Exercises

Chapter 1
1.1 The estimated average density of protein-coding genes is 1 x 10^-5 genes/bp. (3 x 10^4 genes / 3 x 10^9 bp)
1.2 The estimated minimum size of a eukaryotic genome is 2.25 x 10^7 bp. (25,000 protein-coding genes x 300 amino acids x 3 bp/1 amino acid)
1.3 (a.) TTT and TTC both encode Phe or CCT, CCC, CCA, CCG all encode Pro. (b.) TTA and CTA encode Leu. Also, TTG and CTG encode Leu. (c.) I am assuming this means a mutation at the third position that does not give the same amino acid, in which case TGT will give Cys and TGG will give Trp. (d.) This is a little tricky, a mutation in the second position will not generate similar amino acids. However, a mutation in the second position can produce a synonymous mutation when encoding for a STOP (TAA and TGA).
1.4 (a.) yes. TTT (Phenylalanine) to TAT (Tyrosine) or TTC (Phenylalanine) to TAC (Tyrosine). (b.) yes. AGT(Serine) to either CGT, AGA, or AGG for Arginine. Also, AGC (Serine) to CGC, AGA, or AGG for Arginine. (c.) 3 (d.) This is TGG to CAG, so the intermediate codon must be CGG (Arg) as TAG is a STOP codon.
1.5 Changing cytosine to uracil at the second position of the codon could generate a variety of codon changes. These changes may include Serine to Phenylalanine or Leucine, Proline to Leucine, Threonine to Isoleucine or Methionine, or Alanine to Valine.
1.6 A single base pair deletion (or addition) would change the reading frame of the entire rest of the coding sequence. This is called a "frame shift mutation" and it would change the amino acids coded. A deletion in a structural ribosomal gene would also probably be as detrimental as a single bp deletion in an exon because changing the amino acid sequence will change the native structure of the protein and thus it's functional ability. (Side Note from other user: While changing the amino acid sequence will change the native structure of the protein and its functional ability we may want to consider that a gene for a structural RNA such as a tRNA, is composed of ribonucleic acid - it will not involve amino acids. Therefore I think that a single bp mutation in the gene encoding the tRNA may have varied effects. If this mutation is within the anti-codon loop (where mRNA will interact with the tRNA) or the acceptor stem (where the amino acid will attach) the mutation will most likely be highly detrimental. This would be because these two mutations would prevent a codon in a future protein-encoding mRNA from being translated correctly. When considering other regions of the tRNA molecule, I imagine or would like to think that if the mutation were in a less critical area of the tRNA (an area that is not responsible for binding to the rRNA or mRNA or an amino acid) the effects could potentially be less obvious and possibly not observed?)
1.7 I'm not completely sure on this one but I think you would get 35 bp at the beginning and end of the 200 bp fragment with double read and not really know the middle. With the signle end read you are about to get the entire thing? Help on this exercise would be great!
Possible help: A single read would result in knowing the first 35 bp of one strand of complementary DNA. A paired end read would result in knowing the first 35bp from the 5' end of both complementary DNA strands. Does anyone else have input on this problem? Text & figures on page 18 of the book seemed helpful.
Possible help2: I agree, a single read would result in the first 35 bp from one 5' end and the pair-end read in 35 bases complementary to the 5' end of both DNA strands. Also, since the sequences obtained from the pair-end read are not complementary, one could assume the DNA strand is >70 bases.
1.8 section p13.3 on chromosome 16 has the most gene rich area, followed by q22.1.
1.9 The alpha globin gene cluster is at the upper tip of chromosome 16 inthe p13.3 region.
1.10 The globin region contains 89Kbp (211Kbp - 122 Kbp, globin region on chromosome 16). There are 7 globin genes in that stretch so 7genes/89 Kbp = 0.079 genes/Kbp.
I am not sure of this answer, has anyone else tried this problem or does anyone have a different take on it?
1.13 The haemoglobin molecule would be symmetrically divided with a alpha and beta subunit in both halves. If the alpha and beta subunits were identical the protein could be divided in half in any manner. 
1.15 Genomics can be very applicable to conservation biology because the researcher can learn about the specific genomic qualities of an endangered species and preserve it to possibly reintroduce it into a related population (if it goes extrinct at some point). Understanding the genomics of a species can help you better manage the species because you could imagine being able to identify disease susceptiblities by scanning the genome. Once genomes are understood, a library of genomic biodiversity can be created than could theoretically be useful for reintroduction of genetic material in the future.
In the event of potential extinction, genetic disease could be bred out of the species via careful and intentional mating. This would requre human intervention at the earlier stages of 'extinction' to ensure that the healthy organisms of a species mate with each other and the genetically diseased do not. This is the basic concept of Eugenics.