January 24, 2018
Homework Chapter 3+6
1. autosomal dominant
2. x-linked recessive
3. Yes, there is complementation. The cross between the 2 affected parents yields an unaffected offspring. Thus, the mutant phenotype of the parents must be from different homozygous recessive mutations that complement each other to produce a wild-type genotype/phenotype.
a. One underlying factor that could complicate the interpretation is penetrance. In the 2nd generation, the 2 males have an affected parent and affected children so they almost certainly have the disease, however they are phenotypically normal. This is a case of non-penetrance.
b. Yes, the pedigree is compatible with a gene that is genetically imprinted. The affected traits are not present in the second generation but reappear in some of the third generation offspring. It is maybe due to maternal imprinting in the second generation that allowed for healthy traits but the maternal imprinting was turned off for the 3rd generation.
c. Assuming that disease follows an autosomal dominant inheritance pattern, individual IV-1 would have a genotype of “Aa” while the normal male would have a genotype of “aa”. The genotype of offspring would be “Aa or aa”. So, there is a 50% chance that individual IV-1 would have affected children.
5. One possible explanation is a X-linked dominant inheritance with variable expression. From generation II, the male offspring is not affected but one of the female offspring is affected. This disease can’t have recessive inheritance, which would result in only affected male offspring and carrier female offspring. This inheritance also results in affected and stillbirth females. We cannot say that it is fatal to females because some of the affected females survived birth. So, the variable expression of this gene must be the cause of stillbirth and live birth. Greater manifestation of the gene expression leads to stillbirth while lower expression leads to affected live birth.
a. autosomal dominant; no chromosomal location found
b. autosomal recessive; 16q23.1
7. This pedigree suggests a pseudoautosomal inheritance. The shift from a X-linked pattern to a Y-linked inheritance pattern hints that the gene, originally on the X-chromosome, recombined with the Y-chromosome.
9. X-linked ischthyosis is an X-linked recessive skin disorder. The 2 parents have a genotype of Xx and XY. Based on their genotype, they have a 25% chance of a healthy female, 25% chance of a healthy male, 25% chance of a female carrier, and 25% chance of a male with XLI. The first generation approximates the genotype predictions from the parent genotype. Moreover, an X-linked recessive disorder affects mostly males. This is evident in the 3rd generation- affected male offspring from carrier mother. Thus, we can say that the pedigree is consistent with an X-linked recessive pattern of inheritance for the ichthyoids.
10. Authors did a quantitative PCR for STS gene-copy number analysis to identify that there was a deletion and a SNP array analysis for the size of deletion in the patient with XL1.
11. 8 mutations were identified in the promoter and introns of the UGT1A1 gene. Most of the UGT1A1 mutations are due to single nucleotide substitutions thus large number of mutations in the coding sequence were reported to explain this fact.
a. Tm= 4(10)+2(6)= 52
b. Tm= 4(19)+2(20)= 116
a. 55.4 ºC
b. 65.2 ºC
(a) The website analyzes the Tm of the DNA oligos with the addition of sodium and a specific oligo concentration. Calculating the Tm without sodium is not possible using this website. Sodium cations shield negatively charged phosphates, eliminating the repulsive force between the two antiparallel strands. (b) For the larger primer, the simple formula used to calculate Tm is not accurate. Thus, we see differences between the calculated Tm and the Tm from the website.
15. The DNA polymerase amplifies the DNA from 5′ to 3′ direction thus, a 3’OH is necessary for the DNA polymerase to synthesize the DNA in the 5′-3′ direction. If the 3′ end is deoxygenated, a new 3′-5′-phosphodiester bond can’t be created, and DNA synthesis would stop. The 5′ sequence allows for high degree of specificity for the primer to anneal to and for the sequence to be amplified.
16. I am not completely confident to state that sample 3 has an equal amount of gene X as sample 1. The band for sample 3 seems to be larger and brighter than the band for sample 1, which suggests that sample 3 might have a greater amount of gene X than sample 1. However, if sample 3 has an equal amount of gene X as sample 1, there is evidence to support for the usage of proper PCR protocol since there are no bands in the negative control line. This suggests that there was no contamination during the preparation or loading phase of PCR.
b. a probe with the same sequence as the RNA sample however, with thymine instead of uracil. The DNA strand is complementary to the RNA strand so the probe to detect DNA should have the same sequence as the RNA strand. DNA also uses thymine instead of uracil so, the DNA probe should substitute the uracil in RNA sequence to thymine.
a. permits efficient, high-level expression of your recombinant protein
b. Allows for in vitro transcription in the sense orientation and sequencing through the insert
c. Selection of vector/transformants in E. coli
d. selection of stable transfectants in mammalian cells
e. Efficient transcription termination and polyadenylation of mRNA
f. Allows efficient, high-level expression of the neomycin resistance gene and episomal replication in cells expressing SV40 large T antigen
g. Efficient transcription termination and polyadenylation of mRNA
h. High-copy number replication and growth in E. coli
19. RT negative control determines whether there is any contamination. If some of the RNA was not converted to cDNA and remains in the tubes, a band in the RT negative control line can help the researcher identify that the tubes are contaminated with 2 different types of strands.
20. This new approach allows mutant generation at a mutagenic frequency of close to 100% and also has numerous advantages over those other newly developed methods.
a. One of the main problems with previous protocols is primer dimerization. However, this primer design prevents primer dimerization since the target gene is amplified via two separate PCR amplifications. This helps to overcome the problems associated with primer pair self-annealing attributed to completely or partially complementary primers.
b. Primer pairs consist of two flanking (A-up and B-down) primers and two internal (A-down and B-up) primers. The two internal primers comprise two major parts: a 50-complementary region containing the target mutant sequences and an 8-bp minimum 30 -template annealing region.