Answer: this is an incomplete question but pressure is said to decrease with height and increase with depth.
This means pressure at the top of the head is lower than pressure at the feet.
It is taken that blood pressure at the arms is roughly 10% lower than blood pressure at the legs.
Explanation:
A researcher follows a protocol to test the activity of a mitochondrial extract containing all of the soluble enzymes of the matrix. Because the mitochondrial extract was dialyzed, the protocol lists low molecular weight cofactors that must be added to the extract in order to catalyze the oxidation of acetyl-CoA to CO2. The list does not include lipoic acid, a known cofactor of the citric acid cycle. Why is lipoic acid omitted from the list of cofactors to add back to the extract? O O O O O The added TPP can substitute for lipoic acid in the pyruvate dehydrogenase complex. Lipoic acid is covalently attached to the pyruvate dehydrogenase complex. The disulfide bond in lipoic acid prevents diffusion through the dialysis membrane. The Kd of lipoic acid binding to pyruvate dehydrogenase is extremely low. The oxidation of acetyl-CoA to CO2 does not require lipoic acid in vitro.
Answer:
Lipoic acid is covalently attached to the pyruvate dehydrogenase complex.
Explanation:
The pyruvate dehydrogenase complex has lipoic acid bound covalently to it.
This explain why dialysis is not effective in separating the lipoic acid from the enzyme
Answer:
The answer is: Lipoic acid is covalently attached to the pyruvate dehydrogenase complex
Explanation:
It can be said that via a convalent amide bond lipoic acid binds to the terminal lysine residue found in the lipophilic domains of the enzyme. The importance of lipoic acid is that it acts as a cofactor in the pyruvate dehydrogenase enzyme complex.
Read the following information and then answer follwoing questions:
Heartburn is an esophageal symptom characterized by a burning sensation behind the breastbone. It may be related to gastric hyperacidity. There is little doubt that emotional disturbance, excitement, and nervous tension result in hyperacidity, which in turn may lead to heartburn. A gesture that usually accompanies heartburn is movement of the open hand up and down the breastbone. This is in contrast to the stationary tightly clenched fist of angina pectoris. Another important aspect of heartburn is that it is usually relieved, if only temporarily, by taking antacids. Antacids are used in other instances as well, particularly in cases of peptic ulcer. Pain is the outstanding symptom of peptic ulcer, a disorder that is chronic, periodic, almost always characterized by gnawing or burning, and coinciding with digestion. Nausea, emesis, and anorexia are uncommon.
1. Where does the burning sensation occur in heartburn?
2. Which factors lead to heatburn?
3. What is the use of antiacids?
4. What are the symptoms of peptic ulcer?
Answer:
Following are the answers for the given questions.
Heart burn is a burning sensation behind the Breast bone. Breast bone is another name for Sternum.Gastric hyper acidity is the main cause of heart burn. The other factors that would likely lead to hyper acidity of stomach are emotional disturbance, excitement and nervous tension that ultimately leads to heart burn.Antacids are medicines that are mainly used for symptomatic treatment of heart burn and peptic ulcers.Pain is the main and outstanding symptom of peptic ulcers. In chronic cases, peptic ulcer presents with gnawing and burning sensation after food ingestion. Uncommon presentations of chronic peptic ulcers are Nausea, Emesis (vomiting) and Anorexia (loss of appetite)Explanation:
The answers had been taken from the given paragraph in the asked question.
Let’s suppose that weight in a species of mammal is polygenic, and each gene exists as a heavy and light allele. If the allele frequencies in the population are equal for both types of alleles (i.e., 50% heavy alleles and 50% light alleles), what percentage of individuals will be homozygous for the light alleles in all of the genes affecting this trait, if the trait was determined by the following number of genes?
Two
Three
Four
Final answer:
Using the Hardy-Weinberg principle, the percentage of individuals homozygous for the light alleles in a population with equal allele frequencies is 6.25% for two genes, 1.56% for three genes, and 0.39% for four genes.
Explanation:
When assessing the population's genetic structure, biologists focus on the frequencies of the resultant genotypes to infer phenotype distribution. Assuming equal frequencies of heavy and light alleles for a polygenic trait (50% each), the Hardy-Weinberg principle can be applied. Using this principle, the probability of an individual being homozygous for the light alleles can be calculated using p² + 2pq + q² = 1, where p and q represent the frequencies of the heavy and light alleles, respectively. In this case, the frequency of homozygous light alleles (qq) will be q².
For two genes, the probability an individual will be homozygous for the light alleles at both loci is q² for one gene times q² for the other gene:
q² × q² = q⁴ = (0.5)⁴ = 0.0625 or 6.25%.
For three genes, the calculation becomes:
q² × q² × q² = q⁶ = (0.5)⁶ = 0.015625 or 1.56%.
For four genes, the calculation would be:
q² × q² × q² × q² = q⁸ = (0.5)⁸ = 0.00390625 or 0.39%.
The percentage of individuals homozygous for light alleles are approximately 6.25% for two genes, 1.56% for three genes, and 0.39% for four genes.
In this scenario, the frequency of alleles for the mammal's weight is equally divided between heavy and light alleles, each at 50%. If we denote the frequency of the light allele as q (where q = 0.5), we can use the formula for homozygous recessive genotype frequency (q²) to determine the frequency of individuals homozygous for the light alleles in each gene.
Calculation for Two Genes:
For two genes, the probability of an individual being homozygous for light alleles in both genes is calculated by:
Calculate q² for one gene: (0.5)² = 0.25Since each gene is independent, the frequency of homozygous light alleles for both genes is: (0.25)² = 0.0625 or 6.25%Calculation for Three Genes:
For three genes, follow the same steps:
Calculate q² for one gene: (0.5)² = 0.25
For three genes, the frequency is: (0.25)³ = 0.015625 or 1.56%Calculation for Four Genes:
Similarly, for four genes:
Calculate q² for one gene: (0.5)² = 0.25For four genes, the frequency is: (0.25)⁴ = 0.00390625 or 0.39%