Answer:
c. Loud noise
Explanation:
Final answer:
The fear-producing stimulus in the 'Little Albert' study was the loud noise (option c), which was used as the unconditioned stimulus to elicit a natural fear response in the subject.
Explanation:
In the Little Albert study, the fear-producing stimulus used as an unconditioned stimulus (UCS) was the loud noise. Initially, Little Albert was not afraid of a white rat, introduced to him as a neutral stimulus. However, when researchers paired the sight of the white rat with the loud sound of an iron rod being clanged, Albert began to cry. This loud noise functioned as the UCS because it elicited a fear response naturally, without any prior conditioning. After several pairings of the white rat (which became the conditioned stimulus, or CS) with the loud noise (the UCS), Albert began to show fear of the rat by itself. Thus, the UCS in the Little Albert experiment was the loud noise, option c.
Martin Luther King Jr. went to jail many times for acts of civil disobedience, such as ignoring court orders prohibiting marches and boycotts. Some people, even those who agreed with him, criticized him for disobeying the law. Today, however, most people view these actions as a form of?
Answer:
The correct answer is - positive deviance.
Explanation:
This action of the people is clearly the form of positive deviance. Positive deviance is the action of a certain group or individual in every community whose abnormal or uncommon strategist and behaviors help them to identify the most probable than their peers.
In this scenario, the supporters of Martin Luther King Jr. criticized him for his disobeying the law that is considered as positive deviance.
Thus, the correct answer is - positive deviance.
In today's perspective, Martin Luther King Jr.'s acts of civil disobedience can be viewed as a form of protest against unjust laws and policies.
Explanation:In today's perspective, Martin Luther King Jr.'s acts of civil disobedience, such as ignoring court orders prohibiting marches and boycotts, are widely viewed as a form of protest against unjust laws and policies. These actions were part of the larger civil rights movement aimed at achieving equality and ending racial discrimination. King and other civil rights leaders believed that standing up against unjust laws was necessary to bring attention to and ultimately change these unfair practices.
In Maize, spotted kernels are made up of pigmented and colorless cells, the colorless cells in a spotted kernel can result from [choose ALL that apply, note that incorrect selections will deduct points]:
O The absence of Ac in the genome and the absence of Ds in the C gene in the colorless cells.
O The absence of Ac in the genome and the presence of Ds in the C gene in the colorless cells.
O The presence of Ac in the genome and the absence of Ds in the C gene in the colorless cells.
O The presence of Ac in the genome and the presence of Ds in the C gene in the colorless cells.
O The presence of Ac in the C gene in the colorless cells.
O The absence of Ac and Ds in the genome and a wildtype C gene.
Answer: the correct statements are;
-The presence of Ac in the genome and the presence of Ds in the C gene in the colorless cells.
-The presence of Ac in the C gene in the colorless cells.
Explanation: Ds moved from a site near the centromere into the C gene located close to the telomeric ends
In its new location, Ds prevents the expression of C (pigment) gene
The inactivation of C gene explains colorless parts of the kernel .
When the antibiotic valinomycin is added to actively respiring mitochondria, several things happen: the yield of ATP decreases, the rate of O2 consumption increases, heat is released, and the pH gradient across the inner mitochondrial membrane increases. Explain these observations by listing the causal order of events that occur beginning with the addition of valinomycin to the respiring mitochondria and ending with the release of heat. Note: some events listed do not occur and should not be placed Valinomycin added The rate of ATP synthesis decreases The rate of electron transfer and oxygen consumption increases The pH gradient across the The valinomycin-K complex moves K ion out of the mitochondrial matrix. The electrical potential across the The rate of ATP hydrolysis increases The electrical potential across the mitochondrial membrane decreases. Valinomycin binds K" ion. The valinomycin-K' complex moves into the mitochondrial matrix. The pH t across the Heat released
Valinomycin binds to K+ ions, disrupting the potassium gradient and causing a decrease in ATP production. Increased efforts to maintain energy result in enhanced electron transport and oxygen consumption, along with heat generation.
When valinomycin is added to actively respiring mitochondria, it complexes with potassium ions (K+) and facilitates their transfer across the mitochondrial membrane. This disrupts the osmotic pressure and affects the proton gradient. This sequence of events begins with valinomycin binding to K+ ions on the surface of the mitochondria and forming the K+-valinomycin complex. The complex becomes highly soluble in the nonpolar interior of the membrane, allowing it to cross the membrane. Once inside, K+ is released, and the valinomycin is free to return and bind to another K+ ion, continually disrupting the normal potassium gradient. This action increases the permeability of the membrane to K+, leading to the dissipation of the electrical potential across the membrane. This reduction in electrical potential affects the proton gradient, which in turn impairs the ATP synthase activity, leading to a decrease in ATP production. Without the efficient generation of ATP, the rate of electron transport and oxygen consumption increases as the mitochondria try to maintain energy production, while the dysfunctional gradient causes the proton pumps to work harder. The extra work results in the generation of heat. Meanwhile, the pH gradient across the inner mitochondrial membrane increases due to the altered balance between proton influx and efflux.
If the hydrostatic pressure of blood flowing through the glomerular capillaries and forcing fluid through the leaky endothelium was 60mmHg, the pressure caused by the lack of proteins in the fluid inside Bowman's capsule was 32 mmHg, and the pressure caused by the inability of the capsule to expand in volume was 18mm Hg, what would the net filtration pressure (in mmHg) be?
Answer:
The net filtration pressure is 10 mmHg.
Explanation:
for calculating net filtration we will plus the pressure inside boman's capsule with pressure caused by inability of capsule expansion and then subtract them from total pressure in leaky epithelium.
net filtration pressure= total blood pressure in leaky epithelium - total blood pressure in capsule
as we know that total blood pressure in capsule is 32 + 18 = 50 mmHg
therefore
net filtration pressure= 60 - 50
net filtration pressure= 10 mmHg
Which of the following best describes the change in kdr genotype frequencies over time in A. gambiae?a)From pre-2006 to post-2006, the frequency of the r/r genotype increased slightly while the frequencies of +/+ and +/r genotypes decreased dramatically.b)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically.c)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes remained relatively constant.d)From pre-2006 to post-2006, there was little change to the r/r, +/+, and +/r genotype frequencies.
Answer:
b)From pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically.
Final answer:
The frequency of the resistant r/r genotype in Anopheles gambiae mosquitoes increased dramatically post-2006, while the susceptible +/+ and +/r genotypes decreased dramatically, indicating evolution due to insecticide selection pressure.
Explanation:
Population genetics examines the changes in allele frequencies and genotype frequencies in a population over time. According to the provided reference, the change in kdr genotype frequencies in Anopheles gambiae mosquitoes indicates that the frequency of the resistant allele (denoted as 'r') and the corresponding homozygous genotype (r/r) increased dramatically post-2006, which suggests that the use of anti-vector interventions may have selected for this resistance allele at a higher rate. On the contrary, the frequencies of the susceptible homozygous genotype (+/+) and heterozygous genotype (+/r) decreased dramatically over this time period.
The correct description of these changes would most likely be that from pre-2006 to post-2006, the frequency of the r/r genotype increased dramatically while the frequencies of +/+ and +/r genotypes decreased dramatically. This reflects a population undergoing evolution due to selection pressure from the use of insecticides, as alleles that confer resistance to these chemicals are favored and become more common within the population.
People with full blown sickle-cell disease will experience harmful effects such as, red blood cells break down, clump, and clog blood vessels. Broken cells can accumulate in the spleen. This can lead to many symptoms such as physical weakness, heart failure, pain, and brain damage. Having so many symptoms can be explained by __________.
a. the polygenic nature of sickle-cell disease
b. side effects of the drugs used to treat sickle-cell disease
c. a bacterial infection interacting with the sickle-cell allele
d. the sickle-cell allele can impact many different body systems
Answer :a. the polygenic nature of sickle-cell disease
Explanation:
SS is a genetic disorder in which more than two genes control a single character. Contrary to single gene to a trait. It exhibit an abnormal clinical heterogeneous traits for Mendel inheritance with multiple factors exerting influence on an individual disease outcome. Therefore multiple complications arises with symptoms
Answer:d. the sickle-cell allele can impact many different body systems
Explanation:sickle cell anaemia is the shortage of red blood cells.red bloods cells functions to carry oxygen through out the body aided by the haemoglobin.when however a person has sickle cell anaemia,his red blood cells become abnormal.they are no longer flexible and round ,they become stiff and sickle-shaped.when this changes in the red blood cell occurs,it is no longer flexible to pass through the blood vessels,causing the blockage of the vessels and the red blood cells break apart easily ,causing its shortage in the body.as body organs need oxygen to survive, people with sickle cell anaemia tends to have organ failures,pain crises , infections etc.
Which of the following behaviors is not an inherited behavior?
Answer:
Humans eating with utensils is not an inherited behavior.
Explanation: We aren't born knowing how to eat with utensils. This is a skill that we have to be taught, therefore it is a learned behavior.
Inherited behaviors are instinctual and passed down from parents to offspring through genes, while non-inherited behaviors are learned or acquired through experience. Learning to ride a bike, as well as acquiring language, playing an instrument, and solving puzzles are examples of non-inherited behaviors.
Explanation:Inherited behaviors are those that are passed down from parents to offspring through genes. They are instinctual and do not need to be learned. Non-inherited behaviors, on the other hand, are learned or acquired through experience.
One behavior that is not inherited is learning to ride a bike. Riding a bike is a skill that needs to be learned through practice and instruction, rather than being programmed into our genes.
Other examples of non-inherited behaviors include language acquisition, playing an instrument, and solving puzzles. These behaviors are acquired through learning and practice.
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In German cockroaches, the curved wing allele (cv) is recessive to the normal allele wing (cv ). Bill, who is raising cockroaches in his dorm room, finds that the frequency of the allele for curved wings in his cockroach population is 0.6. In the apartment of his friend Joe, the frequency of the allele for curved wings is 0.2. One day Joe visits Bill in his dorm room and several cockroaches jump out of Joes' hair and join the population in Bill's room. Bill estimates that 10% of the cockroaches in his dorm room now consist of individual roaches that jumped out of Joe's hair. What will be the frequency of roaches with curved wings among cockroaches in Bill's room?
The new frequency of curved-wing allele in Bill's cockroach population after incorporating cockroaches from Joe's population would be 0.56, assuming no other changes or selective pressures.
Explanation:To determine the frequency of the curved wing allele after Joe's roaches joined the population in Bill's room, we must first determine the combined frequency of the curved wing allele from both populations. Considering that the allele frequency in Joe's roaches was 0.2 and it made up 10% of the new combined population, and Bill's roaches had a frequency of 0.6 and composed 90% of the population, we can calculate a new overall frequency as follows: (0.6 * 0.9) + (0.2 * 0.1) = 0.54 + 0.02 = 0.56.
So, assuming no other changes or selective pressures, this would be the new frequency of the curved wing allele in Bill's roach population after Joe's roaches joined.
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Final answer:
The new frequency of the curved wing allele in Bill's cockroach population after incorporating roaches from Joe's population is 0.56.
Explanation:
To calculate the new frequency of the curved wing allele (cv) in Bill’s cockroach population after the introduction of cockroaches from Joe’s population, we need to account for the change due to the roaches from Joe’s hair.
Initially, Bill’s population had a frequency of 0.6 for the curved wing allele. Joe’s had a frequency of 0.2. After 10% of Bill's population is replaced by Joe's cockroaches, the new frequency (pnew) of the curved wing allele in Bill’s population can be calculated as:
pnew = (0.9 × Bill’s original frequency) + (0.1 × Joe’s frequency)
pnew = (0.9 × 0.6) + (0.1 × 0.2)
pnew = 0.54 + 0.02
pnew = 0.56
Therefore, the new frequency of cockroaches with curved wings among cockroaches in Bill's room will be 0.56.
Toemas’s syndrome is a autosomal dominant disease, homozygous lethal, seen at a frequency of 1 in 10,000 cats. Felines with this disease have a very strong desire to have their backs scratched. Assume the locus is in Hardy Weinberg Equilibrium.
A male cat with Toemas's syndrome and an unaffected female cat have a newborn kitten. What is the probability that the kitten is male AND he has Toemas’s syndrome? Assume mating is at random.
(Hint: recall your work on matings for two independent loci!)
Answer: 50% or 25% according to the genotype of the male cat.
Explanation:
The female cat is unaffected, and since this is an autosomal dominant diease, it means this cat does not have any affected alleles. Thereby, its genotype is aa. This is because dominant alleles show their effect even if the organisms only has one copy of the allele. The male cat is affected with this syndrome so its genotype is either AA or Aa. For this disease to manifest itself, only one affected allele is needed. Although both alleles can be affected so the individual can be homozygous dominant or heterozygous.
To find this we have to do a punnett square. Here, we use the gametes produced by each individual. Gametes are sex cells, egg or sperm, which only have one allele of each gene.
Assuming that the cross is between aa and AA, the gametes produced by "aa" are "a", and the gametes produced by "AA" are "A" So we are having a cross between an "a" gamete and a "A" gamete. 100% of the offspring will be Aa so it is a heterozygous, which will develope this syndrome. And, males have one Y chromosome and one X chromosomes, thereby the chances of having a male kitten is 50%. Then, to conclude, there is a 50% chances of having a male kitten with this syndrome.
And, assuming that the cross is between aa and Aa, the gametes produced by "Aa" are "A" but also "a" So we are habing a cross between an "a" gamete and "A" or "a" gamete (See picture) In this example, there is a 50% chances of having a kitten with the syndrome, because only one genotype produced (Aa) has the affected allele. And, as we said before, since there is a 50% chances of having a male kitten, we can said that there is a 25% chances of having an affected male. Because to calculate both probabilities so that they occur simultaneously, it is necessary to multiply them. That is, 0.5 x 0.5 = 0.25 x 100 = 25%.
Consider the way that muscle is stimulated to move and the way that the heart is stimulated to move. Based on the positions of the electrodes, what might some challenges be to measuring ECG signals during exercise?
Challenges in measuring ECG signals during exercise include motion artifacts caused by body movement and skin impedance due to perspiration. Also, the placement and number of electrodes used in a standard ECG reading can impact the accuracy of the results.
Explanation:Several factors present challenges in acquiring accurate ECG readings during exercise. One primary issue stems from the locations of the electrodes used to capture the electrical signals produced by the heart's contractions. As the body is in motion, movement artifacts can affect the quality of the readings, introducing noise that can make interpretation of the results more difficult.
Another potential challenge is the increased likelihood of experiencing skin impedance, or resistance to the electrical current readings when the body is generating more sweat. This issue may interfere with the quality of signal transmission between the skin and the electrode and increase the likelihood of inaccuracies in the ECG readings.
Additionally, the placement of the electrodes may also create difficulties in acquiring accurate ECG readings. Standard ECGs use 3, 5, or 12 leads placed on the chest. The voltage between the right arm and the left leg is often graphed, and the coordination of this potential with arterial blood pressure serves as a functioning monitor. Moreover, continuous ambulatory electrocardiographs utilize a portable Holter monitor that continuously tracks the heart's electrical activity for an extended period.
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All of the following can reduce crop water use, except for which one? Group of answer choices Applying irrigation water at times of day when evaporation is low. Leaving the soil surface bare to improve infiltration. Selecting crop varieties that are well-suited to the local climate. Using drip irrigation techniques to apply water only where plants can use it.
Answer:
Leaving the soil surface bare to improve infiltration.
Explanation:
All of the following can reduce crop water use, except Leaving the soil surface bare to improve infiltration.
Leaving the soil bare will do no good to soil's water retention capacity rather it would decrease significantly. This happen because of water runoffs as there would no crops to hold water and the top fertile layer of the soil will runoff along with water.
Answer:
Explanation:
Leaving the soil surface bare for infiltration. When the soil surface is bare it exposes the it's layers to direct sunlight which absorbs all the water content present in the soil and thus is harmful for the soil and crop
The visible colored portion of the eye is the:
a. cornea.b. pupil.c. sclera.d. iris.
Answer:
D. IRIS
Explanation:
Generally in humans and also in most mammals and birds, the iris is a very small circular structure in the eye, responsible for controlling the diameter and size of the pupil and also with the amount of light reaching the retina.
finallly eye color is also known to be defined by that of the iris
What are the conditions that lead to a "beer gut" due to the excess consumption of alcohol? NADH produced from the metabolism of ethanol stimulates the citric acid cycle for glucose-derived acetyl CoA. Excess ethanol metabolism leads to an accumulation of NADH that inhibits fatty acid metabolism. The processing of acetate in the liver becomes inefficient leading to a pH imbalance in liver cells, reducing enzyme efficiency in general. NADH inhibits ketone body formation, stimulating glucose rather than fatty acid metabolism. NADH stimulates citric acid cycle enzymes that stimulates glucose-derived acetyl CoA metabolism.
Answer: Excess ethanol metabolism leads to an accumulation of NADH that inhibits fatty acid metabolism.
Explanation:
Fatty acid metabolism is often activated by limited or absence of NADH, however ethanol (the main constituent of most alcoholic drinks) on metabolism yields several molecules of NADH which rather stimulates belly/abdominal fat production often referred to as "beer gut".
Understanding how alcohol is absorbed and metabolized by the body can provide helpful insights into the potential effects of alcohol.
Please choose the correct statements about alcohol absorption and metabolism.
Select all that apply.
__Alcohol inhibits the release of antidiuretic hormone, which can lead to dehydration.
__Most alcohol is metabolized in the small intestine.
__Alcohol can improve an individual’s sleep.
__Food in the stomach slows the absorption of alcohol.
__Women have about 20 to 30% more alcohol dehydrogenase in their stomachs than men.
Alcohol do decrease the release of Antidiuretic hormone. Its absorption is also slowed down by amount of food in stomach.
Option A and D.
Explanation:Alcohol is very much related with decreasing effect on pituitary gland to release the Anti Diuretic Hormone or ADH which is actually responsible for the reabsorption of water through kidneys. So consumption of alcohol may make a person dehydrated.
Its mainly metabolized in liver by the enzyme alcohol dehydrogenase into aldehydes and then into acids which are further metabolized into several products.
Alcohol has a negative effect on REM sleep which is actually the main part of sleep cycle.
Alcohol dehydrogenase is an enzyme which metabolize the alcohol in liver. In females, the amount of alcohol dehydrogenase is less than what is found in males. So they often have alcohol toxicity much faster than males.
The correct statements about alcohol absorption and metabolism are:
1. __Alcohol inhibits the release of antidiuretic hormone, which can lead to dehydration.__
2. __Food in the stomach slows the absorption of alcohol.__
1.)This is true. Alcohol consumption can inhibit the release of antidiuretic hormone (ADH), also known as vasopressin, from the posterior pituitary gland. ADH normally acts on the kidneys to reabsorb water, but when its release is inhibited, the kidneys excrete more water than usual, leading to increased urination and potentially dehydration.
2.)This is true. The presence of food in the stomach, especially high-protein foods, can slow down the rate at which alcohol is absorbed into the bloodstream. This is because the pyloric sphincter, which controls the passage of food from the stomach to the small intestine, remains closed for a longer period of time when food is present, delaying the emptying of the stomach contents into the small intestine where most absorption occurs.
When a sudden change in the environment, such as a flood or fire, reduces the size of a population, the survivors' collective gene pool will be only a limited representation of what was present before the disaster. This phenomenon is called: a. The bottleneck effect. b. Founder effect. c. Hardy-Weinberg effect. d. Culling effect. e. Genetic load.
Answer: option A - bottleneck effect
Explanation:
Whenever sudden change caused by human or natural disasters occurs in the environment leading to the reduction of organisms or biodiversity present to a lower size, such condition is referred to as BOTTLENECK EFFECT.
Most consumers have a wide variety of choices in terms of the food products they choose to buy. These decisions can be influenced by factors like food advertising or the labels on the products themselves.Choose the correct statements about marketing and labeling.Select all that apply.__Organic foods are free of all pesticides.__All free-range animals have unlimited access to the outdoors.__Most food advertising is for processed foods and packaged foods.__Organic foods have higher levels of nutrients than their conventional counterparts.__The term "natural" on food labels has no standardized definition.
Not all organic foods are free from pesticides and 'free-range' does not necessarily mean unlimited outdoor access. Most food advertising is for processed and packaged foods, organic foods do not always have more nutrients and the term 'natural' has no standardized definition in food labeling.
Explanation:It's true that many consumer decisions related to food purchases are influenced by marketing and labeling. However, not all the statements listed are accurate. Organic foods are not necessarily free of all pesticides. Sometimes, organic farmers use naturally-derived pesticides over synthetic ones but this doesn't mean they are pesticide-free. Similarly, while free-range animals should have access to the outdoors, it doesn't necessarily mean that they have unlimited access.
Furthermore, it's a fact that most food advertising is indeed aimed at processed and packaged foods. This could be because they have a longer shelf-life compared to fresh produce, making them more appealing to grocery stores. As for nutrient levels, organic foods do not always have higher levels of nutrients than their conventional counterparts. The nutrient content can vary greatly depending on a variety of factors, including how the food is grown and harvested.
Finally, the term 'natural' on food labels indeed has no standardized definition. Therefore, it can be used loosely by companies to market their products.
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Under which of the following conditions is a on most likely no evolve abimodaldistribution? o selection favors muniple distinct trait vaues, and similar individuals tend to mate with each other O Traits are strongly corelated and a low carrying capacity keeps the population size small O Environmental conditions change rapidly, and alow muation rate limits the establishment of new variants Selection is weak, and individuals produce large numbers of omspring each generation. the lab, which of the following factors made it more Check al apply: Selectivity in choosing mates The avaiability of smaller seeds Geogaphic barriers separating subpopul Long term weather pattems submit questions D e B a S Web and Windows
Answer:
The correct answer is "selection favors multiple distinct trait values, and similar individuals tend to mate with each other".
Explanation:
A population with a bimodal distribution are groups of organisms with two major traits or "modes", with practically equal amounts of individuals with these two modes. This type of distributions indicate that the population is not homogenous and that more than one trait is favored in the given circumstances. A bimodal distribution is most likely to evolve when selection favors multiple distinct trait values, and similar individuals tend to mate with each other. In this case natural selection favors more than one value, and due to the individuals with one value tend to mate to each other, only two major traits are favored in the population.
Final answer:
Disruptive selection and nonrandom mating are the conditions most likely to result in a bimodal distribution of traits within a population. These conditions can lead to distinct, adapted phenotypes and facilitate speciation.
Explanation:
A bimodal distribution in a population's traits is most likely to evolve when two conditions are met: selection favors multiple distinct trait values, and there is nonrandom mating, where similar individuals preferentially mate with each other. This scenario describes disruptive selection. Disruptive selection can lead to the formation of two distinct phenotypes within a population, each adapted to different environmental niches or conditions. Gene flow and random mating are factors that can prevent disruptive selection from leading to divergent populations; thus, when geographical barriers or mate selectivity are present, they facilitate the establishment of a bimodal distribution.
What is the function of the lacZ gene?
a. This gene encodes an enzyme, galactoside permease, which transports lactose into the cell.
b. This gene encodes an enzyme, b-galactosidase, that cleaves lactose into two glucose molecules.
c. This gene encodes an enzyme, b-galactosidase, which cleaves lactose into glucose and galactose.
d. This gene encodes the repressor of the lac operon.
Answer:
the answer is c
Explanation:
The lacZ gene encodes an enzyme called β-galactosidase, which is responsible for splitting lactose (a disaccharide) into readily usable glucose and galactose (monosaccharides). The lacY gene encodes a membrane protein called lactose permease, which is a transmembrane "pump" that allows the cell to import lactose.
The lacZ gene encodes the enzyme β-galactosidase, which breaks down lactose into glucose and galactose, playing a crucial role in lactose metabolism in bacteria such as E. coli.
Explanation:The function of the lacZ gene is to encode an enzyme called β-galactosidase, which is responsible for cleaving the disaccharide lactose into its monosaccharide components, glucose and galactose. This process is an essential step in the metabolism of lactose by bacteria such as E. coli. When lactose is present in the environment, E. coli can use the lac operon system to express the lacZ gene, along with other genes necessary for the utilization of lactose as a carbon source, such as the lacY gene for lactose permease and the lacA gene for transacetylase.
6. Before direct proof of DNA being the genetic material, several lines of indirect evidence had suggested DNA as the appropriate macromolecule. Which ONE of the following statements supports the hypothesis that DNA is the genetic material?
a. RNA is more stable than DNA or protein.
b. Molecular composition of DNA is variable in different cells of an organism, whereas the molecular composition of RNA and protein does not change in different cells of an organism.
c. DNA consists of only four bases while proteins are comprised of twenty different amino acids.
d. protein content is most abundant in cells just prior to division.
e. Gametes have 1/2 the amount of DNA as somatic cells.
Answer:
e. Gametes have 1/2 the amount of DNA as somatic cells.
Explanation:
a. RNA is more stable than DNA or protein
-the reverese is the case, DNA has double helix structure as a result of phosphodiester bonds in the sugar-phosphate backbone plus 2 or 3 Hydrogen bonds between the nitrogenous bases.
DNA is more stable
b. Molecular composition of DNA is variable in different cells of an organism, whereas the molecular composition of RNA and protein does not change in different cells of an organism.
with the exception of gametes which contain half the amount of DNA necessary, each Cells have equal amount of DNA when compared to variable amount in cells RNA and proteins due to the different degree of expression and cell differentiation
c. DNA consists of only four bases while proteins are comprised of twenty different aminoacids.
This evidence lacks basis of why DNA is the genetic material.
d. Protein content is most abundant in cells just prior to division.
This evidence lacks basis of why DNA stands as the genetic material in individual organisms
e. Gametes have 1/2 the amount of DNA as somatic cells.
-this Argument is firmly supported since half of parent DNA is donated to its offspring forming the total genome of a somatic cell
What is hepcidin?
(A) A substance produced in the spleen that stores iron
(B) An intestinal mucosa protein that assists in iron turnover
(C) A factor in meats that enhances iron absorption
(D) A substance in legumes that interferes with iron absorption
(E) A liver-derived hormone that helps regulate iron absorption and transport
Answer:
The answer is E
Explanation:
A liver-derived hormone that helps regulate iron absorption and transport.
What is the chance that the 4th child from the couple indicated by the arrow will have a girl who suffers from the disease?
Explanation:
Tho i can see no arrow but there is probability of 75% that a fourth child suffers the disease of the parent
What causes the mucus plug to break?
Answer:
The correct answer will be- widening of the cervical canal to prepare itself for delivery of the child.
Explanation:
The mucus plug is the collection of the thick and acidic substance called mucus produced by the cervix during pregnancy period. The mucus seals the cervical canal acting as a barrier to microorganism and unwanted infections to protect the uterus carrying a child.
The mucus plug breaks down when the cervical canal prepares itself for the delivery of the child. Thus, mucous breakdown acts as a precursor for the delivery of the child.
Thus, widening of the cervical canal to prepare itself for delivery of the child is the correct answer.
The mucus plug can break and be expelled approximately 1-2 days prior to the onset of true labor. This is known as the 'bloody show' and is a natural sign that labor is approaching.
Explanation:The mucus plug, which accumulates in the cervical canal during pregnancy, can break and be expelled prior to the onset of true labor. This typically happens approximately 1-2 days before labor begins. The breaking of the mucus plug is often accompanied by a small amount of blood and is known as the 'bloody show.' The loosening and expulsion of the mucus plug is a natural sign that labor is approaching.
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In Drosophila, an autosomal gene determines the shape of the hair, with B giving straight and b giving bent hairs. On another autosome, there is a gene of which a dominant allele I inhibits hair formation so that the fly is hairless (i has no known phenotypic effect). a. If a straight-haired fly from a pure line is crossed with a fly from a pure-breeding hairless line known to be an inhibited bent genotype, what will the genotypes and phenotypes of the
F1
and the
F2
be? b. What cross would give the ratio 4 hairless : 3 straight : 1 bent?
Answer:
a. F1 genotype : IiBb F1 phenotype : Hairless
F2 genotype and phenotype :
I_B_ (hairless), iiB_ (straight), iibb (bent)
b. IiBb X iiBb
Explanation:
a. Fly 1 : pure breeding straight haired : iiBB
Fly 2 : pure breeding hairless with bent genotype : IIbb
Fly 1 X Fly 2 : iiBB X IIbb = IiBb (F1)
All the offspring will be hairless since there is one dominant I allele which will inhibit hair production regardless of the type of allele of other gene.
Two F1 flies will be crossed to obtain F2 flies. IiBb X IiBb :
I_B_ : hairless = 9
I_bb : hairless = 3
iiB_ : straight hair = 3
iibb : bent hair = 1
Hairless : straight : bent will be in 12 : 3 : 1 ratio.
b. The offspring of this cross :
hairless : I_ _ _
straight : iiB_
bent : iibb
There is a bent offspring which has both the recessive alleles of both the genes. Hence it must have obtained the two alleles from the two parents. So their genotype is _i_b and _i_b.
Half of the offspring are hairless as given by the ratio. This is only possible if the cross is between Ii and ii fly:
I i
i Ii ii
i Ii ii
So the genotype is Ii_b and ii_b.
The remaining hairy flies are in 3:1 ratio of straight to bent. This is only possible if the cross is between Bb and Bb fly:
B b
B BB Bb
b Bb bb
So the final genotype of the flies is IiBb and iiBb.
How does latitude affect radiation?
Answer:
When the sun's rays strike Earth's surface near the equator, the incoming solar radiation is more direct (nearly perpendicular or closer to a 90˚ angle). ... At higher latitudes, the angle of solar radiation is smaller, causing energy to be spread over a larger area of the surface and cooler temperatures.
Explanation:
Answer:
At higher latitudes, the angle of solar radiation is smaller
Explanation:
which of these is not a basic function of a cell
obtaining energy
storing energy
destroying energy
releasing energy
Answer: destroying energy
Explanation:
The cell is the basic unit of life. It consists of several components such as
Nucleus,
Mitochondria: it helps in obtaining energy through lipid oxidation within its layer
Cytosol: It helps in releasing energy from glucose breakdown (glycolysis)
Vacuoles, endoplasmic reticulum also help in storing energy in cells of adipose tissues.
But, NONE of the cell components helps in destroying energy, since energy is needed and usually well-regulated.
Which step of cross-bridge cycling is considered the power stroke?
a. when the myosin cross-bridge forms
b. when ATP binds to the myosin head and releases it from the thin filament
c. when the phosphate is released and the myosin cross-bridge moves to the right in the animation
d. when ATP is hydrolyzed and the myosin head moves to the left in the animation
Answer:
c. when the phosphate is released and the myosin cross-bridge moves to the right in the animation.
Explanation:
Power stroke refers to the step whereby ADP and phosphate are released from the myosin, allowing the mysoin head to bend.
The largest fenestrae are seen in __________ capillaries, while the smallest fenestrae are seen in __________ capillaries.
Answer:
Sinusoidal and fenestrated .
Explanation:
Capillaries may be defined as the small blood vessels and carries the blood between the venules and the areterioles. These are the smallest blood vessels present in the body.
Sinusodial capillaries are also known by the name of the discontinuous capillaries. These capillaries has large fenestrae and mainly found in liver and bone marrow. Fenestrated capillaries consist of pores in their endothelial cells and has small fenestrae.
Thus, the answer is sinusoidal and fenestrated .
Write a paragraph of 3-5 sentences describing 1) whether the statement is correct or incorrect and 2) the reasoning for your answer. Your answer should demonstrate your understanding of the physiological concepts underlying the statement and be written professionally. An imbalance wherein osteoblast activity significantly exceeds osteoclast activity would lead to the development of osteoporosis over time.
Answer:
Incorrect
Explanation:
In the bone modelling process, osteoblasts a responsible for bone formation and osteoclasts are responsible for bone resorption. In the young adult during growth, the formation of bone exceeds the resorption of bone to maintain bone mass. Therefore,
As we get older, bone formation decreases due to reduction in osteoblastic activity and numbers and bone resorption increases as a result of sex hormone deprivation, this will result in decreased bone mass (osteoporosis). Therefore, osteoclastic activity would have to exceed osteoblastic activity in order or osteoporosis to develop
A federally covered investment adviser (IA) instructs its investment adviser representatives (IARs) to add the initials "RIA" (registered investment adviser) after their names on their business cards.
The SEC prohibits the use of "RIA" because it may suggest that an IAR has obtained a certain level of competence.
Answer:
The correct answer is option [C] "This is prohibited by the SEC because the use of "RIA" implies that the IAR is competent".
Explanation:
The missing options of this question are:
[A] This is permitted because the IAR is registered.
[B] This is permitted because the SEC lacks authority to object to the content of business cards.
[C] This is prohibited by the SEC because the use of "RIA" implies that the IAR is competent
[D] This is prohibited by the SEC because "RIA" may be confused with "IAR."
The correct answer is option [C] "This is prohibited by the SEC because the use of "RIA" implies that the IAR is competent".
The federally covered investment adviser (IA) must not instruct its investment adviser representatives (IARs) to add the initials "RIA" (registered investment adviser) after their names on their business cards. An IAR cannot use the initials RIA, according to Rule 206(4)-1 under the Investment Advisers Act of 1940. Using the initials implies that the IAR is competent and this action is misleading for the clients and prospective clients.
A cricket has a diploid genome of 1.2 billion base pairs. Assume that there are about 200,000 individuals in the population and that the mutation rate is one per 100 million base pairs. About how many new mutations will occur in each generation of the population of crickets?
A. 1.2B. 12C. 200,000D. 1 millionE. 2.4 mllion
The number of new mutations per generation in the entire population of crickets will be about 2.4 million, calculated by multiplying the total number of individuals, the mutation rate, and the total number of base pairs in a diploid genome.
Explanation:The number of new mutations in a population per generation can be calculated by multiplying the mutation rate per base pair, the total number of base pairs in the diploid genome, and the total number of individuals in the population.
In this case, if the mutation rate is one per 100 million base pairs, and the cricket's diploid genome contains 1.2 billion base pairs, then the number of mutations per individual per generation will be 1.2 billion (base pairs) divided by 100 million (base pairs), which gives a result of 12 mutations per individual per generation.
If there are about 200,000 individual crickets in the population, then the total number of new mutations per generation in the entire population will be 12 (mutations per individual) multiplied by 200,000 (individuals), which yields a total of about 2.4 million mutations. So answer E, 2.4 million, is correct.
Learn more about Genome and Mutation Rate here:https://brainly.com/question/33617928
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