Other genetic polymorphismsThe between-genotype difference for the homozygotes (A/A vs. G/G) for the b2-adrenergic receptor Arg16Gly gene for the cardiovascular death endpoint reached significance (P=0.05) (Fig. 2). Selected other comparisons approached significance (aldosterone synthase promoter C-344T, C/Cvs. T/T, stroke, P=0.06; type 1 angiotensin receptor A11 66C, A/C vs. A/A, primary composite, P=0.07; bradykinin 2 receptor I/D, I/D vs. I/I, primary composite, P=0.07; and G protein b3-subunit C825T, C/T vs. C/C, MI, P=0.07). However, most of these did not seem biologically plausible because the effects were seen in heterozygotes only, and after adjusting for multiple testing of the many endpoints (comparison with Bonferroni-adjusted P value of 0.001), no between-genotype differences were significant, suggesting that these results are most likely chance findings.Genotype-by-treatment differences between losartan and atenolol for the cardiovascular endpoints were not significant (P values 0.06–0.98, Supplementary Table 4, Supplemental Digital Content 4, http://links.lww.com/FPC/A77), with the ex- ception of cardiovascular death for the b2-adrenergic receptor Gln27Glu genotype (P=0.02) (Supplemental Table 4d, Sup- plemental Digital Content 4, http://links.lww.com/FPC/A77). Again, this is likely a chance finding because the P values were not less than the Bonferroni-adjusted P value of 0.001.
Question
Other genetic polymorphismsThe between-genotype difference for the homozygotes (A/A vs. G/G) for the b2-adrenergic receptor Arg16Gly gene for the cardiovascular death endpoint reached significance (P=0.05) (Fig. 2). Selected other comparisons approached significance (aldosterone synthase promoter C-344T, C/Cvs. T/T, stroke, P=0.06; type 1 angiotensin receptor A11 66C, A/C vs. A/A, primary composite, P=0.07; bradykinin 2 receptor I/D, I/D vs. I/I, primary composite, P=0.07; and G protein b3-subunit C825T, C/T vs. C/C, MI, P=0.07). However, most of these did not seem biologically plausible because the effects were seen in heterozygotes only, and after adjusting for multiple testing of the many endpoints (comparison with Bonferroni-adjusted P value of 0.001), no between-genotype differences were significant, suggesting that these results are most likely chance findings.Genotype-by-treatment differences between losartan and atenolol for the cardiovascular endpoints were not significant (P values 0.06–0.98, Supplementary Table 4, Supplemental Digital Content 4, http://links.lww.com/FPC/A77), with the ex- ception of cardiovascular death for the b2-adrenergic receptor Gln27Glu genotype (P=0.02) (Supplemental Table 4d, Sup- plemental Digital Content 4, http://links.lww.com/FPC/A77). Again, this is likely a chance finding because the P values were not less than the Bonferroni-adjusted P value of 0.001.
Solution
This passage discusses the results of a genetic study. The researchers found a significant difference between two homozygous genotypes (A/A and G/G) of the b2-adrenergic receptor Arg16Gly gene in relation to cardiovascular death. However, this significance was only found at a P value of 0.05.
Other comparisons were made between different genotypes of various genes, some of which approached significance (P values ranging from 0.06 to 0.07). However, the researchers noted that these results did not seem biologically plausible as the effects were only seen in heterozygotes. Furthermore, after adjusting for multiple testing, none of the between-genotype differences were significant. This suggests that these results may be due to chance.
The researchers also looked at genotype-by-treatment differences between two drugs, losartan and atenolol, for cardiovascular endpoints. These differences were not significant, with P values ranging from 0.06 to 0.98. The only exception was for the b2-adrenergic receptor Gln27Glu genotype, which had a P value of 0.02. However, this is likely a chance finding as the P value was not less than the Bonferroni-adjusted P value of 0.001.
Similar Questions
There were no associations between ACE I/D or 12 other polymorphisms of hypertension susceptibility genes and clinical outcomes or treatment effect in White patients with hypertension and LVH in the LIFE study. This suggests that the observed effects of losartan versus atenolol in the LIFE study do not depend on these genotypes, at least among these White patients studied in the Scandinavian countries. sum up in one sentonce
Abnormal blood lipid levels play a significant role in the development of coronary artery disease (CAD). The discovery of genomic variations that confer a protective effect against CAD has prompted investigations into the biological impact of these protective alleles. Researchers have observed that specific single-base variations in the ANGPTL4 gene are common among patients with a low occurrence of CAD. ANGPTL4 is located on chromosome 19 and codes for angiopoietin-like 4, a multifunctional glycosylated protein.One of the functions of angiopoietin-like 4 is the regulation of lipid metabolism. Specifically, angiopoietin-like 4 inhibits the activity of the enzyme lipoprotein lipase (LPL). Generally, active LPL is a homodimer consisting of two individual subunits that are each approximately 450 residues in length and weigh 50-kDa. LPL functions on the surface of vascular endothelial cells to degrade the triglycerides of chylomicrons and very-low-density lipoproteins (VLDLs) in the blood. Elevated triglyceride levels have been correlated with increased CAD risk.To identify common loss-of-function mutations in ANGPTL4, DNA samples from 43,000 individuals with CAD and 78,000 individuals without CAD were genotyped. A detailed analysis of ANGPTL4 genomic variants was performed, and the results of individuals without CAD were mapped in Figure 1. The lipid profiles of ANGPTL4 mutation noncarriers were also compared to those of mutation carriers (Figure 2).Figure 1 Loss-of-function mutations observed in the ANGPTL4 gene of participants without CAD (Note: * indicates mutation resulting in a truncated protein; † indicates substitution of a cytosine for its complementary base.)Figure 2 Mean plasma lipid levels for ANGPTL4 mutation carriers and noncarriers. (Note: LDL = low-density lipoprotein; HDL = high-density lipoprotein; error bars = standard deviation.)To further analyze the effect of ANGPTL4 on blood lipid levels, ANGPTL4 knockout mice and mice that constitutively express ANGPTL4 were created. Researchers hypothesized that two of the four biological profiles shown in Table 1 would be observed in these mice.Table 1 Predicted Biological Profiles of ANGPTL4 Mouse Models Question 48A scientist studying the sequence of the LPL homodimer would most likely expect that:A.ribosomes must translate a total of 1,350 mRNA nucleotides into amino acids to form one 100-kDa LPL homodimer.B.ribosomes must translate a total of 2,700 mRNA nucleotides into amino acids to form one 100-kDa LPL homodimer.C.ribosomes must translate a total of 1,350 mRNA nucleotides into amino acids to form one 50-kDa LPL homodimer.D.ribosomes must translate a total of 2,700 mRNA nucleotides into amino acids to form one 50-kDa LPL homodimer.
Hypernatremia commonly occurs in _____Group of answer choicesDecreased production of antidiuretic hormone (ADH)Syndrome of inappropriate ADH (siADH)Decreased aldosteroneNephrotic syndrome
Q3. Two groups of people have volunteered to take part in a genetic study. Group 1 (n = 126) are volunteers with no history of Type I Diabetes in their immediate family, whilst Group 2 (n = 183) have all been diagnosed with Type I Diabetes. A genotyping study was undertaken on these volunteers using 25,786 SNPs selected due to their proximity to key immune genes. Researchers are looking to identify any SNP genotypes which may increase the risk of Type I Diabetes. In your answer, consider the reference SNP allele as A and the alternate SNP allele as B, using the genotypes AA, AB and BB.For an individual SNP, what test would be appropriate for this comparison? [1 mark]Define H₀ and Hₐ for the genotype at each individual SNP. [2 marks]If there was no true difference in any genotypes between the two groups, how many p-values would you expect to see < 0.05? [1 mark]Using Bonferroni’s method, what would a suitable cutoff value be to consider a SNP as being associated with an increased risk of Type I diabetes, i.e., to reject H₀ [1 mark]Given the following genotype table, would you reject or fail to reject H₀? Provide your working and a full explanation. [3 marks]Group AA AB BBControl 25 60 41T1D 21 55 103
genotype of ii will result in which blood group phenotype?Group of answer choicestype AB bloodtype A bloodtype O bloodtype B bloodNext
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