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Chapter 3 exercises

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3.1) Given one parent has the genotype AB/AB and the other parent has the genotype ab/ab. I would predict that the offspring would have some combination of Aa and Bb - meaning that I think all offspring would have the genotype Ab/aB. (This is assuming that recombination only occurs during meiosis - this being the case even if the parents chromosomes underwent recombination, the gametes would still be AB for one parent and ab for the other parent.)

3.2) The extent of linkage disequilibrium is: D = pa1b1 x pa2b2 - pa1b2 x pa2b1, = 0.2 x 0.45 - 0.15 x 0.2, = 0.06. This implies that the system is close to equilibrium, but that chromosomes a1b2 and a2b1 are more prevalent than thought.

3.4) I would predict that it would not be likely for a patient to transmit leukaemia to their offspring. I would predict this because I would expect the mutation to be localized to the cancerous region - in this case the bone marrow. The mutation should not occur in all the cells in the individual?????

Translocation can happen during gametogenesis and also in somatic cells. In this case, it is in somatic cells and hence is not passed on to the offspring.

3.5) 2q genes in human should be linked (somewhat) because they are on the same chromosome. The same genes in chimpanzee are split onto two different chromosomes so you would not see linkage groups within those genes. (page 88) A good page about the differences between the human and chimp genome can be found here .

3.7) I am not sure if I am understanding the question correctly, but if the question were asking to compare haplotype blocks measured in base pairs to one measured by centimorgans, I would say the haplotype block measured in centimorgans would vary more than the haplotype block measured in base pairs.

The haplotypes would vary more if they were measured in centimorgans because there is a variable distance between linkage groups. They tend to be in clusters with vast "gene deserts" in between where no crossing over would occur.

3.8) If these SNPs are on the Y chromosome there should only be 1 haplotype. Provided that the 10kB length enables different haplotypes to form, if these SNPs are on a diploid chromosome there should be 2^10 haplotypes. (The book mentions if a sequence is less than 100 kB it should stay intact, that is why I am saying "Provided that the 10kB length enables different haplotypes to form").

3.12) I calculated that there are 1.089x10^10 km of DNA per average human. This figure is larger than Pluto's orbit, which is ~5.9x10^9km.

(3.2x10^9bp/cell)(10^13cells/human)(3.4x10^-9m/10bp)(1km/1000m)=1.089x10^10 km/human

3.13) This continuation of the helical DNA structure on a ssDNA segment is not structurally correct because DNA does not keep the helical shape as its first nature. The helical shape is partially formed by the hydrogen bond patterning between complementary nucleotides. With out the complementary strand, a single strand of DNA has no pattern to adhere to.

3.15) a.) Provided that coverage is the average number of times each base appears in fragments, I would suspect that the minimal coverage is 1?? Minimal coverage looks like the maximal coverage in this picture is eight?? c.) the average length of coverage is ??? (3.15 questions for whatever reason did not seem clear). N=27, L=500bp and if G=5000bp then C= NL/G ==> 2.7 which would be average coverage.

3.16) The coverage to be expected would be: ( 300,000 x150 bp)/ (3.2 x 10^6 kb) = 14.06.

3.17) If the amoung numts is proportional according to size of genomes, then there would be aproximatel 658 kb of numts in the 4.7 Gb long mammoth genome. This is because in the 3.2 Gb human genome, 1005 segments * 446 bp = 448,230 b, which is aproximately 448 Kb. This estimate might not be accurate since the averge leng of reads per numt might be different in the mammoth genome.

3.18) By primer n-1 in its third round and primer n-2 in its second round.

3.19) Base at position 3 is C. In this sequencing you have a 7 base probe. The first two is a combination of 16 dinucleotides followed by 6 that can base pair with any base.And at the 3' end is a colored probe specific to the combination of the dinucleotide. Every reaction covers seven base and the first two bases are called. The calling a base depends on the previous base read and called. Since the color is green it could be dinucleotide combination of AC,CA,GT or TG.

3.20) Assuming that shotgun reads are haplotype data, fragment paired end reads are double stranded and both are thought as "raw" data, then the total data produced was 50.628 Gb, equivalent to 15.82 folds of the human genome (which is aprox 3.2 Gb long).

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