Answer:
of a cycle where scratching will cause even more of an itchy sensation
Answer:
Positive feedback.
Explanation:
Positive feedback is a process in which the end product of an action cause more of that action to occur in a feedback loop.
Inflammation is thhe local reaction of bodily tissues to injury caused by physical damage , infection or due to any allergic reaction. Injured tissue mast cells release histamine, which causes surrounding blood vessels to dilate and increase permeability. This allows fluid and cells of immune system to leak from bloodstream through vessel walls and migrate to site of tissue injury or infection where they fight infection and heal injured tissues.
Which of the following is true of the opioid fentanyl? It is used primarily in conjunction with surgical anesthesia. It is found in the brain tissues and have potent psychoactive effects. It stimulates the brain area controlling nausea and vomiting. It makes the respiratory centers less responsive to carbon dioxide levels in the blood.
Answer: it is used primarily in conjunction with surgical anesthesia
Explanation: it is one of the synthetic opioid , it is used intravenously as anesthesia and to treat pain. It is given with a muscle relaxant and a sedative hypnotic. The effect is quick in the body and central nervous system and can last less than 2 hours.
Answer:
It is used primarily in conjunction with surgical anesthesia.
Explanation:
Fentanyl opioid is an injectable or orally applied medication used to treat pain or used in anesthesia. Its effect occurs very fast providing the quick aliveio of the pain the loss of sensation of the anesthetized place, generally in about an hour or two.
Despite having a great medicinal value, this medication is highly dependent and can leave those who consume it quickly addicted and causing serious damage to your body, even leading to death.
This medication should not be used without a doctor's prescription. It can cause drowsiness, nausea, constipation, pressure drop, addiction, respiratory depression, among others.
1. True/False: DNA replication is considered "semi-conservative" because only half of the chromosomes are copied during S phase of interphase.
Group of answer choices
True
False
2.The monomers of DNA consist of
Group of answer choices
a. deoxyribose, phosphate, and a nitrogenous base
b. deoxyribose, phosphate, and hydrogen bonds
c. ribose, phosphate, and nitrogenous bases
d. ribose, phosphate, and hydrogen bonds
e. none of the above are correct
Answer:
1) The statement is false
2) Option A) deoxyribose, phosphate, and a nitrogenous base
Explanation:
1) In semi-conservative DNA replication:
- a parent double-stranded DNA splits in two.
- Each strand is then read by the enzyme, DNA polymerase, to ensure accurate synthesis of a new daughter strand
- the newly synthesized strand contains nucleotides that are complimentary to free nucleotides present in the parent strand.
Thus, because the parent strand is retained in the newly synthesized DNA, DNA replication is described as semi-conservative
2) DNA consists of several repeating units of polydeoxynucleotides where each is made up of a nitrogenous base (Adenine, thymine, cytosine or guanine) linked to
deoxyribose sugar by an N-glycosidic linkage, and then the sugar linked to a phosphate group by phosphoester bond.
Answer: 1. True.
Option A. deoxyribose, phosphate and nitrogenous base.
Explanation:
DNA replicates in semi conservative because when a parent helix replicates, it produce two daughter helixes which contain one of the two helical strand of the parent. During S phase in interphase, DNA is unwind by enzyme helicase and it is duplicated. The two single strands is use as a template to form identical strands.
2. The monomers of DNA are also called nucleotides. They include 5 carbon sugar(deoxyribose, Nitogenous base and phosphate group.
The roots of plants are located underground where light does not penetrate; therefore, photosynthesis cannot occur. Where do plant cells located in the roots obtain energy for their metabolic needs? [2 pts] aerobic respiration – the root cells metabolize the sugar produced during photosynthesis oxidative phosphorylation – the root cells directly convert phosphate in the soil to ATP fermentation – the root cells ferment the sugar produced during photosynthesis anaerobic respiration – the root cells metabolize sugars absorbed from the soil
Answer:
The correct answer is: aerobic respiration -the root cells metabolize the sugar produced during photosynthesis
Explanation:
From the germination of seeds, roots will depend exclusively on energy in the form of photosynthates supplied from the aerial parts of the plant. Photosynthesis takes place in the leaves. Sugar is produced which is transported to all the other plants via phloem tube.Roots absorbs air from the air spaces present between the soil particles and is able to carry out aerobic respiration. The sugar is metabolized and energy is produced.
Plant roots obtain energy for their metabolic needs primarily through aerobic respiration, a process where the glucose produced from photosynthesis is metabolized using oxygen to produce ATP.
Explanation:The roots of plants obtain energy for their metabolic needs primarily through aerobic respiration. While photosynthesis primarily occurs in leaf cells where sunlight is accessible, the energy stored in the form of glucose travels to all parts of the plant, including roots. During aerobic respiration in the root cells, this glucose is metabolized, or broken down, to generate ATP, the energy currency of cells. This process requires oxygen, which is delivered to the roots from air spaces in the soil and from aboveground parts of the plant.
Beyond aerobic respiration, plants do not typically use oxidative phosphorylation to directly convert soil phosphate to ATP. Anaerobic respiration and fermentation processes can occur in the absence of oxygen, but these are not the main source of energy for root cells. Instead, these processes tend to be more common in microorganisms and in plant or animal cells under conditions of oxygen deprivation.
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Sugar loaded into the source end of the phloem draws water into the sieve tubes by osmosis, raising the pressure. What happens to the water at the sink end?
Answer:
it is pushed out harder and faster
Explanation:
Increase genetic variation among species came about as a result of
Answer: It is as a result of Genetic Mutation.
Explanation:
Mutation and sexual reproduction result in increased genetic variation.
Mutation is the change in genetic content or DNA of an organism. It can be as a result of mutagens or errors during DNA replication. When there is mutation in reproductive cells such as sperm or egg, it can be pass to the next generation. The new traits will become permanent if it is subject to natural selection.
Sexual reproduction involves the transfer of new genes from the parents and to the offsprings which lead to new phenotypes and these can help the organism to adapt to it's environment.
You are about to give your first speech in front of your speech class and are nervous about getting up in front of everyone. As you anticipate getting up in front of the class, you begin to feel butterflies in your stomach and a pounding in your chest. Using this example, explain how the sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react to this situation.
Answer:
Sympathetic division is responsible for the nervousness, pounding in the chest and the butterfly movement in the stomach while parasympathetic division overest a and digest the nervousness.
Explanation:
The nervousness experienced when someone is about to speak in front of a class is caused by the sympathetic system which have divergent effects as many different effector organs are activated together for that same purpose. I.e More oxygen are inhaled and delivered to skeletal muscle. The respiratory, cardiovascular, and musculoskeletal systems are all activated at the same time causing an unstable reaction to the central nervous system...
On the other hand, the parasymthetic division causes the central nervous system to rest and digest at a very slow rate..
At a point in time the central nervous system triggered an Homeostasis action which posses a balance between the two divisions.I.e symthetic and parasymthetic and then brings about a balance to the body reaction at that same point.
The sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react to the stress of giving a speech. The sympathetic division triggers the fight-or-flight response, causing the pounding in your chest and butterflies in your stomach. The parasympathetic division helps to calm your body down after the event.
Explanation:When facing a nerve-wracking situation like giving a speech, the sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react. The sympathetic division is responsible for your body's fight-or-flight response, which increases heart rate and causes the release of adrenaline. This leads to the pounding in your chest and the feeling of butterflies in your stomach.
The parasympathetic division then helps to calm your body down after the stressful event, bringing heart rate and breathing back to normal.
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Imagine you are a food science researcher. You are interested in developing a method to slow down fruit oxidation. When fruit flesh (e.g. apple) is exposed to air, it is oxidized and turns brown. This is because the enzyme diphenol oxidase catalyzes the oxidation of phenols (which are colorless) in the fruit to dark-colored quinone products.
Answer: Wash or soak your fruits with salt solution or any solution that contains ascorbic acid such lemon juice, Citric acids as well as solutions containing peptides. They have been proven to inhibit the enzyme (ascorbic acid and salt) and also inhibit the browning effect of quinone (dipeptides, salicylic acids).
Explanation: The discolouration in fruits only occurs when oxidation reaction takes place. With food products, fats could be oxidized leading to rancidity, if the pigments are oxidized, discolouration takes place (majorly browning effect). Oxidation in food reduces the desirability as well as the nutritive value in food.
The browning effect is due to the action of the enzymatic activity of poly-phenol oxidase on polyphenols thereby producing quinones as a by product. Hence to inhibit this activity, your major strategy would be either enzymatic inhibition or reducing the quinone (by utilising reducing agents capable of converting the undesired quinone to a colourless by product).
Ascorbic acid (found in Lemons) is mostly used as an antibrowning agent due to its ability to reduce quinones to diphenols (colourless). Most carboxylic acids such as citric acids found in oranges, tartaric acid (in grapes) also act as PPO inhibitors by lowering its pH or by acting as a ligand at the enzyme active site.
Other browning inhibitors include Peptides and this is because of the thiol groups present in the amino acid called Cysteine. Cysteine is known to be involved with a nucleophilic attack on quinones thereby forming a colourless product which in turn attacks the browning effect. Hence any product that contains peptides (e.g honey) can exhibit browning or enzymatic inhibition of PPO.
To slow down fruit oxidation, one can use ascorbic acid by applying lemon juice to inhibit the oxidase enzyme, refrigeration to slow chemical reactions, optimal oxygen packaging, and avoiding metal containers that catalyze oxidation.
To address the schoolwork question regarding the development of methods to slow down fruit oxidation, it is necessary to understand the chemical processes involved. When fruit flesh, such as an apple, is exposed to air, it undergoes enzymatic browning due to the presence of diphenol oxidase, which catalyzes the oxidation of phenols to quinones, resulting in a brown color. The presence of iron compounds in the apple contributes to this reaction.
Several strategies can prevent or slow down this process. Applying lemon juice, rich in ascorbic acid (an antioxidant), to the cut fruit can inhibit enzymatic browning by denaturing the oxidase enzyme. Refrigeration can also slow chemical reactions such as ripening and browning, as the cold temperature reduces enzyme activity. Moreover, storing fruits in special packaging with an optimal oxygen concentration can prevent excessive oxidation without inducing off-flavors. Antioxidants like ascorbic acid, BHA, and BHT are also widely used as preservatives in food to guard against oxidation. Lastly, keeping the fruit away from metal containers that can catalyze oxidation is another preventive measure.
One of the student in your study group is struggling with the idea that mammals and vertebrates are separate designation.To illustrate the two term,you describe an animal that has the expected mammalian characteristics.What should you include in your description of this animal that would make clear that the animal is specifically a mammal and not just a vertebrate (check all that apply)
A) Hair
B) Notochord at some point during development
C)Cranium
D)Four limbs
E)specialized teeth
Answer:
A) Hair
E) specialized teeth
Explanation:
The Mammals are the group of animals which possess mammary glands. The mammals can be distinguished from the vertebrates as they possess certain features which are unique to them like:
1. The presence of hairs on body which provides protection to the skin used as camouflage and provides insulation.
2. They possess teeth in their lower and upper jaws which can be replaced once in their lifetime.
3. The presence of a four-chambered heart.
4. Presence of mammary glands.
Thus, the selected option is correct.
Mammals are the groups that belong to the vertebrates and phylum Chordata. Hair and specialized teeth should be included in the description of the animal.
What are mammals and vertebrates?Mammals are the class that belongs to vertebrates and are characterized by the presence of mammary glands for feeding the young ones.
Vertebrata is the subphylum of the Chordata phylum. They are characterized by the presence of the backbone in their body and includes mammals, amphibians, birds, reptiles and many others.
Mammals can be distinguished from vertebrates by the presence of the hair on their body that protects them from the heat, provides insulation and camouflage.
They are also distinct from the vertebrates in having a lower and upper set of the jaw that can be replaced once in their lifetime by the permanent set of teeth.
Therefore, option A. hair and E. specialized teeth are correct.
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In one experiment, scientists raised mice in germ-free conditions so the mice lacked intestinal microbes. The mice were fed a low-fat diet rich in complex plant polysaccharides, such as cellulose, that are often called fiber.
When the mice were 12 weeks old, the scientists transplanted the microbial community from the intestine of a single "donor" mouse into all of the germ-free mice. Then they divided the mice randomly into two groups and fed each group a different diet.
Group 1 (the control group) continued to eat a low-fat, high-fiber diet.
Group 2 (the experimental group) ate a high-fat, high-sugar diet.
Identify the components of this experiment by matching the terms with the appropriate category below:
Factors being tested Factors controlled Factors to be measured
(independent variable) (Kept consistent) (dependent variable)
1. initial composition of the microbial community
2. age of the mice
3. change in body fat
4. diet
5. weight gain
6. final composition of the microbial community
Answer:
The answers are:
1. initial composition of microbial community (controlled variable)
2. age of the mice (controlled variable)
3. change in body fat (dependent variable)
4. diet (independent variable)
5. weight gain (dependent variable)
6. final composition of the microbial community (controlled variable)
Explanation:
Before I go ahead to explain each of the choices of classification, let us define each of the different types of variable.
1. Independent variable: this is a variable that is under the control of the experimenter. It is made up of the factors that are being tested. In our example, the independent variables is; diet. The aim of the experiment is to study the effect of the type of diet on the mice. The diet can also be varied and assigned at will by the experiment.
2. Dependent variable: a dependent variable is one whose outcome is reliant on the effect of the changes in the independent variable. It is the factor to be measured. In our example, the factors that obey these definitions are; change in body fat and weight gain. these two factors depend on the types of diet taken by the mice, hence they are the factors to be measured.
3. Controlled variable: these are variables that are kept constant throughout the experiment to rule out biases, because they produce the same effect on the dependent variable as the independent variable and must be keep constant or uniform in the test groups. in our example, the factors that are controlled include; initial and final compositions of the microbial communities and the age of the mice, because the microbial composition if not uniform among the groups can cause biases in how the diets are processed, and the age of the mice also must be uniform to avoid age-related differences in the processing of the diets
One character in peas that Mendel studied was yellow versus green seeds.
A cross between a homozygous yellow line (YY) and a homozygous green line (yy) will result in F1 plants that are heterozygous (Yy) for this trait and produce yellow seeds.
When an F1 plant undergoes meiosis, what gamete types will it produce, and in what proportions?
Use your understanding of Mendel's law (or principle) of segregation to label each gamete possibility as "correct" or "incorrect.a. incorrect
b. correct
c. incorrect
d. incorrect
e. incorrect
f. incorrect
Answer:
1/2 will be Y
1/2 will be y
Explanation:
In the process of meiosis haploid gametes are formed which means one gamete contains half the information from the parent cell (one set of 23 chromosomes). Because of Mendel's law (or principle) of segregation, if the parent has genotype heterozygous (Yy), it will be divided in two gametes in the following way:
Yy will be broken in to Y + y.
One gamete will be Y the other gamete will be y
Your BFF is a red-headed, left-handed, blue-eyed girl. She finds out that you are taking Human Genetics and wants you to explain to her the genetic basis of her three traits. So, your task is to provide her with the most up-to-date explanation, to the extent that is possible. You have to do so in 400-500 words.
Answer:
Red hair, left handed, and blue eyes are recessive traits.
Any trait has its two allele, dominant and recessive. Dominant trait allele if present with the recessive allele, then the phenotype will be of dominant allele.
Suppose A = dominant allele for brown color hair, a = recessive allele for red color hair.
As mentioned above red color hair is recessive in nature. for the recessive traits to show, the alleles should be recessive only.
Aa (brown hair) (parent 1) is crossed with Aa (parent 2) (brown hair). They both have gametes A and a.
Their children will have genotype AA, aa, and Aa. AA = brown color, aa = red color hair, and Aa= brown color hair.
Same case goes with the left hand and blue eyes traits.
So, the girl parents must be having recessive alleles for all the traits, and she must have inherited all the recessive allele only.
Explanation:
First of all, you should ask your friend if any of her parents also have these characteristics. In this you must explain the concept of heredity, which is the genetic phenomenon that allows us to be able to inherit genetic information from our parents, through DNA. That's because each of us has half the DNA of each of our parents, so we have characteristics in common with them.
After explaining heredity, you should talk about your friend's blue eyes. Blue eyes are characterized by a small amount of melanin, which is the brown pigment that colors our eyes, skin and hair. The more melanin, the darker the eyes. What determines the amount of melanin in the eyes is DNA. If your friend's parents have light eyes, it means they have little melanin and that trait was passed to her through DNA.
Red hair, on the other hand, is the result of a genetic mutation that is passed from parents to children through DNA. People with red hair have a mutation in the gene responsible for the production of melanin. This mutation causes the gene to produce very low levels of melanin for the hair, but it produces large amounts of pheomelanin which is a reddish pigment.
However being left-handed is not related to heredity. Many scientists believe that what determines whether a person is left-handed or not is a set of genes that choose the left or right side of the body as dominant. However, other scientists believe that this is actually linked to a matter of skill and custom.
b) WRITE a brief (between 150-200 words) email to a scientific colleague that explains (IN YOUR OWN WORDS) how the cholera bacterium causes diarrhea AT THE CELLULAR LEVEL and what role OSMOSIS plays in this.
Answer:
The bacterium Vibrio cholerae is the primary cause of cholera disease that mainly infects the small intestine and primarily leads to the dehydration of the body.
Explanation:
The genetic analysis reveals that the aforementioned causative bacteria surpass the acidic conditions of the stomach and eventually reaches the intestinal wall and attaches to it. This is followed by the production of toxic protein by the bacterium. This protein is taken inside the cell via receptor mediated endocytosis followed by its binding to the host protein Arf6. This binding leads to the production of cAMP that results in the dehydration process. This mechanism leads to excessive accumulation of chloride ion in the intestine preventing the entry of sodium ion.
These two ions are associated with the creation of water-salt environment in the intestine that leads to tremendous diarrhea via the process of osmosis.
Hence, we can say that cholera bacterium affects the individuals at the cellular level and osmosis plays a vital role in the process.
You are observing a specimen under the microscope. You have located a cell and brought it into sharp focus using the 10x objective (below left). However, when you switch to the 40x objective, the cell is no longer visible (below right)! What has happened and how would you correct this? [2 pt; G5.2]
Answer:The cell was no longer visible because the image wasn't centered before switching to higher objective of 40X.
Explanation:
While using microscopes, the image will approximately remain in focus if you adjust the focus during change of magnification. Therefore, there is need to always adjust the focus during change in magnification.
If object is not centered before switching to a higher power objective, object will not be visible.
To avoid this, center the object before switching to a higher power objective. This will help you find the object after switching the objective. And it is also important to use fine focus and not coarse focus.
Answer:
When you shift from low power(10x) to high power(40x), the 40X objective lense moves directly over the specimen and this in turns results in the following changes:
•increase in the magnification of a specimen
•decrease in the light intensity which makes the image appears dim
•decrease in the area of field of view, i.e the specimen will appear larger
•decrease in depth of field, i.e. you won't be able to view an entire surface of the specimen.
And the only way to correct this is to shift back to 10X,centre the specimen and then again switch back to 40X
How do scientists mark the end of a geologic era and beginning of the next?
Answer:
Scientists mark the beginnings and the ends of eras by observing extinctions and appearances of species in the fossil record. The appearance of various species of ammonites, belemnites, trilobites, foraminifera, nannofossils and magnetic events also mark the boundaries of many of the defined eras.
Explanation:
Answer:
Scientist usually mark the end of a geologic era and the beginning of the next by mass extinction or drastic changes to the earth.
Explanation:
hope this helps
10. Blood returning to the mammalian heart in a pulmonary vein drains first into the
A. Left ventricle
B. Left atrium
C. Right atrium
D. Vena cava
E. Right ventricle
Answer: Option B.
Left atrium.
Explanation:
Pulmonary veins are veins that transmit oxygenated blood from the lungs to the heart. There are four largest pulmonary veins,two from the lungs that drain into the left atrium of the heart. The blood that is drained into the left atrium is pumped into the left ventricle through biscupid valve. The pulmonary vein function in respiration by receiving oxygenated blood in the alveoli and return it to left alveoli.
18. Which of the following is NOT a factor in the movement of water into tissues?
A. active transport
B. salinaity gradient
C. fluid pressure
D. diffusion
E. osmotic gradient
Answer:D
Explanation:diffusion involved movement of molecules ( gas,) from high concentration to low concentration without a semipermeable membrane. Whereas movement of water require active transport through specialized xylem tissues.
Dr. Albertson performed a lumbar laminectomy, 2 vertebral segments, for decompression on Grace James on September 15. One month later, as originally planned, Dr. Albertson brought Grace back into the OR to implant an epidural drug infuser with a subcutaneous reservoir. Code both procedures.
Answer:
The procedure is epidural drug administration.
Explanation:
The patient (Grace) must have suffered back pain as a result of lunar laminectomy performed on her and epidural drug administration was introduced.
In epidural drug administration, epidural injection are used to treat radicular pain from herniated discs, spinal stenosis, chemical disc, chronic pain secondary to post surgery syndrome.
The injection is given in theatre conditions.
Final answer:
Dr. Albertson performed a lumbar laminectomy on two vertebral segments and later implanted an epidural drug infuser, but without specific medical coding resources, exact CPT codes for these procedures cannot be provided.
Explanation:
The question pertains to procedural medical coding for two separate surgeries performed by Dr. Albertson. The first surgery is a lumbar laminectomy which is a procedure that entails removing part of the vertebral bone called the lamina. The second surgery involves the implantation of an epidural drug infuser with a subcutaneous reservoir. These procedures are coded differently in medical coding systems like CPT (Current Procedural Terminology) codes used for billing purposes. The codes would be specific to the number of vertebral segments operated on during the laminectomy and the specific type of infusion device implanted. However, without access to the specific coding books or software, I am unable to provide the exact codes for these procedures.
Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advantage relative to b, so that the fitnesses of the three genotypes are BB = 1, Bb = 1, and bb = 1-s. In this case, s = 0.50, so that bb homozygotes have 50% fitness of heterozygotes and BB homozygotes. If the population has the following genotypic counts prior to selection of BB = 500, Bb = 250, and bb = 250, what is the expected change in the frequency of B after one generation with selection? Please give your answer to two decimal places.
Answer:
0.09
Explanation:
Before selectionTotal number in population = 1000
Genotype frequencies
Genotype frequency of BB = 500/1000 = 0.5
Genotype frequency of Bb = 250/1000 = 0.25
Genotype frequency of bb = 250/1000 = 0.25
Allele frequencies
Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625
Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375
After selectionWe are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).
Now the total number of individuals is 500 + 125 + 250 = 875.
Genotype frequencies
Genotype frequency of BB = 500/875 = 0.57
Genotype frequency of Bb = 250/875 = 0.29
Genotype frequency of bb = 125/875 = 0.14
Allele frequencies
Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715
Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285
Change in frequency of B after 1 generation0.715 - 0.625 = 0.09
Final answer:
The expected change in the frequency of allele B after one generation of selection, given the provided genotypic counts and selection coefficients, is an increase by approximately 0.09.
Explanation:
First, we can calculate the initial frequencies of alleles B and b. Each BB individual contributes two B alleles, and each Bb individual contributes one B allele. The total number of B alleles in the population is (2 × 500) + (1 × 250) = 1250, and similarly, the total for b alleles is (2 × 250) + (1 × 250) = 750. The total number of alleles is 2000. Therefore, the frequency of B (p) is 1250/2000 = 0.625 and the frequency of b (q) is 750/2000 = 0.375.
After selection, the new frequencies of the alleles can be determined. The fitness values are wBB = 1, wBb = 1, and wbb = 0.5. We need to adjust the number of each genotype by its fitness:
Adjusted BB = 500 × 1 = 500Adjusted Bb = 250 × 1 = 250Adjusted bb = 250 × 0.5 = 125The new frequency of B after selection (p') will be the sum of the alleles from BB and Bb genotypes after fitness adjustment over the total adjusted alleles.
New frequency of B (p') = (500 × 2 + 250) / (2 × (500 + 250 + 125))p' = (1250) / (2 × 875)p' = 0.7143 (to four decimal places)The change in the frequency of B after one generation of selection is the difference between the new frequency and the original frequency.
Change in frequency of B = p' - p = 0.7143 - 0.625 = 0.0893 or to two decimal places, 0.09.
You can make a cell extract that is able to perform glycolysis in vitro (in a test tube) if glucose is added. Arsenate is a potent inhibitor of triose phosphate dehydrogenase, the enzyme required for the 6th step in glycolysis. If both arsenate and glucose are added to the cell extract, what happens?
a. ATP levels decrease.
b. Both ATP and pyruvate levels decrease.
c. Both ATP and pyruvate levels increase.
d. ATP levels increase.
Answer: B
Explanation:
If glucose and arsenate are both added to the cell extract, at first glycolysis will start.
In step one of glycolysis, glucose is phosphorylated to glucose-6-phosphate catalyzed by hexokinase which splits the ATP into ADP, and the Pi is added on to the glucose.
In step 3 of glycolysis, fructose-6-phosphate is further phosphorylated to fructose 1,6-bisphosphate. The enzyme is phosphofructokinase. This again involves hydrolysis of another ATP molecule.
A total of two ATP is used.
Step 6 in glycolysis reaction which involves generation of 2 ATP's molecules is inhibited by arsenate. Hence all other glycolytic reaction would not take place. Therefore no ATP is produced and pyruvate is not produced also.
ATP level decreases because ATP is only used up but no ATP is gained from the inhibited pathway. Also the inhibition of the step 6 enzyme cut short the pathway and pyruvate the end product of the pathway is not formed.
4. Which of the following statements are accurate?
A. The heartbeat in a human heart is initiated by the atrioventricular mode.
B. During systole, the heart's ventricle(s) is/are relaxing and filling with blood.
C. During systole, blood will be pumped from the heart into the veins of the body
D. The pulse is used to determine one's heart rate.
E. The nervous system and certain hormones can alter pacemaker activity.
Answer: options D and E
Explanation:
The heartbeat in a human is mostly initiated by the sinoatrial node (SA node also known as the heart's natural pacemaker. Made up of a specialized bundle of cell that receives an impulse causing contraction of the atria wall allowing blood to flow into the ventricles. During systole, the ventricles are not relaxed but contracted to allow for bloodflow out of the ventricles to the aorta which is the largest artery in the human body.
A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
Hello. This question is incomplete. The full question is:
"A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively.
Which of the following is a possible genotype for the mother?
a. IAIB b. IAIA c. ii d. IAi e. IBIB"
Answer:
d. IAi
Explanation:
As we know, there are four types of blood, which are: A, B, AB and O.
There are three genes responsible for determining the blood type in humans, these genes are called IA, IB and i.
The genes IA and IB are dominant in relation to gene i, but they do not present dominance among themselves. The "IA" gene refers to type A blood, the "IB" gene refers to type B blood, while the "i" gene refers to type O blood. In addition, bloods can be negative or positive in relation to whether or not they have the Rh suit.
Based on this, we can say that in relation to the question above, the only option that can represent the genotype of a woman who has positive type A blood is IAi.
Which of the actions below is preformed by all cells to maintain homeostasis
A.They divide rapidly into many specialized cells
B.They use light energy to produce food molecules
C.They obtain energy from nutrients and remove waste products
D.They combine with another cell to increase in size
Answer: Option C.They obtain energy from nutrients and remove waste products
Explanation:
Homeostasis is the maintenance of a stable internal environment. Hence, cells allow processes within its cytoplasm that produces energy by degrading macromolecules like carbohydrates or lipids while expelling by-products such as gases, salts etc out of its cell membrane.
Answer:C
Explanation: they maintain a balance between the nutrients obtained from food and waste product generated during digestion of food.
Which of the following is true of leukocytes? A. They are very similar in appearance and function as red blood cells. B. Leukocytes are smaller than and more numerous than red blood cells. C. Like red blood cells, mature leukocytes also lack a nucleus and other intracellular organelles. D. Leukocytes always remain in circulation throughout the body. E. The leukocyte is a major component of the body’s defenses against disease.
Answer: the correct option is E. (The leukocyte is a major component of the body’s defenses against disease)
Explanation: the leukocyte also called the white blood cells is one of the major types of the hematocytes. There are different types of white blood cells which are classified based on their functions and appearance, they include:
- neutrophils,
- eosinophils,
- basophils,
-lymphocytes, and
- monocytes.
The major function of the white blood cells is to protect the body against disease by phagocytosis of invading microorganisms capable of causing infections.
Final answer:
The correct answer is E. Leukocytes, also known as white blood cells, are a major component of the immune system and they have a nucleus and intracellular organelles, unlike red blood cells. They are larger, less numerous than red blood cells, and can leave the circulation to participate in defending the body against diseases and infections.
Explanation:
Among the statements provided about leukocytes, or white blood cells (WBCs), the correct answer is E. The leukocyte is a major component of the body’s defenses against disease. Leukocytes are indeed significantly different from erythrocytes, or red blood cells, in several ways:
Contrary to erythrocytes, leukocytes do contain a nucleus and other intracellular organelles, making them the only complete cell type among the formed blood elements.
Leukocytes are larger than erythrocytes and are far less numerous in the bloodstream, typically with a count of 5000 to 10,000 leukocytes per μL as compared to millions of red blood cells in the same volume.
While red blood cells remain in the circulatory system, leukocytes often leave the bloodstream to perform their functions, which include protecting the body against infections and cleaning up debris.
There is a variety in leukocyte types, including neutrophils, basophils, eosinophils, lymphocytes, monocytes, and macrophages, each with specific roles in the immune response.
It is these characteristics that enable leukocytes to play a pivotal role in the immune system by fighting off infections and other threats to the body's health.
Suppose you want to know the proportion of A_B_ among the total progeny, you may first estimate which genotypes represent A_ and which genotypes represent B_. Clearly AA and Aa for the first genotype A_.
Answer and Explanation:
Complete question:
"What is the expected proportion of progeny with the following broad genotypes each having a distinct phenotype in a typical Mendelian dihybrid cross:A_B_ : A_bb : aaB_ : aabb. Be able to derive the above ratios directly by studying alleles for one gene at a time (individual Punnett) and applying the product rule.
Example: (1 AA: 2Aa: 1aa) (1BB : 2Bb: 1bb) Suppose you want to know the proportion of A_B_ among the total progeny, you may first estimate which genotypes represent A_ and which genotypes represent B_. Clearly AA and Aa for the first genotype A_. And BB and Bb for the second genotype".
Answer:
In a typical Mendelian dihybrid cross, we have:
Parental) AaBb x AaBb
Gametes) AB Ab aB ab
AB Ab aB ab
Punnet Square) AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBB aaBB aaBb
ab AaBb Aabb aaBb aabb
F1 phenotypic frequencies)
9/16 A-B-
3/16 aaB-
3/16 A-bb
1/16 aabb
But if you do not want to do the dihybrid cross, you might just perform the cross for each gene and then apply the product rule. This is:
Hybrid Cross for gene A)Parental) Aa x Aa
Gametes) A a A a
Punnet Square) A a
A AA Aa
a Aa aa
F1 genotypic frequencies)
1/4 AA
2/4 Aa
1/4 aa
F1 phenotypic frequencies)
3/4 A-
1/4 aa
Hybrid Cross for gene B)Parental) Bb x Bb
Gametes) B b B b
Punnet Square) B b
B BB Bb
b Bb bb
F1 genotypic frequencies)
1/4 BB
2/4 Bb
1/4 bb
F1 genotypic frequencies)
3/4 B-
1/4 bb
To estimate the progeny phenotypic proportions (A-B-, aaB-, AAB-, aabb), from the genotypic proportions of progeny, you can sum the genotypic frequency of homozygote plus heterozygote, to get the proportions A- and B-.
AA=1/4Aa=2/4
A- = AA+Aa= 1/4 + 2/4 = 3/4
BB= 1/4Bb= 2/4
B- = BB + Bb = 1/4 + 2/4 = 3/4
Then multiply these proportions to get the progeny proportions, like this:
A-B-= (AA+Aa) x (BB + Bb) = 3/4 x 3/4 = 9/16
A-bb = (AA + Aa) x bb = 3/4 x 1/4 = 3/16
aaB- = aa x (BB + Bb) = 1/4 x 3/4 = 3/16
aabb = aa x bb = 1/4 x 1/4 = 1/16
Note: You can get the same results using directly the phenotypic proportions of each cross.
Based on your results in Part A, which of the following statements most appropriately reflects the relative effectiveness of the vancomycin treatment, teixobactin treatment, and the control?
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
Neither teixobactin or vancomycin were effective at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly less effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Answer:
Option C is correct,which states that Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin reduced the number of MRSA colonies about as effectively as vancomycin relative to the control. Statement 1
Explanation:The most appropriate statement reflecting the relative effectiveness of vancomycin treatment, teixobactin treatment, and the control is:
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
This means that both vancomycin and teixobactin were similarly effective in reducing the number of MRSA colonies compared to the control group. Therefore, teixobactin was not significantly more or less effective than vancomycin.
Learn more about Teixobactin Against MRSA here:https://brainly.com/question/36741462
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In poodles, black fur is dominant to white fur. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the BEST conclusion?
Answer:
The black poodle is true breeding (homozygous) for the black fur colour.
Explanation:
Let the allele for fur colour be represented by B.
Black fur (B) is dominant over white fur (b).
Black poodle (B _) is crossed with white poodle (bb):
B _ x bb
Progeny (all black) = Bb, Bb, _b and _b.
Since all the progeny are black, _b = Bb (remember, B is dominant over b)
Hence, B_ is also BB.
The best conclusion is that the black poodle is true breeding (homozygous) for the black fur colour.
Answer:
Explanation: all the puppies in the F1 generation are black because homozygous black (BB) is crossed with a homozygous white(bb) and since the black fur is dormant all the puppies will be black ie heterozygous black (Bb).
Botulinum toxin is taken up by axon terminals and inhibits the ability of a motor neuron to stimulate a skeletal muscle motor unit to contract by preventing the release of acetylcholine from the axon terminal. Explain how this prevents skeletal muscle contraction 2. Explain why injecting Botox (derived from botulinum toxin) into a superficial facialmuscle reduces the appearance of deep facial wrinkles called muscle lines Hint: Superficial facial muscles are attached at one end to skin
Answer:
1. An essential neurotransmitter, which instigates the contraction of muscles by transmitting the nerve impulses towards the neuromuscular junction present at the skeletal muscles is known as acetylcholine. Thus, when the discharging of acetylcholine from the terminals of axons is inhibited by the administration of botulinum toxin, the tendency of the motor neuron to perform the contraction of skeletal muscles also gets prevented.
2. A neurotoxin generated with the application of botulinum bacterial species is termed as botox. The administration of this botox or botulinum toxin into the superficial facial muscle declines the appearance of muscle lines or deep facial wrinkles, due to this it also has an application as cosmetic medicine.
The mechanism of the working of this toxin is that it prevents the discharging of the neurotransmitter acetylcholine from the axon terminals of the neuromuscular junction present at the skeletal muscles. It temporarily paralyzes the facial muscles and prevents its contraction, which eventually prevents the appearance of wrinkles.
Explain why the underproduction of albumin by a cirrhotic liver contributes to excessive filtration, which leads to ascites. Refer again to the forces that determine net filtration pressure in the hepatic capillaries, and to the function of albumins
Answer:
Albumin is produced mainly by the liver. Albumin is the major protein that is acts as the carrier protein for steroids, thyroid hormones and fatty acids in the blood.
Albumin stays in blood vessels as they are too large to move across the capillary walls, in blood vessels by contributing to osmotic pressure or colloid osmotic pressure that maintain water volume. So, the albumin is the protein that is involve primarily on the water reabsorption.
Bacteria are grown in N15 over time, then switched to growth in N14 medium. After three divisions (60 min) what is the percentage of bacterial cells that have N15 labeled DNA? (A) 100% (B) 50% (C) 25% (D) 10% (E) 5%
Answer:
C. 25%
Explanation:
A bacterial cell gives rise to two daughter cells by one division. If a single bacterial cell enters into cell division, it will form a total of 2^3= 8 cells after three rounds of cell division. The two DNA strands of the parent DNA duplex having N15 will be present in two bacterial cells out of the total 8 cells. This would occur since the process of DNA replication forms a new DNA duplex with one parental strand and one new strand. So, after three round of cell division 25% cells will have DNA with N15 (2/8 x 100= 25%)
Final answer:
After three divisions in 14N medium starting with 15N-labeled DNA, due to semi-conservative replication, 25% of bacterial cells will contain DNA with 15N label per Meselson and Stahl's experiment.
Explanation:
The experiment by Meselson and Stahl with E. coli growing in heavy nitrogen (15N) and then switching to light nitrogen (14N) demonstrates the semi-conservative model of DNA replication. After the first division in 14N medium, each DNA molecule contains one strand with 15N and one with 14N, making it fifty percent 14N. By the principle of semi-conservative replication, after three divisions, there will be one more division than there are strains that have any 15N isotopes. Thus, one quarter of the bacterial cells should contain 15N since those are the ones that did not undergo the last division and still contain one original heavy strand. This makes the answer (C) 25%.
The recent hominin fossil finds from Ileret, Kenya, negate the conventional view held since 1960 that H. habilisand H. erectusevolved one after the other. Instead, they lived side by side in eastern Africa for perhaps half a million years.A. TrueB. False
Explanation:
The recent finding of the fossils which showed that Homo erectus and Homo habilis lived side by side in eastern Africa for perhaps half a million years challenged the conventional way that these two species evolved one after the other(H.habilis 1.44 million years old and H.erectus 1.55 million years old)
The fossils were found in Kenya and took years to prepare the specimens for study and to be sure of the identification of the species, the scientists saidUniversity of Utah geologists determined the dates of the fossils from volcanic ash depositsThe most recent Homo habilis that had been known was about the same age as the earliest Homo erectus, said Daniel Lieberman, a professor of biological anthropology at Harvard University, “Now we have extended the duration of the habilis species, and there’s no doubt that it overlaps considerably with erectus”The fact that the two hominid species lived together in the same lake basin for so long and remained separate species, Meave Leakey said in a statement from Nairobi, “suggests that they had their own ecological niche, thus avoiding direct competition”