Answer:
Improved food sources
Explanation:
Artificial selection is also known as selective breeding. Its a deliberate process of breeding plants and animals for better breeds.
It's also called genetic engineering.
This in a way impacts the society positively as crops with better yield are produced. Example is wheat that has more grain or in animals, cow that produces more milk.
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.
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.
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.
hope i helped...............mark me brainliest?
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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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
During the winter months, people who live near the North Pole will have *blank* than people who live near the equator.
A.shorter nights
B.longer nights
C.longer days
Answer:
B Longer Nights.
Explanation:
My teacher lived in Sweden and the days were short and the nights were long.
Answer:
B longer night
Explanation:
Because land closer to the poles spins slower.
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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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.
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
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.
Observe: An organelle is a cell structure that performs a specific function. Observe the samples below under the highest magnification. Click the Show labels checkbox to label the organelles. List the organelles and approximate size of the cells in each sample. Sample Organelles Estimated size (μm) Mouse skin Fly muscle Maple leaf Elodea Fungus What do all of these samples have in common? In eukaryotic cells, genetic material is contained inside a distinct, membrane-bound nucleus. Plant and animal cells are classified as eukaryotes. Observe: Click on the cow and observe E. coli under the highest magnification. Notice the microscope magnification is larger for this organism, and notice the scale bar is smaller. What is the approximate size of E. coli? What organelles are present in E. coli? What organelle is missing from E. coli? E. coli is an example of a bacteria. Bacteria are classified as prokaryotic cells because their DNA is not contained in a membrane-bound nucleus. Compare: Look at the Sand/silt sample under the microscope. Turn on Show labels. Does sand/silt have any internal structures? Do you think sand or silt is alive? Explain.
All samples mentioned have cells. However, E. coli is a prokaryote and doesn't have its genetic material in a membrane-bound nucleus. It's also smaller than eukaryotic cells. Sand/Silt, meanwhile, aren't living and don't have internal structures.
Explanation:The Mouse skin, Fly muscle, Maple leaf, Elodea, Fungus, Sand/silt and E. coli all share a common trait - they are made up of cells. However, the E. coli is different from the rest because it is a prokaryotic cell. This means that E. coli does not have its DNA contained within a membrane-bound nucleus.
The size approximation of E. coli, like many bacterial cells, can be roughly 1-2 μm in length and 0.25-1 μm in diameter. Due to the fact that E. coli is a prokaryote, it does not have many of the organelles usually found in eukaryotic cells. Therefore, the E. Coli would not have the more complex cell structures, notably the nucleus.
In contrast, Sand/Silt does not have any internal structures as they are not living. They are environmental elements made up of small rock and mineral particles. Because of this, they do not need internal structures or organelles to perform functions.
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Eukaryotic cells like mouse skin, fly muscle, maple leaf, etc. have organelles and are larger in size while E.coli is a prokaryotic cell and smaller in size because it lacks a membrane-bound nucleus. Sand/Silt exhibits no internal structures as it's not a living organism.
Explanation:The samples from the mouse skin, fly muscle, maple leaf, elodea, and fungus all contain organelles and are larger in size because they are eukaryotic cells, with the genetic material stored in a distinct, membrane-bound nucleus. On the other hand, E. coli is a prokaryotic organism, which means it is smaller in size and does not contain a distinct, membrane-bound nucleus, so it's missing this organelle. Therefore, it might have ribosomes, a cell membrane, and cytoplasm but won't have a nucleus, mitochondria, or other specialized organelles found in eukaryotic cells.
As for the sand/silt samples, these are abiotic components, implying they are non-living. Therefore, they don't have any internal structures or organelles because they are not comprised of cells.
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. 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
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.
Which of the following is a secondary air pollutant?
Carbon dioxide.
Answer:
Carbon Dioxide is a Primary Pollutant so definitely not that
Explanation:
Carbon Dioxide is a greenhouse gas emitted from combustion.
Nitrogen dioxide is a secondary air pollutant formed through the reaction of nitrogen oxide and volatile organic compounds in the presence of sunlight.
Explanation:The correct answer is nitrogen dioxide. Nitrogen dioxide is a secondary air pollutant because it is not directly emitted, but forms in the atmosphere through the reaction of nitrogen oxide (NO) and volatile organic compounds (VOCs) in the presence of sunlight.
Here is the step-by-step explanation:
Nitrogen oxide (NO) is emitted from various sources such as vehicles, industrial processes, and power plants.Volatile organic compounds (VOCs) are also emitted from sources like automobiles, gasoline, solvents, and certain industrial processes.In the presence of sunlight, nitrogen oxide and volatile organic compounds react to form nitrogen dioxide (NO2).Nitrogen dioxide is a harmful air pollutant that contributes to the formation of smog and respiratory problems.
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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 Coriolis effect causes surface ocean currents to
Answer:
Bend and curve around the planet
Explanation:
Answer:
To bend and curve the planet.
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.
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.
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.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
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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Two genes determining a trait In primroses, the dominant allele of gene K is necessary to synthesize blue flower pigment. Blue pigment synthesis is inhibited by a dominant allele of gene D. In other words, plants with the genotype K- D- will not produce pigment (and their flowers will be white) because of the presence of the D allele. If you set up a testcross for (Kk Dd) plants what is the ratio of blue to white offspring in the progeny?
Answer:
The ratio of blue to white offspring in the progeny is 4 blue : 12 white.
Explanation:
Available data:
The dominant allele K is necessary to synthesize blue flower pigment K is inhibited by the dominant allele D Plants with the genotype K- D- will not produce pigment (and their flowers will be white)Cross: testcross for (Kk Dd) plants
Parental) KkDd x kkdd
Gametes) KD kD Kd kd
kd kd kd kd
Punnet square) KD Kd kD kd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
Whenever D is present, it inhibits the expression of the K gene, so every plant with the dominant D allele will be white. This plants´ genotype is kkD- or K-D-.Whenever D is absent and K is present, every plant with genotype K-dd will be blue.The recessive form for K and D genes will express white-flowered plants, with genotype ddkkF1) Progeny genotype: 4/16 KkDd, white-flowered plants
4/16 Kkdd, blue-flowered plants
4/16 kkDd, white-flowered plants
4/16 kkdd, white-flowered plants
The ratio of blue to white offspring in the progeny is 4 blue : 12 white.
Blue-flowered plants: 4 Kkdd
White-flowered plants: 4 KkDd + 4 kkDd + 4 kkdd
Final answer:
In the testcross of a primrose plant with genotype Kk Dd, half of the offspring will have blue flowers and half will have white flowers, resulting in a 1:1 ratio.
Explanation:
The testcross for (Kk Dd) plants in primroses is designed to determine the genotypic makeup of the progeny when the parent is crossed with a homozygous recessive (kk dd). For these primroses, blue pigment is only expressed when the dominant K allele is present and not inhibited by the dominant D allele. Thus, a plant will only have blue flowers if it has at least one K and is homozygous recessive for the D allele (dd).
To determine the ratio of blue to white flowers in the progeny, we use a Punnett Square to cross the genotypes Kk Dd with kk dd. Each parent contributes one allele for each gene to their offspring. Here's the cross:
Kd Kd kd kdIn this scenario, the offspring with a genotype of Kk or KK (where the K allele is present) will potentially have blue flowers, while those with the genotype kk will lack the blue pigment and have white flowers. Since the dd genotype is necessary for the expression of blue pigment and all offspring from this cross would be dd due to the homozygous recessive kk dd parent, we only have to consider the K allele. Every offspring has the recessive dd genotype, therefore their flower color is solely dependent on the K allele.
Because all offspring have the dd genotype allowing for blue pigment expression, we look at the Kk allele pairs to determine flower color. As the Punnett Square shows, all offspring are either Kk or kk. Half of the offspring (Kk) will have blue flowers, and the other half (kk) will have white flowers, resulting in a 1:1 ratio of blue to white flowered plants.
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.
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
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:
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%
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.
Transposition _________________ requires rolling circle replication is the process by which proteins are transposed within cells packages new genes into virions may result in the inactivation of chromosomal genes at the insertion site is a form of homologous recombination
Answer:
may result in the inactivation of chromosomal genes at the insertion site may result in the inactivation of chromosomal genes at the insertion site
Explanation:
A transposon is a sequence of the DNA which can change its position on the chromosomes and can insert wherever it wants.
This process of changing the position by the transposons or jumping genes was examined by Barbara Mcclintock and referred this as transposition.
When the DNA is integrated into the DNA, it inactivates the genes at the insertional site and disrupts the function of the gene.
Thus, the selected option is correct.
Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were black. When these progeny were intercrossed, they produced an F2 consisting of 91 black, 39 yellow, and 30 chocolate. (a) Propose an explanation for the inheritance of coat color in Labrador retrievers. (b) Propose a bio- chemical pathway for coat color determination and indi- cate how the relevant genes control coat coloration.
The coat color inheritance in Labrador retrievers follows an example of epistasis where one gene impacts the phenotype of another. The ee recessive genotype results in yellow labs, while black labs are represented by Ee or EE. A biochemical pathway comprises the E gene which regulates the synthesis of eumelanin for black or brown color.
Explanation:The inheritance of coat color in Labrador retrievers is a good example of epistasis, which is a phenomenon where one gene affects the phenotype of another gene. In terms of genetics, yellow labs are homozygous for a recessive allele (ee) while black labs are either Ee or EE. Here, E prevents epistasis. For the chocolate labs, they are bb either Ee or EE. Thus, the black color expression requires at least one dominant E and B, chocolate requires at least one dominant E and two recessive bs, while yellow occurs when there are two recessive es, irrespective of B gene.
As for the biochemical pathway, the E gene codes for an enzyme called tyrosinase-related protein 1, which helps to synthesize eumelanin, a pigment responsible for black or brown color. When the E gene is not present (represented by ee), pheomelanin, a pigment responsible for yellow color is synthesized.
Learn more about Epistasis in Labrador retrievers here:
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One hypothesis that has been proposed to explain the development of viruses suggests that viruses were beneficial because they allowed for rapid gene transfer, especially in prokaryotic cells, and therefore increased genetic diversity. If most bacteriophages were temperate, then this could have increased horizontal gene transfer between prokaryotic cells and potentially could have increased fitness. With respect to this hypothesis, why would it matter whether the bacteriophages were temperate or virulent?
a. Temperate phages are capable of lysogeny and their genetic material can be integrated into the host genome, causing damage rather than increasing fitness.
b. Virulent phages are capable of lysogeny and their genetic material can be integrated into the host genome, causing damage rather than increasing fitness
c. Temperate bacteriophages are capable of both horizontal and vertical gene transfer, allowing a wide range of ways to increase genetic diversity. Virulent bacteriophages are not capable of either.
d. Temperate phages don't immediately kill their hosts.
Answer:
Temperate phages don't immediately kill their hosts.
Explanation:
There exist phages that have shown the capability to alternate lytic and lysogenic life cycles, where don't kill the host cell. This feature allows a phage to reproduce without killing the host cells, thereby having higher probabilities of replicating its genetic material
The enzyme glucose 6-phosphate dehydrogenase catalyzes the first step of the Pentose Phosphate Pathway (we will study this pathway after the exam). Based strictly on the KM values of the three potential substrates listed below, which of them would best interact with this enzyme and be converted to product?
Compound KM
fructose 6-phosphate 1.5 mM
glucose 6-phosphate 0.05 mM
galactose 6-phosphate 0.02 mM
Answer and Explanation:
By michaelis mentos equation
Rate of reaction, [tex]V_0 = \frac{V_{max}[S]}{K_m + [S]}[/tex]
where [tex]K_m[/tex] is michaelis mentos constant
For same value of Vmax and [S] the rate of reaction will be higher for that as lowest(minimum) value of Km
Here for galactose-6-phosphate the Km value is minimum (Km = 0.02 mm)
So "Galactose-6-phosphate" will best interact with this enzyme and be converted to product