Answer:
d. Measure thyroid-stimulating hormone level
Explanation:
The most suitable next step in management of this patient is to measure a thyroid stimulating hormone, or thyrotropin level.
This patient's apparent treatment refractory dyslipidemia is most likely contributed by undiagnosed thyroid disease. Despite sticking to the treatment with atorvastatin, his LDL AND total cholesterol are raised.
Additionally, there are some clinical signals that give an idea about the diagnosis of hypothyroidism especially fatigue and constipation.
Hypothyroidism is concerned with and is a very common secondary cause of dyslipidemia and thyrotropin level measures are indicated.
Hyperlipidemia may show better results to thyroid stimulating hormone.
Hence option D is the right answer.
Answer:
a.Many mitochondrial genes resemble proteobacteria genes, while the genes in the chloroplast resemble genes found in some photosynthetic bacteria.
c.Mitochondria and chloroplasts both have their own circular DNA and 70S ribosomes that are similar to those found in bacteria.
d.Mitochondria and chloroplasts replicate by a process similar to mitosis.
Explanation:
Endosymbiotic theory states that mitochondria and chloroplast which are organelles of eukaryotic cells were once independently living micro-organisms but with due course of time eukaryotic cells engulfed them and they become an integral part of these eukaryotic cells.
The resemblance between mitochondrial genes with those of proteobacteria and chloroplast genes with photosynthetic bacteria strongly support endosymbiotic theory. Apart from this, the presence of their own DNA that too circular just like prokaryotic microbes and 70 S ribosomes also support this theory. Also just like prokaryotic cells, before cell division mitochondria and chloroplasts undergo replication by means of a process known as binary fission.
I think the answer is c because each polymer has 1 monomer
Answer:
F1: Aa Bb Cc: 4.5 lbs
F2: Too many to list: Between 3 lbs and 6 lbs
Explanation:
F1: Aa Bb Cc, will be the genotypes and will weigh 4.5 lbs, the reason for that is because each capital letter allele adds half a pound to the base weight of 3 lbs.
F2: There are many different genotypes that can be present in the second generation, for example (AA BB CC, AA BB Cc, AA BB cc, etc.) each of which will allow the plant to grow to a certain height. The majority of phenotypes will be 4.5 lbs. There will be one who is AA BB CC, and another one person who has the code, aa bb cc.
The ICD-10-PCS code is 30240G4.
Firstly, select "Administration" (section 3) because the procedure is the administration of bone marrow to the patient. Secondly, select "Circulatory" (section 30) because it is being administered in a vein (circulatory system). Third, select "Transfusion" (section 302) because it is being done a transfusion of bone marrow. Then select "Central Vein" (section 3024) because that's the place of administration in the circulatory system. Lastly, go to "Bone Marrow" (section 30240G) as that is what's being transfused and then choose "Transfusion of Allogeneic Unspecified Bone Marrow into Central Vein, Open Approach" (<span>30240G4) because it is not specified what type of transfusion it is.</span>
