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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.

Question

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.

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Solution

The answer is B. Ribosomes must translate a total of 2,700 mRNA nucleotides into amino acids to form one 100-kDa LPL homodimer.

Here's the step-by-step reasoning:

  1. The question states that an active LPL is a homodimer consisting of two individual subunits that are each approximately 450 residues in length and weigh 50-kDa.

  2. A homodimer means it is made up of two identical proteins. Therefore, to form one 100-kDa LPL homodimer, we need two 50-kDa subunits.

  3. Each amino acid is coded by a triplet of nucleotides in the mRNA. Therefore, to get the number of nucleotides, we multiply the number of amino acids (or residues) by 3.

  4. So, for one 50-kDa subunit with 450 residues, we have 450 residues * 3 nucleotides/residue = 1350 nucleotides.

  5. Since a 100-kDa LPL homodimer is made up of two 50-kDa subunits, we multiply the number of nucleotides for one subunit by 2.

  6. Therefore, 1350 nucleotides/subunit * 2 subunits = 2700 nucleotides are needed to form one 100-kDa LPL homodimer.

This problem has been solved

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Describe the role of the coronary arteries.

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