Answer:
72/485
Explanation:
this question was just on my bio test i took
Enzymes interact with many different substrates a. true or b. false
Answer:
false
Explanation:
because enzyme has specific active site
Answer:
True
Explanation:
Enzymes works by binding with chemical reactants called substrates. There may be one or more substrates for each type of enzyme, depending on the particular chemical reaction.
Glycolysis and gluconeogenesis are both pathways the body utilizes for energy production. Although both pathways employ common enzymes, they are not the reverse of one another.
They are also not usually used simultaneously.
Which pathway(s) is(are) enhanced during starvation?
gluconeogenesis
glycolysis and gluconeogenesis
glycolysis only
During starvation, the body enhances gluconeogenesis, not glycolysis, to maintain blood glucose levels by generating glucose from non-carbohydrate sources.
Explanation:During periods of starvation, the body enhances gluconeogenesis to generate glucose from non-carbohydrate sources, such as pyruvate, lactate, glycerol, and gluconeogenic amino acids. This pathway is crucial for maintaining blood glucose levels and providing energy, especially to the brain and other organs that rely heavily on glucose. Gluconeogenesis predominantly occurs in the liver and, to a lesser extent, in the kidneys and involves converting these precursors back into glucose. This process is particularly vital during starvation or fasting when dietary glucose is not available.
Glycolysis, on the other hand, is the pathway by which glucose is broken down into pyruvate, yielding small amounts of energy. Though both pathways share other enzymes and can conceptually be seen as reverses of each other, they are regulated independently to prevent both pathways from being active simultaneously, a phenomenon known as the 'futile cycle.' Therefore, during starvation, gluconeogenesis is enhanced, not glycolysis.
Darwin was curious about the distribution of living things which is the study of
A. Evolution
B. Variation
C. Hybridization
D.Biogeography
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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A color blind man without hemophilia (both X-linked traits) marries a woman who is a carrier for both traits. What is the probability they will have a son with both color blindness and hemophilia? Hemophilia and color blindness are unlinked genes.
a. 1/1
b. 1/16
c. 1/8
d. 1/4
e. 3/16
Answer:
Option D.
Explanation:
Both genes are linked to X chromosome. They are recessive diseases and a heterozygous gene, which means that two different forms of a particular gene are inherited, one from each parent.
CB : Color blindness carrier, healthy
H: Hemophilia carrier, healthy
h: Disease
cb: Disease
Mom: X(CB/H) X(cb/h) (carrier)
Dad: X(cb/H) Y (color blind man, and dominant for hemophilia - healthy)
Let's make a table
MOM/DAD | X(CB/H) X(cb/h) |
X(cb/H)Y | X(cb/H) X(CB/H) | Healthy girl for H, carrier for cb
X(cb/H)Y | X(cb/H) X(cb/h) | Healthy girl for H, sick for cb
X(cb/H)Y | Y X(cb/h) | Boy sick for cb and h
X(cb/H)Y | Y X(CB/H) | Boy healthy for CB and H
1 probable case, over 4 possible cases
Which of the following statements about regulation of eukaryotic gene expression is INCORRECT?
A) The presence of a nuclear membrane separating transcription and translation in eukaryotes led to the evolution of additional mechanisms of gene regulation.
B) In eukaryotes, most structural genes are found within operons.
C) Eukaryotic mRNAs are generally more stable than prokaryotic mRNAs.
D) The rate of degradation of mRNAs is important in regulation in eukaryotes.
E) Posttranslational regulation of histones is unique to eukaryotes.
Answer:
The answer is B
Explanation:
In eukaryotes, most structural genes are found within operons.
The incorrect statement is 'B) In eukaryotes, most structural genes are found within operons.' This is because eukaryotic genes are generally not organized in operons, unlike prokaryotic genes.
Explanation:The statement which is incorrect about the regulation of eukaryotic gene expression is 'B) In eukaryotes, most structural genes are found within operons.' This is because eukaryotic genes are generally not organized in operons. The concept of the operon, a functional unit of DNA containing clustered genes that are under control of a single regulatory signal or promoter, is more commonly associated with prokaryotic genetics.
The other options represent correct statements about eukaryotic gene expression. For instance, the nuclear membrane in eukaryotes does indeed enable additional layers of gene regulation. Additionally, the stability of eukaryotic mRNAs and the rate of their degradation are important factors in gene regulation. Lastly, posttranslational regulation of histones is a unique feature of eukaryotic cells.
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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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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 .
What are the correct outputs, during the citric acid cycle, from one molecule of glucose?
Answer:
no clue
Explanation:
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.
Aniridia is a human condition in which the eye has no iris. The protein encoded by the gene responsible for Aniridia is very nearly identical to the protein product of the fly eyeless protein. What experiment could provide evidence that the two genes are functionally equivalent? A. Introduce the aniridia mutation into Drosophila embryos to look for iris formation. B. Sequence the eyeless and aniridia DNA sequence and regulators. C. Use the Eyeless mRNA as a probe in other invertebrate and non-mammalian species. D. Introduce the mouse aniridia wild-type sequence into the fly to see whether flys eyes develop. E. Introduce the mouse wild-type sequence into the fly egg to see whether mouse eyes develop.
Answer:
Option B. Sequence the aniridia and eyeless DNA sequence and regulators.
Explanation:
DNA sequencing:
It is the method of determining the sequence of nucleic acid. DNA sequence includes the method to determine the order of nucleotide sequences. Through the process of sequencing aniridia and eyeless DNA are sequenced and analysed on Gel electrophoresis. Sequencing similarity shows evidence that the two genes are functionally equivalent.
Best choice:
B. Sequence the eyeless and aniridia DNA sequence and regulators.
The experiment could provide evidence that the two genes are functionally equivalent - D. Introduce the mouse aniridia wild-type sequence into the fly to see whether fly's eyes develop.
Aniridiais a human condition in which the eye has no iris.
The protein is encoded by the PAX6 gene.this gene is responsible for Aniridia by mutation is very nearly identical to the protein product of the fly eyeless protein.if we introduce the mouse aniridia wild-type sequence or gene into a fly and check the result we would be able to find evidence that the two genes are functionally evidentif fly develops eyes then both are functionally equivalent.Thus, The experiment could provide evidence that the two genes are functionally equivalent - D. Introduce the mouse aniridia wild-type sequence into the fly to see whether fly's eyes develop.
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If an important cytoplasmic determinant were missing from an animal zygote, what would you expect to happen as the zygote developed into a multicellular organism? If an important cytoplasmic determinant were missing from an animal zygote, what would you expect to happen as the zygote developed into a multicellular organism? The animal would lack certain amino acids. The animal would lack ribosomes. DNA polymerase would be unable to replicate the animal’s genome. The animal would develop normally. One of the animal’s major body axes would fail to develop normally.
Answer: Option E - One of the animal’s major body axes would fail to develop normally.
Explanation:
Cytoplasmic determinants help the organs of the embryo to develop.
Hence, absence of cytoplasmic determinants would result in one of the animal’s major body axes failing to develop normally.
the urea cycle rids in the body of excess nitrogen by converting it to a non-toxic form that can be excreted in the urine.it utilize one nitrogen atom from carbonyl phosphate and one nitrogen atom from asparate.
1.which enzyme releases urea a product? a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
2. which enzyme requires ATP a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
3- Which enzyme is located inside the mitochondrion? a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
4- which intermediates of the urea cycle must cross the mitochondrial membrane? a) argininsuccinate b) ornithine c) arginine d) citrulline
Answer:
Explanation:
Number 1: The answer is A; Ariginase is responsible for the release of urea as product.
Number 2: The answer is B; Argininosuccinate synthetase requires ATP.
Number 3: The answer is D; Arginase is located in the mitochondrion
Number 4: The answer is B; Ornithine produced in the cytosol must first cross the inner mitochondrial membrane into the mitochondrial matrix where it is carbamylated
Multicellular organisms have nervous systems, made up of nerve cells, which help to generate behaviors. The nerve cells throughout the body communicate with each other in order to help the nervous system carry out its function.
Which of the following behaviors requires help from the nervous system?
A.
swallowing food
B.
driving a car
C.
sensing heat
D.
all of these
Answer:
D
Explanation:
Which of the following is false?A. If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection.B. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce.C. Individuals who are heterozygous HbA/HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-ridden regions in Africa.D. In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
Answer:
D
Explanation:
A. If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection.
This is true. If the disease reduces fertility, the likelihood of the affected individual passing it on to children is decreased. If there is no fitness advantage, it will slowly be reduced in the population by chance, because natural selection is not acting positively upon it
B. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce.
This is true, individuals carrying two copies of the sickle cell allele might be more resistant to malaria, but they also have a range of health problems that often cause early mortality or severely reduced quality of life, cancelling out the beneficial features of this allele.
C. Individuals who are heterozygous HbA/HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-ridden regions in Africa.
This is true. Malaria persists in certain regions of Africa because the conditions are favourable for the mosquito that carries the disease. The heterozygous HbA/HbS genotype is prevalent in African population because it protects individuals from malaria, giving them a fitness advantage. Therefore, the natural selection acts positively upon the sickle cell allele in these areas, and over time, it becomes more prevalent
D. In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
This is false. Without the selection pressure of malaria, there is no advantage of having a copy of the sickle cell allele over the normal allele.
In regions where malaria does not occur, individuals who are heterozygous
HbA/HbS have a fitness advantage over those who are hom ozygous for the
normal hemoglobin allele (HbA) is False.
The individuals who have heterozygous HbA/HbS only have an advantage in the aspect of malaria as it makes them less susceptible to the disease.
This is the only advantage it offers and it doesn't offer any special advantage
such as more fitness or increase in fertility.
Individuals with sickle cell die early thereby preventing natural selection as
they are mostly unable to reproduce before death.
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In his studies of alcoholic fermentation by yeast, Louis Pasteur noted that the sudden addition of oxygen (O2) to the previously anaerobic culture of fermentation grape juice resulted in a dramatic decrease in the rate of glucose consumption. This "Pasteur Effect" can be counteracted by the addition of 2,4-dinitrophenol (DNP), an uncoupler of oxidative phosphorylation. (6 marks) (a) Why would the yeast cells consume less glucose in the presence of oxygen? (b) Why would DNP counteract or prevent the Pasteur Effect?
Answer:
A. 38 mol of ATP is yielded during aerobic oxidation of glucose while only 2 ATP is yielded during anaerobic oxidation of glucose. This will cause the yeast cell to need 19 times more glucose anaerobically and to need less glucose aerobically.
(b) DNP is going to block the major step in aerobic ATP production causing the yeast cells to use about the same amount of glucose as anaerobic cells when compared to yeast cells without DNP present
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
The basis for the excitation of excitable cells is an electrical charge on the plasma membrane. True or False
Answer:True
Explanation:
The reversal of charges in the membrane of cells e.g neuron cells results in electrical transmission.
Sodium diffuses in with the positive charges to replace the negatively charged internal environment of the neuron axon, to produce Action potential,provided that the deoplarisation produced by the reversal of charges is upto the threshold level.
Answer:
True
Explanation:
The basis for the excitation of excitable cells is an electrical charge on the plasma membrane. The activity of the nervous system is reflected in a variety of electrical and chemical signals that arise in the receptor organs, the nerve cells and the effector organs, including the muscles and secretory glands. What literally happens is that the electrical charge difference (voltage) called the membrane potential, which occurs across the plasma membranes of all cells are generated by the passive diffusion of ions such as Na+, K+, Ca2+ and Cl− through highly selective molecular pores in the cell surface membrane called ion channels. As we all know that Ion channels play a role in membrane excitation as central as the role of enzymes in metabolism.
The opening and closing of specific channels shape the membrane potential changes and give rise to characteristic electrical messages.
What is a major difference between DNA polymerase I and DNA polymerase III?
(A) DNA polymerase I synthesizes DNA on leading strands and DNA polymerase III synthesizes DNA on lagging strands
(B) DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
(C) DNA polymerase I repairs DNA and DNA polymerase III synthesizes DNA in the 3׳ to 5׳ direction
(D) DNA polymerase I synthesizes DNA in the 5׳ to 3׳ direction and DNA polymerase III synthesizes on lagging strands
Answer:
(B) DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
Explanation:
Both the enzymes DNA polymerase I and DNA polymerase III involved in the process of DNA Replication with specialised functions. DNA polymerase I synthesize DNA on lagging strand where it degrades RNA primer and replace it with DNA. On the other hand, DNA polymerase III synthesize DNA from 5' to 3' end on the leading and lagging strand but stops at the RNA Primer.
Answer:
DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
Explanation:
Which of the following characteristics would NOT increase or decrease the fitness of the organisms that possess it? a. Certain mice in a population carry genes that cause their fur to be darker than the fur of other mice in the population that do not possess those genes. b. Certain fruit flies in a population carry genes that increase their ability to survive longer without food than other fruit flies in the population that do not possess that gene. c. Certain sterile mules in a population carry genes that allow them to carry heavier loads than other sterile mules in the population that do not possess those genes. d. Certain humans in a population carry genes that cause their eyesight to deteriorate at a more rapid rate than other humans in the population that do not possess those genes. e. Certain male birds in a population carry genes that increase the size of their tail relative to the size of the tails of the male birds in the population that do not possess those genes.
Answer:
maybe for me is answer is c
The characteristic related to sterile mules does not affect fitness since they cannot pass on their genes. Other characteristics either increase or decrease fitness based on survival and reproduction advantages or disadvantages.Option c is correct.
Let's analyze each characteristic with respect to how they affect fitness:
Certain mice in a population carry genes that cause their fur to be darker than the fur of other mice in the population that do not possess those genes. This characteristic could affect fitness, depending on the environment. Dark fur might offer camouflage in certain habitats, increasing survival rates.Certain fruit flies in a population carry genes that increase their ability to survive longer without food than other fruit flies in the population that do not possess that gene. This characteristic would likely increase fitness, as it provides a survival advantage.Certain sterile mules in a population carry genes that allow them to carry heavier loads than other sterile mules in the population who do not possess those genes. This characteristic does not affect fitness, as the mules are sterile and cannot pass on their genes.Certain humans in a population carry genes that cause their eyesight to deteriorate at a more rapid rate than other humans in the population that do not possess those genes. This characteristic would likely decrease fitness, as poor eyesight could affect survival and reproduction.Certain male birds in a population carry genes that increase the size of their tail relative to the size of the tails of the male birds in the population that do not possess those genes. This characteristic might affect fitness positively or negatively, depending on the trade-offs between attracting mates and evading predators.Therefore, the characteristic that would NOT increase or decrease the fitness of the organisms that possess it is related to sterile mules (option c).
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.
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In the real world, many organisms don't "fit" clearly into either the r-strategist or K-strategist category for Select one:
a. population reproductive strategies.
b. human growth factors.
c. population growth curves.
d. intrinsic reproduction.
Answer:
a. population reproductive strategies.
Explanation:
Strategies K and r are closely related to reproductive strategy, habitat selection and the ability to disperse.
The organisms that reproduce by strategy r are small organisms that reach maturity in a short time, have short life periods, have numerous offspring (many of which fail to reach adulthood), devote little or no energy to breeding of the youngest of the species, they do not have mechanisms to limit their reproduction to the carrying capacity of their habitat, and tend to be opportunistic invading new areas and adapting to them easily. In this group are most insects, plants that reproduce by spores, turtles, toads and rabbits. The population of these species considered strategic r depends largely on how quickly they reproduce, and not on the carrying capacity of the habitat.
K strategists, on the other hand, are larger, mature very slowly, tend to live for a longer period of time, their offspring are more resistant to diseases, have few young, dedicate time and energy to raising children , they have mechanisms to limit their reproduction and adjust it to the carrying capacity of their habitat, and they remain in a particular habitat without invading those of other species.
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.
Which option best represents the ideal characteristics of drugs for use in low- and middle-income countries?
a. Easy to transport and store, injectable, heat sensitive, single dose
b. Bulky, oral administration, heat sensitive, multiple doses
c. Easy to transport and store, oral administration, heat stable, single dose
d. Easy to transport and store, oral administration, heat sensitive, single dose
Answer:A
Explanation: The answer is A
Because it Increases the security and accuracy of stored and dispensed medication.
Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey�s tail was only 6 cm while Erma�s was 30 cm. Their F1 piglet offspring all grew tails that were 18 cm. When inbred (F1X F1), an F2 generation resulted in many piglets (Erma and Harvey�s grandpigs) whose tails ranged in 4 cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). Most had 18 cm tails while 1/64 had 6 cm and 1/64 had 30 cm tails.a) Explain how tail length is inherited by describing the mode of inheritance, indicating how many gene pairs are at work, and designating the genotypes of Harvey, Erma, and their 18 cm offspring.b) If one of the 18 cm F1 pigs were mated with the 6 cm F2 pigs, what phenotypic ratio would be predicted if many offspring resulted?
Answer:
a) Tail length shows quantitative mode of inheritance. It is controlled by three pair of genes and each allele contributes a small effect. Length of tail varies according to the cumulative effect shown by all the alleles present. Let the three genes controlling the trait be A,B and C. Dominant allele contributes 5 cm of tail length whereas recessive allele contributes 1 cm of tail length.
Harvey = aabbcc = 1*6 = 6 cm
Erma = AABBCC = 5*6 = 30 cm
When Harvey and Erma mate : AABBCC X aabbbcc : AaBbCc (F1)
AaBbCc = (3*5) + (3*1) = 18 cm
b) AaBbCc ( 18 cm ) X aabbcc ( 6 cm ) = 8 type of offspring will be produced in 1:1:1:1:1:1:1:1 ratio :
abcabc, Abcabc, aBcabc, abCabc, ABcabc, aBCabc, AbCabc, ABCabc
In the "Little Albert" study, the fear-producing stimulus used as an unconditioned stimulus was the:_________a. White rat b. Fear of the rat c. Loud noise d. Fear of the noise
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.
____________and ____________are chromosomal mutations that may activate cellular oncogenes by moving them to new regulatory sequences where they become over-expressed.(a) Duplications; deletions(b) Inversions; translocations(c) Duplications; inversions(d) Deletions; translocations(e) Aneuploidy; deletions
Answer:
(d) Deletions; translocations
Explanation:
Deletion mutations: It is the removal of a number of nucleotides from a DNA sequence. Deletions that result in frameshifts, that is, disrupt the 3-nucleotide reading frame can cause changes in regulatory sequences inducing over-expression of certain genes.
Translocation mutations: Are caused the the rearrangement of genes in chromosomes that are usually in close proximity. Encountering new enhancers and/or promoters can cause oncogene activation.
Calculate the average ml of oxygen molecules in 100 ml blood in the athletes at low altitude and then in athletes in high altitude while training at high altitude. Use the following information: 1.39 ml of oxygen per gram of hemoglobin.
Answer:
[tex]27.8[/tex] mL of oxygen in [tex]100[/tex] ml of blood
Explanation:
It is proven that a healthy human being has in general [tex]20[/tex] grams of hemoglobin in [tex]100[/tex] milliliters of blood.
It is given that -
Amount of oxygen found in [tex]1[/tex] grams of hemoglobin is equal to [tex]1.39[/tex] milliliters of oxygen.
Thus, the total amount of oxygen in milliliters would be equal to product of total weight of hemoglobin and the total amount of oxygen in ml in one gram of hemoglobin.
[tex]= 20 * 1.39 \\= 27.8[/tex] mL
Place the following events in activation of a B cell by a microbe in the order in which they would occur.
Several rounds of division occur, producing many B cells that all express the same B cell receptor.
A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.
The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.
Second messenger signaling is activated in the B cell.
Gene expression is altered in the B cell, and the B cell begins to divide.
Answer:
The proper order is as follows-
1) A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.
2) Gene expression is altered in the B cell, and the B cell begins to divide.
3) Several rounds of division occur, producing many B cells that all express the same B cell receptor.
4) The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.
5) Second messenger signaling is activated in the B cell.
The activation process of a B cell by a microbe includes differentiation from a lymphocyte stem cell, antigen binding by the B cell receptor, activation of signaling pathways, alteration of gene expression, and rounds of division producing B cells with identical receptors.
Explanation:The activation of a B cell by a microbe involves a series of steps which must occur in a specific order for effective immune response. Listing these steps in the correct sequence provides a clear understanding of B cell activation:
A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.Second messenger signaling is activated in the B cell.Gene expression is altered in the B cell, and the B cell begins to divide.Several rounds of division occur, producing many B cells that all express the same B cell receptor.These steps are critical for the body's adaptive immune system to mount a specific response against the antigen presented by the microbe. The ultimate goal of these steps is the production of plasma cells and memory B cells that can produce antibodies specific to the antigen and protect the host from future infections by the same pathogen.
Evolutionary geneticists carefully genotype a population of saber-toothed tigers and find the following genotypes:
250 tigers are A1/A1, 500 tigers are A1/ A2, and 250 tigers are A2/A2.
Tragically, an asteroid lands in the middle of the population, killing 50% of each genotype.
A) What will the genotype frequencies be in the next generation?
Answer:
Explanation:
since 50% of each genotype was tragically destroyed, the new population will be as follows;
A1A1 = 125, A1A2 = 250, A2A2 = 125
The new genetic frequencies are as follows;
A1A1 = 125/500 = 0.25
A1A2 = 250/500 = 0.50
A2A2 = 125/500 = 0.25