2.1) a.) Several sources suggest about 3.1 bp are present in the human genome. So for two random people they will have 2.07 x 10^6 to 3.1 x 10^6 different basepairs (3.1 billion/1000 to 3.1 billion/1500).
2.2) It is possible to have calico cats because in each cell one x-chromosome is chosen at random for complete inactivation by the RNA molecule produced by the Xist gene, creating a Barr Body (named after the geneticist who discovered it Murray Barr). The Xist gene on the other X chromosome is inactivated by cytosine methylation allowing the chromosome to have regular activity. This allows the female calico to have a mosaic of coloring schemes depending on which X chromosome is the Barr Body. Kangaroos cannot have calico patterns because in female marsupials the paternal X chromosome is always silenced giving rise to only one fur color.
On a separate but interesting note: Male calico cats are very rare, about 1:3,000. To exhibit calico color schemes, two X chromosomes are required because there are not color genes on the Y chromosome. The male cats who have calico patterns have XXY chromosomes (also known as Kleinfelters Syndrome). This allows them to randomly silence one X chromosome, but renders them sterile. (http://en.wikipedia.org/wiki/Calico_cat#Genetics).
2.3) (a) Dysfunction of a metabolic enzyme: Phenylketonuria (PKU). This metabolic disease is caused by an enzyme deficiency of phenylalanine hydroxylase, which is responsible for converting phenylalanine (phe) to tyrosine (tyr). Lack of the enzyme causes phe and phenylpyruvate, a ketone, to build up to toxic levels in the blood. This toxic build up can have detrimental effects such as mental retardation and seizures. There are a number of known mutations appearing on all 13 exons of the gene that affect catalytic activity, a few that affect regulation and very few that affect enzyme assembly. This disease can be managed with a phenylalanine free diet. Trials for enzyme replacement therapy are in the works.
(b) 'Read through' mutation produces longer proteins: B-Thalassaemias. The loss of a termination codon leads to an unstable extended polypeptide chain of the beta subunit of hemoglobin genes creating gama-beta fusion proteins. Similar to Sickle Cell Anemia, the misshapen protein does not easily flow through capillaries and can lead to painful vaso-occlusive episodes.
(c) Proteins with lower stability: Sickle Cell Anemia. A single base pair substitution that changes a hydrophilic charged glutamic acid (Glu) to a hydrophobic Valine (Val) on the beta hemoglobin subunit. This causes 'cell puckering' (sickle shaped cells) in deoxygenated red blood cells. This sticky pocket on the sickled RBC's causes formation of hemoglobin polymers. The inability for RBC's to flow smoothly through capillaries, like B-Thalassaemias, can cause vaso-occlusive episodes which can lead to organ failure, RBC lysis causing severe anemia and jaundice. The blood cell lysis can be particularly useful in battling malaria by causing RBC's infected with the parasite to lyse and in turn kills the parasite by removing its food source.
(d) Increase in risk factor of disease not known to be associated with primary function of protein: Late onset Alzheimer's. If a person contains at least 1 E4 allele of the ApoE gene they have an increased risk for developing Alzheimer's. The ApoE protein is responsible for removing cholesterol from the blood.
(e) Enhanced probability of developing cancer: BRCA1 & BRCA2 tumor suppressor gene mutations. Mutations in the BRCA genes are associated with an increased risk in developing breast and ovarian cancer. These tumor suppressor genes are required for chromosome stability and participate in DNA double strand break repair. Mutations result in the introduction of an early stop codon. This produces a truncated and partially non-functional protein.
(a) I would expect the mutation to not be in the coding region but rather in a regulatory region because the phenotype is constituative activity or persistence of the enzyme.
(b) I think the data suggests that the mutation arose in northern Europe and radiated outward because of the extremely high percentage of Swedes and Finns that are lactose persistent.
a.) If a 100 kb region is likely to stay intact a 10 kb region should definitely stay intact. Therefore there should only be 1 haplotype.
b.) If the region were on a diploid chromosome the haplotype likely would not change.
(a) Figure 2.13 shows a protein truncation test. The mutant triplet (m1,2,3) shows a difference in movement through the gel (due to charge and/or size differences) when compared to the wildtype protein (w1,2,3)
(b) There is a bigger difference between W3 and M3 than between W2 and M2 based on traveling distance in the gel.
It might be contraindicative to prescribe MAOA to a child experiencing anxiety in a maltreatment case because the genotype would probably be not thought about when diagnosing the child and his/her situation. It may be that the child has low levels of MAOA and needs the supplement.