Answer:
2.) Aegilops tauschii contributed two sets of chromosomes, and T. turgidum contributed four sets of chromosomes.
Explanation:
The mating of diploid and tetraploid species is one breeding method which has proven majorly satisfactory with the largest and most prominent mass producers of seedlings for uniform high-quality progeny, has been the mating of the tetraploid of good form with the free and prolific diploid.
The result yields a good progeny. This is due to the fact that the tetraploid parent influences twice as much as the diploid parent do. It's traits, form and color are more pronounced. This also implies that the diploid contributes two chromosomes and the tetraploid contributes four chromosomes. When hexaploid wheat was produced synthetically by He and co-workers, they mated diploid Aegilops tauschii, and the tetraploid species, T. turgidum. This implies that Aegilops tauschii contributed two chromosomes and T. turgidum contributes four chromosomes.
Aegilops tauschii, the diploid species, contributed two chromosomes, while the tetraploid species T. turgidum contributed four chromosomes to create a hexaploid wheat.
Explanation:The relative contribution of each parent to the genome of the hexaploid offspring involves Aegilops tauschii, the diploid species contributing two chromosomes, and T. turgidum, the tetraploid species, contributing four chromosomes. Thus, the correct statement would be 'Aegilops tauschii contributed two chromosomes, and T. turgidum contributed four chromosomes'. The hexaploid wheat produced synthetically by He and co-workers is a result of the genetic material from the diploid species Aegilops tauschii and the tetraploid species T. turgidum.
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Yeast cells are recovered from a fermentation broth by using a tubular centrifuge. Sixty percent of the cells are recovered at a flow rate of 12 l/min with a rotational speed of 4000 rpm. Recovery is inversely proportional to flow rate. a. To increase the recovery of cells to 95% at the same flow rate, what should be the rpm of the centrifuge?
Answer:
[tex]\\ \omega = 5033.22 \ rpm[/tex]
Explanation:
Given that:
[tex]R_1 = 0.6\\\\R_2 = 0.95\\\\\omega_1 = 4000 \ rpm\\\\\omega_2 = ???\\\\Q_1 = 12\ l/min\\\\Q_2 = 12\ l/min[/tex]
For a continuous centrifuge ; solid recovery R ∝ [tex]\frac{\omega^2}{Q}[/tex]
where;
ω = angular velocity of the centrifuge in rpm
Q = flow rate
SO;
[tex]R_1 \alpha \frac{4000^2}{12}[/tex]
[tex]R_2 \alpha \frac{\omega^2}{12}[/tex]
[tex]\frac{R_1}{R_2} = \frac{4000^2}{12}[/tex] ÷ [tex]\frac{\omega^2 }{12}[/tex]
[tex]\frac{R_1}{R_2} = \frac{4000^2}{12} * \frac{12}{\omega^2}[/tex]
[tex]\frac{R_1}{R_2} = \frac{4000^2}{\omega^2}\\\\\frac{0.6}{0.95} = \frac{4000^2}{\omega^2}[/tex]
[tex]{\omega^2} = \frac{4000^2*0.95}{0.6}\\\\\\{\omega} = \sqrt{ \frac{4000^2*0.95}{0.6}}\\\\\\ \omega = 5033.22 \ rpm[/tex]
The rotational speed when the recovery of cells is increased to 95% at the same flow rate is; ω₂ = 5033.22 rpm
We are given;
Flow rate; Q₁' = Q'₂ = 12 l/min
Rotational speed 1; ω₁ = 4000 rpm
Recovery 1; R₁ = 60% = 0.6
Recovery 2; R₂ = 95% = 0.95
Now, the formula for recovery of tubular centrifuge is;
R = kω²/Q'
where k is a constant of proportionality
For R₁, we have;
0.6 = k * 4000²/12
k = (0.6 * 12)/4000²
When R₂ is 0.95, we have;
0.95 = ((0.6 * 12)/4000²) * ω₂²/Q₂'
⇒ 0.95 = ((0.6 * 12)/4000²) * ω₂²/12
0.95 * 4000²/0.6 = ω₂²
ω₂² = 25,333,333.33
ω₂ = √25,333,333.33
ω₂ = 5033.22 rpm
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2. Chimpanzees and gorillas more closely resemble each other anatomically than either resembles humans. For example, the hands of chimpanzees and gorillas are structurally similar and quite different from human hands. Genetic-distance data suggest, however, that humans and chimpanzees are more closely related to each other than either is to gorillas. Assuming that the genetic-distance data are correct, give two explanations for the observed anatomical similarity between chimpanzees and gorillas.
Answer:
Human evolution in the hominid family occurred faster than between Chimpanzees and gorillas. Neutral mutations may result in sequence divergence between organisms that are structurally similar (in this case, chimpanzees and gorillas)
Explanation:
Human and chimpanzee share higher sequence identity than predicted in the hominid family since observed mutations are necessarily associated with functional divergence
Answer:
Biologists categorize humans, together with solely a number of alternative species, as nice apes (species within the family Hominidae). The family Hominidae embrace 2 distinct species of Pan troglodytes (the Pan paniscus, pygmy chimpanzee, and therefore the common Pan troglodytes, Pan troglodytes), 2 species of ponged (the western gorilla, great ape, and therefore the eastern gorilla, ponged graueri), and 2 species of orang (the Malaysian orangutan, orangutang, and therefore the Sumatran orangutan, Pongo abelii).
Both chimpanzees and gorillas had little C4 genes. The human factor was long thanks to Associate in Nursing ERV. Apparently, orangutans and inexperienced monkeys had constant ERV inserted at precisely the same purpose. This is often particularly vital as a result of humans are imagined to have a more modern common root with each chimpanzees and gorillas and solely a lot of distantly with orangutans. However constant ERV in mere the same location would imply that individuals and chimps had the more modern common root. Here may be a sensible case wherever ERVs don't line up with the expected biological process progression. Notwithstanding, they're still delayed as proof for common ancestry.
Significantly, not all of the information support chimp-human mutual ancestry as nicely as evolutionists generally recommend. Above all, once scientists created a careful contrast among human, chimpanzee, and ponged genomes, they found a major variety of genetic indicators wherever humans harmonized gorillas a lot of carefully than chimpanzees! So, at 18
In this Gizmo, only one crossover is allowed in each segment. In reality, crossovers can occur at almost any point along the chromosome. How do the random distribution of chromosomes and crossovers create more variation in the resulting gametes?
Answer:
Meiosis and Genetic Variation.
Explanation:
When homologous chromosomes form pairs during prophase I of meiosis I, crossing-over can occur. Crossing-over is the exchange of genetic material between non-sister chromatids of homologous chromosomes. It results in new combinations of genes on each chromosome.
When cells divide during meiosis, homologous chromosomes are randomly distributed during anaphase I, separating and segregating independently of each other. This is called independent assortment. It results in gametes that have unique combinations of chromosomes.
In sexual reproduction, two gametes unite to produce an offspring. But which two of the millions of possible gametes will it be? This is likely to be a matter of chance. It is obviously another source of genetic variation in offspring. This is known as random fertilization.
Final answer:
Crossovers during meiosis create recombinant chromosomes with unique gene combinations, while independent assortment results in millions of potential genetic combinations of chromosomes in gametes. Together, these mechanisms contribute to the vast genetic diversity seen in offspring.
Explanation:
The random distribution of chromosomes and crossovers create more variation in resulting gametes through a couple of different mechanisms. During meiosis, chromosomal crossover occurs when non-sister chromatids of homologous chromosomes exchange segments of DNA, creating recombinant chromosomes with unique combinations of genes. This recombination process can occur multiple times along the chromosome, vastly increasing genetic diversity.
Furthermore, during metaphase I of meiosis, the random orientation of homologous chromosomes, known as independent assortment, leads to the production of genetically unique gametes. Each pair of chromosomes can align in multiple ways, with the potential for millions of different combinations when considering all chromosome pairs.
All of these mechanisms ensure that each gamete has a unique genetic makeup, which, when combined during fertilization, results in a zygote with a genetic combination that is one of trillions of possibilities. This is why individuals within a species exhibit such genetic variability.
The cycling of carbon between the living and non-living parts of the Earth system is an important result of which of the following processes?
A: Plants and animals exchanging gases with the atmosphere
B: Condensation of water vapor leading to cloud formation
C: Herbivores digesting plant materials to get nutrients
D: Methane entering the atmosphere from volcanic activity
The carbon cycling between the living and non-living components of the Earth's system is primarily due to the exchange of gases between plants and animals and the atmosphere. During photosynthesis, plants absorb carbon dioxide and convert it into glucose. The animals that consume these plants then respire, releasing the carbon back into the atmosphere.
Explanation:The cycling of carbon between the living and non-living parts of the Earth system is largely the result of plants and animals exchanging gases with the atmosphere. This process is a part of the carbon cycle, where plants absorb carbon dioxide from the atmosphere during photosynthesis and convert it into glucose. These plants are then consumed by animals, which in turn respire, releasing the carbon back into the atmosphere in the form of carbon dioxide. Thus, choice 'A' is the correct one.
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A novel antibiotic is under investigation for determination of safety margins. Concerns have been raised regarding potential hearing loss due to neurotoxicty. Based on early clinical trials, the median effective dose (ED50) is 10 mg/kg. Followup studies have found that median toxic dose (TD50) that is associated with hearing loss is 80 mg/kg. What is the therapeutic index for this novel antibiotic
Answer:
10
Explanation:
Therapeutic index determines the length of how safe a drug is by using quantitative approach. The therapeutic index is the ratio of the median toxic dose(TD50) or the dose that produces toxicity to the to the effective dose (ED 50) of a drug. In the present study the ED 50 is calculated as 10 mg/kg and TD 50 is calculated as 80 mg/kg.
So the therapeutic index for this novel antibiotics = [tex]\frac{median \ toxic \ dose }{median \ effective \ dose}[/tex]
[tex]=\frac{80}{10}[/tex]
= 10
Imagine that you have two people using breathing gear under water. Diver 1 is using an old style "rebreather" that removes carbon dioxide before the mixture is "rebreather" and therefore decreases overall carbon dioxide concentrations in the blood. Diver 2 is using regular SCUBA gear and is breathing air with the normal amount of carbon dioxide that does not lower blood levels of carbon dioxide. Which diver would have a higher blood pH and which would feel more an "urge to breathe"
a. Diver 2 would have higher blood pH and feel the "urge to breathe" sooner.
b. Diver 1 would have higher blood pH, while Diver 2 would feel the "urge to breathe" sooner.
c. Diver 2 would have higher blood pH, while Diver 1 would feel the "urge to breathe" sooner.
d. Diver 1 would have higher blood pH and would feel the "urge to breathe" sooner.
e. There would be no difference in blood pH or breathing between the two divers.
Answer
:b. Diver 1 would have higher blood pH, while Diver 2 would feel the "urge to breathe" sooner.
Basically higher blood pH indicate low acidity, but higher alkalinity.Therefore the higher the concentration of CO2 in the blood the lower the acidity of the blood and higher the pH.
Therefore Diver 1 with rebreather, that removes CO2, and reduced the blood CO2 concentration, will have high blood pH. since pH = -log[H+]. and acidosis decreases with increases in pH
The Diver 2 will have high blood saturation of CO2 and therefore low pH. But with urge to breath sooner, because Hb needs to releases CO2 in the lungs , to load Oxygen molecules, Diver 2 , needs to breath sooner for this exchange to take place in the lungs.Unlike Diver 1, whose re breather has helped in the extraction of CO2 and faster loading loading of Oxygen.
Explanation:
The citric acid cycle is a stage of catabolism that oxidizes acetate into carbon dioxide and generates energy. There are eight enzymes involved in the citric acid cycle.
Which enzymes produce coenzyme A as a product?
A. citrate synthase
B. succinic dehydrogenase
C. succinyl‑CoA synthetase
D. fumarase
Answer:
The correct answer is option A. "citrate synthase" and C. "succinyl‑CoA synthetase".
Explanation:
Citrate synthase is the enzyme that catalyzes the first reaction of the citric acid cycle, which is the the conversion of oxaloacetate and acetyl-coenzyme A into citrate and coenzyme A. Succinyl‑CoA synthetase catalyzes the reversible reaction of succinyl-CoA to succinate, letting coenzyme A as byproduct of the reaction.Therefore, these two enzymes produce coenzyme A as a product of their reactions.
Answer:
Citrate SynthaseAconitaseIsocitrate dehydrogenasealpha-Ketoglutaurate dehydrogenaseSuccinate thiokinaseSuccinate dehydrogenaseFumaraseMalate dehydrogenaseExplanation:
The citric acid cycle (CAC) which is also called the tricarboxylic acid cycle (TCA) but this was originally referred to as the Krebs cycle. This is an important in respiration that is designed to carefully release energy from acetyl-CoA in a series of chemical reactions by using eight different enzymes.
Citrate synthase produces coenzyme A
Large animals need specialized organs for gas exchange because ________. a volume increases more rapidly than surface area as size increases b they require more oxygen per unit of volume c surface area increases more rapidly than volume as size increases d they collect more energy and therefore have more energy to allocate to specialized tissues
Answer:
b. they require more oxygen per unit of volume.
Explanation:
Large animals have a specialized organs for exhalation and inhalation of gases due to the requirement of large amount of oxygen for cellular respiration. Respiration is a process in which energy is released in the form of Adenine tri phosphate (ATP) in the mitochondria of the cell with the addition of oxygen. The waste product such as carbondioxide is produced which is removed from the body body through lungs.
Final answer:
Large animals require specialized organs for gas exchange because a. volume increases more rapidly than surface area as size increases as they grow. These organs, along with the circulatory system, ensure adequate oxygen intake and waste removal, essential for survival.
Explanation:
Large animals need specialized organs for gas exchange because a volume increases more rapidly than surface area as size increases. As a cell grows, its volume increases much more than its surface area. Since the surface of the cell is what allows the entry of oxygen, large cells cannot get as much oxygen as they need. This challenge is overcome in large animals that have specialized organs that effectively increase the surface area available for exchange processes, like the lungs, kidneys, and intestines. These organs are complex and have a high surface area relative to their volume, ensuring efficient gas exchange.
Additionally, the circulatory system plays a crucial role in moving materials and heat energy between the surface and the core of the organism, effectively supporting the gas exchange process. Thus, as animals become larger, the importance of having specialized organs becomes essential to maintain a functional surface area-to-volume ratio. This ratio is critical in many aspects of animal development, including the efficiency of muscle mass in supporting skeletons and in the regulation of body heat.
The citric acid cycle is a stage of catabolism that oxidizes acetate into carbon dioxide and generates energy. There are eight enzymes involved in the citric acid cycle.
Which enzymes produce carbon dioxide as a product?
A. malate dehydrogenase
B. isocitrate dehydrogenase
C. α‑ketoglutarate dehydrogenase
D. succinate dehydrogenase
Answer:
Isocitrate dehydrogenase and α‑ketoglutarate dehydrogenase are the two enzymes that produces carbon dioxide as a by-product.
Explanation:
The release of carbon dioxide from any compound is known as decarboxylation. The first step where carbon dioxide is released as a by product is when Isocitrate is oxidized and decarboxylated to α‑ketoglutarate and this reaction is catlyzed by isocitrate dehydrogenase. The second step where carbon dioxide is released as a by product is from decarboxylation of α‑ketoglutarate into succinyl-Co A and this reaction is catalyzed by α‑ketoglutarate dehydrogenase.
The Coriolis effect causes surface ocean currents to
Answer:
Bend and curve around the planet
Explanation:
Answer:
To bend and curve the planet.
10. What characteristic of fungi makes them different from plants?
(10 Points)
They grow only in shade
They have a different cell membrane
They do not feed in the same way
They are prokaryotes, not eukaryotes
The characteristic of fungi that makes them different from plants is that "they do not feed in the same way" as plants.
Fungi differ from plants in that they are heterotrophs, absorbing nutrients externally after digestion with enzymes, and their cell walls are made of chitin instead of cellulose. Unlike plants, fungi are heterotrophs; they cannot produce their own food through photosynthesis since they lack chloroplasts. Fungi absorb nutrients from other organisms after breaking them down externally with digestive enzymes. Additionally, the cell walls of fungi are made of chitin, a tough carbohydrate, instead of cellulose, which is found in plant cell walls. These differences are fundamental and are a part of why fungi are classified in a different kingdom from plants.
Select all the statements that are true during the filling phase of the cardiac cycle.
O Isovolumetric contraction
O Ventricular Systole
O Atria pressure is greater than ventricular pressure
O Ventricular pressure is greater than aortic pressure
O Ventricles are contracting
O The AV valves are open
O All valves are closed
O Ventricular Diastole
O Aortic Valve is open
Answer:
Atrial pressure is greater than ventricular pressure.
The av valves are open
Ventricular diastole
Explanation:
The filling phase occurs during diastole and it has four stages and they are;
Isovolumetric relaxation: this phase start when aortic valve and mitral valve are closed. It uses energy for it to relax.
Rapid filling: this occur when the atrial pressure is greater than ventricular pressure. The mitral valve open and allow blood to flow to left ventricle.
Slow filling: occurs when the left ventricular pressure approach atrial pressure, there by contributing to 5% diastolic volume.
Atrial systole: the left ventricular volume is delivered as the aorta contracts.
Final answer:
During the filling phase of the cardiac cycle, the atrial pressure is greater than ventricular pressure, the AV valves are open, and ventricular diastole occurs as ventricles fill with blood. The statements relating to isovolumetric contraction, ventricular systole, and aortic valve being open are not true during this phase.
Explanation:
During the filling phase of the cardiac cycle, several events occur to allow blood to enter the ventricles:
Atria pressure is greater than ventricular pressure, which facilitates the flow of blood into the ventricles.The AV valves are open, allowing blood to move from the atria to the ventricles without obstruction.Ventricular Diastole takes place as the ventricles relax, creating a pressure gradient that helps fill the ventricles with blood.The following statements are not true during the filling phase and pertain to other parts of the cardiac cycle:
Isovolumetric contraction, which occurs during ventricular systole when all valves are closed and the ventricles contract but do not eject blood because the pressure is not high enough to open the semilunar valves.Ventricular Systole and the ventricles contracting occur after the filling phase, when the ventricles are actively pumping blood out of the heart.Ventricular pressure being greater than aortic pressure applies to the phase when the ventricles are ejecting blood.All valves are closed is a characteristic of isovolumic or isovolumetric phases, not the filling phase.The Aortic Valve is open during the ejection phase when blood is expelled from the left ventricle into the aorta.The ability to taste phenylthiocarbamide (PTC) is a trait controlled by 2 alleles allele? (PTC taster and PTC non-taster). Suppose 36% of a remote mountain village cannot taste PTC and must, therefore, be homozygous recessive (aa) for the PTC non-taster allele. If this population conforms to Hardy-Weinberg expectations for this gene, what percentage of the population must be homozygous (AA) for the PTC taster allele?
a. 32%
b. 16%
c. 48%
d. 60%
e. 40%
Final answer:
Using the Hardy-Weinberg principle, and knowing that 36% of the population is homozygous recessive for the non-taster allele, we calculate the frequency of the taster allele and determine that 16% of the population must be homozygous for the PTC taster allele.
Explanation:
The question asks us to determine the percentage of the population that must be homozygous (AA) for the PTC taster allele in a remote mountain village, given that 36% cannot taste PTC and are therefore homozygous recessive (aa). To answer this, we use the Hardy-Weinberg principle, which states that the allele frequencies in a population will remain constant or in equilibrium from generation to generation provided that only random mating occurs and there are no other evolutionary influences.
According to the Hardy-Weinberg principle, the frequencies of alleles 'A' and 'a' are represented by p and q, respectively, such that p + q = 1. Since the village cannot taste PTC at a 36% rate, this represents q2 (the homozygous recessive genotype frequency). So, q = sqrt(0.36) = 0.6. Consequently, p = 1 - q = 1 - 0.6 = 0.4. Therefore, the percentage of homozygous dominant individuals (p2) is (0.4)2 = 0.16 or 16%.
The correct answer is option b. 16%
What is the time frame in which a virally infected cell releases new viral particles
Answer:
All viruses depend on cells for reproduction and metabolic processes. By themselves, viruses do not encode for all of the enzymes necessary for viral replication. But within a host cell, a virus can commandeer cellular machinery to produce more viral particles. Bacteriophages replicate only in the cytoplasm, since prokaryotic cells do not have a nucleus or organelles. In eukaryotic cells, most DNA viruses can replicate inside the nucleus, with an exception observed in the large DNA viruses, such as the poxviruses, that can replicate in the cytoplasm. RNA viruses that infect animal cells often replicate in the cytoplasm.
Explanation:
Answer:
Viral replication involves six steps: attachment, penetration, uncoating, replication, assembly, and release. During attachment and penetration, the virus attaches itself to a host cell and injects its genetic material into it.
Explanation:
. The following enzyme is required for the
hexose monophosphate shunt pathway:
A Gluconeophosphatase
1 Phosphorylare
1 Aldolase
D Glucose phosphate dehydrogenase
Answer:
D. Glucose phosphate
How does blood travel from your heart to your limbs, such as your arms and legs?
Answer:
Blood always moves through the body by the heart’s pumping action. The muscles in the legs assist blood flow by their action. Aiding these actions, the veins have valves in them that prevent the blood from flowing backward.
Explanation:
Answer:
Blood moves throughout the body because the heart pumps blood
Please help!
This is timed
Please answer under 2 minutes!
Answer:
Sample 3
Explanation:
In sample 3, the adenine is greater than the uracil. The cytosine and the guanine are not equal as well, so the bases do not match up in either case.
Which diagram below illustrates fertilization what would lead to development of a normal human female?
Answer:
1. X + X
Explanation:
The sperm and the egg each carry half of the DNA for the zygote (which is one cell that eventually becomes a baby).
A normal human male has the sex chromosomes XY
A normal human female would have XX
If the Y chromosome is present, the result is a male.
If there either more or less than 2 sex-chromosomes, there is usually some sort of abnormality, like delayed development or the inability to mature and reproduce.
The development of a normal human female happens when a sperm cell carrying an X chromosome fertilizes an egg cell, which always carries an X chromosome, resulting in an XX zygote. Each parent contributes one of their sex chromosomes, and the sperm cell's chromosome determines the embryo's sex.
Explanation:The fertilization which leads to development of a normal human female happens when a sperm cell carrying an X chromosome fertilizes an egg (which always carries an X chromosome). If the sperm carrying an X chromosome fertilizes the egg, the resulting zygote is XX, that will develop into a female human being. The kind of sperm cell (carrying either X or Y chromosome) that reaches and fertilizes the egg first usually determines the sex of the offspring. Each parent contributes one sex chromosome to the fertilized egg. All egg cells carry an X chromosome, while sperm can either carry an X or a Y chromosome. Therefore, it is the sperm cell that determines the sex of the embryo.
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flower color in snapdragon plants is a character that shows incomplete dominance. Plants with a CRCR genotype have red flowers. Plants with a CWCW genotype have white flowers. Plants with heterozygous genotypes have pink flowers. Two plants were crossed. Nineteen of the offspring plants had pink flowers. Twenty-one of the offspring plants had white flowers. What was the genotypes of the parental plants
Answer: The parental plants has heterozygous genotype (CRCW) and homozygous white genotype (CWCW)
Explanation: Analising the offspring, we have 19 pink flowers and 21 white flowers.
From the question, we know that to be white, the individual has to have CWCW genotype. So one of the allele from the parents must be CW.
Yet, to be pink, the plant has to be heterozygous, which means CRCW. So, the other allele must be CR.
Now, since there are white offspring, one of the parents must has the genotype CWCW, otherwise, there would be a red flower plant.
Therefore, one parent is CWCW and the other is CRCW.
To confirm that, we can cross the genotypes, using a Punnett Square:
CW CW
CW CWCW CWCW
CR CRCW CRCW
From the square, the chance of having genotype CWCW is 50%, which is the same as with CRCW. So, half of these parents offspring will be white and half will be pink. Analising the numbers, 19 out 40 is pink, while 21 out of 40 is white. The proportion is approximately 50/50. So, the crossing represents this genealogy and the parents' genotypes are as described.
Final answer:
The genotypes of the parental snapdragon plants for the cross resulting in nineteen pink and twenty-one white offspring were homozygous white (CWCW) and heterozygous (CRCW), demonstrating incomplete dominance.
Explanation:
In the case of snapdragon plants, flower color exhibits incomplete dominance, where the heterozygous genotype results in a phenotype that is a blend of the two homozygous phenotypes. Given that nineteen offspring had pink flowers and twenty-one had white flowers, it can be inferred that one parent was homozygous white (CWCW) and the other parent was heterozygous (CRCW). This is because a cross between a CWCW and a CRCW individual will generate approximately half CWCW (white) and half CRCW (pink) offspring, aligning with the observed outcome.
Monoamine oxidase inhibitors cause an antidepressant effect by
A. facilitating the breakdown of norepinephrine.
B. increasing the breakdown of transmitters within the brain.
C. inhibiting the breakdown of neurotransmitters in the brain.
D. facilitating the breakdown of dopamine.
Answer:
C. inhibiting the breakdown of neurotransmitters in the brain
Explanation:
Monoamino oxidase (MAO) facilitates the break down and removal of neurotransmitters like nor-epinephrine, epinephrine, serotonin in brain which have excitatory effects on neurons. When these neurotransmitters are degraded by MAO, this would lead to decreased excitatory level of neurons and it will become less functionable as seen in most of the patients suffering from depression, panic disorders, social disphobia and some neurodegenerative disease.
Monoaminooxidase Inhibitors prevents the degradation of above mentioned neurotransmitters which leads to increasing their levels and concentration which allows them to work and influence on neurons effected by depression.
A biologist studied a population of squirrels for 15 years. During that time, the population was never fewer than 30 squirrels and never more than 45. Her data showed that over half of the squirrels born did not survive to reproduce, because of competition for food and predation. In a single generation, 90% of the squirrels that were born lived to reproduce, and the population increased to 80. Which inference(s) about this population might be true?
Question options:
a. The amount of available food may have increased.
b. The number of predators may have decreased.
c. The squirrels of subsequent generations should show greater levels of genetic variation than previous generations, because squirrels that would not have survived in the past will now survive.
d. A and B only
e. A, B, and C
Answer:
e.
Explanation:
The correct option is e.
The population of an organism is usually limited by a number of factors such as competition for food, predation, etc. Hence, an increase in the amount of available food may lead to an increase in the population of an organism. In the same vein, a decrease in the number of predator in the environment will ensure that more members of the population survive.
The population of an organism is directly proportional to the genetic variability of that population. That is, the bigger the population, the higher the genetic variability of the population and vice versa. Hence, the squirrel of he subsequent generation should show greater levels of genetic variation than the squirrels of the previous generation.
All the three options A, B, and C are correct.
Answer:
d. A and B only
Explanation:
It is possible to observe a higher proportion of squirrels in a single generation for two reasons:
1-The amount of available food may have increased,
2-The number of predators may have decreased
The statment c. is incorrect because the genetic variation will have the same value, since if population does not receive migration the unique source capable of generating new genetic variability is mutation
1. How is the nervous system similar to other body systems? How is it different?
Answer:
it is similar because it involves multiple body parts.
it is different because it doesn't involve breaking down anything or providing things through the blood.
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Nervous system is the main regulatory system of the body. Nervous system is made up of neurons. These send and receive messages in the form of signal in response to stimuli.
What is Nervous System?
Nervous system uses the specialized cells called the neurons which sends the signals, or messages, all over the body in response to stimuli. The electrical signals travel between the brain, skin, organs, glands and all the muscles of body. The messages helps in moving the limbs and feel sensations, such as pain, touch, temperature, etc.
Nervous system is similar to the endocrine system which is a regulatory system. However, instead of using the electrical impulses for signaling, the system also produces and uses the chemical signals called hormones, which travels through the bloodstream and controls the action of cells and organs.
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1. A fat molecule is composed of two types of smaller molecules, including only one molecule of_______ . 2. A fat molecule also includes one, two, or three__________ molecules. 3. A fat molecule with only one fatty acid is called a_________ . 4. A fat molecule with three fatty acids is called a triacylglycerol, or a________ . 5. A fatty acid includes a carboxyl group (-COOH) at one end as well as a long_________ . 6. The hydrocarbon chains found in fats store a lot of energy. They also make fats________ , or insoluble in water.Fill in above blanks.a. hydrophilic b. fatty acid c. triglyceride d. monoglyceride e. glycerol f. hydrocarbon chain g. hydrophobic
Answer:
1. glycerol
2. fatty acid
3. monoglyceride
4. triglyceride
5. hydrocarbon chain
6. hydrophobic
Explanation:
A fat molecule is made up of two parts with one part been the glycerol backbone and three fatty acid tails. A fat molecule with just one fatty acid chain can be regarded as a monoglyceride and with three fatty acids, you have a triglycerides.
One end of the fatty acid is made up of even number carbon-carbon chains at one end and the -COOH on the other end making it a carboxylic acid. The hyrocarbon chains of the fatty acid is hydrophobic in nature and always cluster up in molecules such as in the biological cell meembrane.
A fat molecule, such as a triglyceride, consists of glycerol and fatty acids. Glycerol is an organic compound with three carbon atoms and three hydroxyl (-OH) groups, while fatty acids have a long hydrocarbon chain with an acidic carboxyl group. The fatty acids are attached to each of the three oxygen atoms in the glycerol molecule with covalent bonds.
Explanation:A fat molecule, such as a triglyceride, consists of two main components-glycerol and fatty acids. Glycerol is an organic compound with three carbon atoms, five hydrogen atoms, and three hydroxyl (-OH) groups. Fatty acids have a long chain of hydrocarbons to which an acidic carboxyl group is attached, hence the name "fatty acid." The number of carbons in the fatty acid may range from 4 to 36; most common are those containing 12-18 carbons. In a fat molecule, a fatty acid is attached to each of the three oxygen atoms in the -OH groups of the glycerol molecule with a covalent bond.
What domain name should replace the question mark in this chart?
A) Archaea
B) Bacteria
C) Eukarya
D) Monera
Answer:D
Explanation: kingdom
The question mark in the chart should be replaced with the domain name 'Plantae' as it is not one of the domains in Woese and Fox's phylogenetic tree, which only includes the Eukarya, the Archaea, and the Bacteria domains.
The domain name that should replace the question mark in the chart is Plantae, as it is not a domain in Woese and Fox's phylogenetic tree. The three domains of life recognized by scientists, as proposed by Woese in 1977, are the Eukarya, the Archaea, and the Bacteria. Plantae is a kingdom within the Eukarya domain, not a distinct domain on its own. Both Archaea and Bacteria are prokaryotes, meaning they do not have a nucleus, while Eukarya are organisms with cells that have nuclei.
"Nitrous acid causes all possible transitions (A<-- -->G, C<-- -->T). If a protein has an alanine at position 65, what amino acid(s) would be likely to replace alanine as a result of nitrous acid mutagenesis?a. isoleucine
b. "valine, threonine"
c. threonine
d. "isoleucine, threonine, valine"
e. leucine
Answer:
Due to nitrous acid causes all possible transitions "isoleucine, threonine, valine" is the amino acid that would most likely to replace.
Explain the benefits of STEM cell therapy. Be specific with examples and give a pretty brief reason.
Answer:
In stem cell transplants, stem cells replace cells damaged by chemotherapy or disease or serve as a way for the donor's immune system to fight some types of cancer and blood-related diseases, such as leukemia, lymphoma, neuroblastoma and multiple myeloma. These transplants use adult stem cells or umbilical cord blood.Explanation:
In Drosophila melanogaster, round eyes (rai+) are dominant to raisin eyes (rai). Round eyes are found in wild-type flies, while raisin eyes are the result of a mutant allele. If a heterozygous parent is crossed with a homozygous recessive parent, then what are the correct genotypic ratios for the F1 progeny?
A. 4:0, round eyes
B. 3:1, raisin eyes:round eyes
C. 1:1, round eyes:raisin eyes
D. 4:0, raisin eyes
Answer:
The correct genotypic ratios for the F1 progeny is 1 round eyes : 1 raisin eyes (Option c)
Explanation:
Available data:
round eyes (rai+) are dominant to raisin eyes (rai)round eyes are found in wild-type flies, raisin eyes are the result of a mutant alleleCross: A heterozygous parent (rai+ rai) is crossed with a homozygous recessive parent (rai rai)
Parental) rai+ rai x rai rai
Gametes) rai+ rai rai rai
Punnet square) rai+ rai
rai rai+ rai rai rai
rai rai+ rai rai rai
F1) Progeny genotype: 2/4= 1/2 rai+ rai, round-eyed individuals
2/4=1/2 rai rai, raisin-eyed individuals
The correct genotypic ratios for the F1 progeny is 1 round eyes : 1 raisin eyes
The correct response to the cross between a heterozygous (rai+/rai) and a homozygous recessive (rai/rai) parent is a 1:1 genotypic ratio in the F1 generation. Therefore, about half the offspring should have round eyes and the other half should show the raisin eyes trait. The correct answer is option C. 1:1, round eyes:raisin eyes.
Explanation:In the scenario you provided, a heterozygous parent for the eye trait (rai+/rai) is crossed with a homozygous recessive parent (rai/rai). In a typical single-factor cross of this type involving one heterozygous and one homozygous recessive parent, the expected genotypic ratio in the F1 generation is 1:1. That means, out of every two offspring, around one should have round eyes (rai+) and one should have raisin eyes (rai).
So, the correct answer is C. 1:1, round eyes:raisin eyes. This illustrates how dominant and recessive traits work in Mendelian genetics to influence the appearance and prevalence of particular traits in offspring.
Learn more about Genetic Inheritance here:https://brainly.com/question/30092183
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Cardiac muscle cells always have two nuclei true or false?
Answer:
False
Explanation:
The majority of cardiac muscle cells also known as cardiomyocytes of myocardiocytes have one nucleus even though they might have as many as four.
what are the two reasons for the change in the population of the giraffes?
Answer:
Explanation:
Giraffe numbers plummeted by a staggering 40% in the last three decades, and less than 100,000 remain today. Habitat loss through expanding agriculture, human-wildlife conflict, civil unrest, and poaching for their meat, pelts, and tails, are among the reasons for the decline.
Final answer:
The giraffe population changed primarily due to overproduction of offspring and differential survival and reproduction, where longer-necked giraffes had a survival advantage and passed this trait to their progeny.
Explanation:
The two main reasons for the change in giraffe populations can be attributed to natural selection mechanisms. First is the overproduction of offspring: Giraffes produced more young than the available food sources could support, which led to a struggle for existence where only some could survive. Second is differential survival and reproduction: Giraffes with longer necks could reach food sources higher up in trees, which gave them a survival advantage, leading to greater reproductive success and passing on of the long-neck trait to their offspring. Over time, this caused a shift in the population towards giraffes with ever longer necks, illustrating the principle that chance variations can significantly impact a species' survival if environmental conditions change.
Humberto measured the length of a stick's shadow from sunrise to sunset during the day. What did he notice from his observations? A. Shadow length increased from sunrise to sunset. B. Shadow length increased from sunrise to noon. C. Shadow length decreased from sunrise to noon. D. Shadow length decreased from noon to sunset.
Answer: C. Shadow length decreased from sunrise to noon.
Explanation: Think about how the sun is positioned in the sky. If it was noon, the sun is above the stick which means little to no shadow at all. If the sun was rising, its hitting one side of the stick, making the shadow longer.