomes which is why the H antigen is not modified on either chromosome. 6. View the video on the "Electrophoresis Concepts page" next week's lab on Brightspace. When the UV is on and the power is on and off, watch what happens to the mass markers of different sizes. Describe the relationship between the masses of the markers and their positions on the gel at any given time. [4 pts] Learning objective 10.2: describe the relationship between molecule size and the distance that it travels across an electrophoresis gel. Answer: When the UV is on, more fragments are able to be seen compared to when the UV is off, it’s harder to see the separation of fragments. Generally, the farther away the mass markers are from the gel, the smaller the fragments are because they are smalle meaning they move faster and further. BE SURE TO READ NEXT WEEK’S LAB prior to coming to lab and to sometime between this lab next week’s lab to log into Brightspace to do your Reading Assessment Questions. Five multiple choice questions, randomly selected from a large question bank.