Answer:
Gram positive cells retains purple stains of crystal violet, gram-negative cells takes up red or pink stain of safranin
Explanation:
Gram staining is a simple method of distinguishing between Gram-positive cells and Gram-negative cells. The method make use of the following steps
Heat fix cells on a glass slide and then:
1. Stain cell with crystal violet for 1 min
2. Rinse with water
3. Decolourize with alcohol for 30 seconds and wash with water
4. flood cells with iodine as mordant
5. Flush with water
6. Counterstain with safranin for 1 min and wash with water
7. View under the microscope
Gram-positive cell possesses a thick peptidoglycan in its cell wall which helps retain the purple colour of the crystal violet, While Gram-negative cells will loose out this colour when flushed with alcohol but takes up the red or pink colour when stained with safranin.
Crystal violet stains Gram-positive cells purple, while safranin stains Gram-negative cells pink in the Gram stain.
Explanation:Crystal violet and safranin are used in the Gram stain to differentiate between Gram-positive and Gram-negative cells. Crystal violet is a primary stain that binds to the thick peptidoglycan layer of Gram-positive cells, making them appear purple.
Safranin, a secondary counterstain, is applied after decolorization and stains the decolorized Gram-negative cells pink. The differences in cell wall structure and the interaction of the stains with the cell walls result in the differential staining of Gram-positive and Gram-negative cells.
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When Elodea cells are placed in a hypertonic solution, they lose water and their plasma membranes pull away from the cell wall. What name do we use to describe this phenomenon
Final answer:
The phenomenon observed when Elodea cells are placed in a hypertonic solution, leading to the loss of water and the plasma membrane detaching from the cell wall, is known as plasmolysis.
Explanation:
When Elodea cells are placed in a hypertonic solution, the phenomenon observed is called plasmolysis. This describes the process where the cell loses water and the plasma membrane detaches from the cell wall, resulting in the constriction of the cell membrane.
Unlike animal cells, plant cells have a rigid cell wall that prevents them from bursting in a hypotonic environment, but in a hypertonic environment, the fluid movement is out of the cell causing it to become flaccid as it loses turgor pressure. Plasmolysis can be observed in plant cells that have been underwatered and become wilted due to the absence of adequate turgor pressure to keep the cells turgid and the plant upright.
In a car accident, the injuries to the driver include a fractured femur. Part of the broken end of the diaphysis of the femur is visibly penetrating through the skin in the thigh area. Which of the following properly classifies this fracture?1. Displaced, complete, open (compound)
2. Nondisplaced, complete, open (compound)
3. Displaced, complete, closed (simple)
4. Nondisplaced, incomplete, closed (simple)
5. Displaced, incomplete, open
Answer:
1. Displaced, Complete, Open
Explanation:
Displaced fracture - in this type of bone fracture the bone breaks into two or more parts, moves and thus the parts become misaligned.
Complete fracture - in this type of fracture the parts of bone become completely separated from rest one.
Open fracture - it is compound fracture where skin breaks and part of the broken bone protrudes out of the skin.
An organism described as 2n=4 has the chromosomes below with genes indicated by letters and centromeres indicated by periods. Select the BEST description of the chromosome aberration present in this organism.
HJM.NPQ HJM.NPQ R.STWXSTZ R.STWXZ
Select one:
a. Reverse Tandem Duplication
b. Terminal Deletion
c. Tandem Duplication
d. Heterobrachial Displaced Duplication
e. Reciprocal Translocation
f. Homobrachial Displaced Duplication
g. Nonreciprocal Translocation
h. Robertsonian Translocation
i. Interstitial Deletion
j. Paracentric Inversion
k. Pericentric Inversion
Answer:
k. Pericentric InversionExplanation:
Such type of the condition in which any abnormality takes place in the chromosomes due to deletion or addition of the nucleotide in the DNA of the chromosomes is known as a chromosomal aberration or chromosomal abnormality. Generally, it takes place due to, mutation process and is classified into different groups such as deletion, insertion, pericentric inversion, and some others.
Inversion is a type of chromosomal abnormality in which the genetic material is inverted and thus caused abnormality. When the inversion process occurs around the centromeres then such type of chromosomal abnormality is called a pericentric inversion.
The chromosomal aberration present in the organism is a Pericentric Inversion. This is a mutation where a segment of a chromosome including the centromere has reversed its orientation.
Explanation:This question pertains to identifying a chromosomal aberration, specifically by examining the genetic sequence of an organism that is described as having a 2n = 4 chromosome count. Observing the provided sequences, we can notice that in the third and fourth sequences of the chromosomes, there's a segment 'STWX' followed by another 'STXZ'. This repetition represents a kind of chromosomal mutation. This particular type of mutation is called a Pericentric Inversion.
A Pericentric Inversion is a chromosomal mutation or aberration where a segment of a chromosome which includes the centromere has reversed its orientation - hence the term inversion.
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The compounds that penetrate the human body best are ... Select one: a. the ones that are insoluble in water but are somewhat soluble in grease. b. the ones that are insoluble in both water and grease. c. the ones that are somewhat soluble in water but not grease. d. the ones that are somewhat soluble in both water and grease.
Answer: Option D.the ones that are somewhat soluble in water and grease
Explanation: ionic solvent (polar solvent) readily dissolve in water while oils dissolve in grease. The most common substance that can dissolve in grease and water is soap.
Soap has both hydrophilic and hydrophobic ends which makes it readily dissolve in water and grease. The hydrophilic ends contains a polar head while the hydrophobic ends contains a non polar hydrocarbon tail
Black hair in rabbits is determined by a dominant allele, B. White hair results when a rabbit is homozygous for the recessive allele, b. Two heterozygote rabbits mate and produce a litter of three offspring.a.What is the probability that the offspring are born in the order white-black-white
Answer:
The answer if 3/64 or 4.69%
Explanation:
B-: Black; bb: white
Two heterozygote rabbits: 2^2 = 4 possible gametes.
P: Bb x BbF1: 1/4 BB; 2/4 Bb; 1/4 bbthere are 3/4 black and 1/4 whiteThe probability that their offspring are given birth following the order white - black - white is:
1/4 white * 3/4 black * 1/4 white = 3/64 = 4.6875%Final answer:
The probability that two heterozygous rabbits will produce offspring in the order of white-black-white is 3/64.
Explanation:
The question pertains to the probability of obtaining offspring with specific phenotypes in a certain order when two heterozygous rabbits mate. Since black hair in rabbits is determined by a dominant allele B, and white hair is the result of being homozygous for the recessive allele b, a heterozygous rabbit (Bb) can produce both black and white offspring. When two heterozygous (Bb) rabbits mate, each offspring has a 3/4 chance of being black (genotypes BB or Bb) and a 1/4 chance of being white (genotype bb).
To calculate the probability of the offspring being born in the order white-black-white, we can use the multiplication rule of independent events:
Probability of first offspring being white (bb): 1/4.
Probability of second offspring being black (BB or Bb): 3/4.
Probability of third offspring being white (bb): 1/4.
Multiplying these probabilities together:
(1/4) x (3/4) x (1/4) = 3/64
So, the probability that the offspring will be born in the order white-black-white is 3/64.
You purchase two identical houseplants and place them side by side on your windowsill. You water both plants equally. One plant, plant a, you leave alone. On the other plant, plant b, you inject florigen into the cells of the apical meristem. Which of the following would you expect to occur? (Select all answer options that apply.)
a. While flowering, the leaves will stop growing on plant b because florigen inhibits leaf meristems.
b. In subsequent years, plant b will only generate flowers in the same place.
c. Plant b will produce a flower at its apical meristem.
d. Once the flower is gone, plant b will grow taller, but only from growth of lateral meristems.
e. Both plants will produce flowers at the same time, but plant b will have more of them.
f. Once the flower is gone, the apical meristem will develop again, and plant b will continue to grow from that stem.
Answer: Option E.
Both plants will produce flower at the same time but plant B will have more of them.
Explanation:
Florigen is a protein or hormone like molecule that help to control or boost flower production in plants. It is a flowering hormone. It is produced in plants leaves and acts on apical meristem or growing tips. The hormone is normally injected at the growing tips or apical meristem where flowers growth will be activated. Since plant B is injected with florigen,it will further trigger flower production on the plant than plant A .
Answer: C
Explanation:
A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY SAD. The researchers begin to characterize the enzyme In the first experiment, with [E_t] at 4 nM, they find that the V_max is 1.6 mu M s^-1. In another experiment, with [E]total at 1 nM and [HAPPY] at 30 µM, they find that v0 is 300 nM•s–1. Based on this second experiment, what is the Km for happyase?
Answer:
a. kcat = Vmax/[Et] = (1.6 \muM/s) / 0.004 \muM = 400 s-1
b. Vmax = [Et] kcat = [1 nM] (400 s-1) = 400 nM/s which is 0.4 \muM/s.
Now we can use the Michaelis-Menten equation if all the units are similar (all molar conc in nM or in this case \muM):
V0 = Vmax*[S] / Km+[S] = 0.3 \muM/s = (0.4 \muM/s) (30 \muM) / (Km+30\muM)
Then solving for KM, we get:
(0.3 \muM/s) (Km + 30 \muM) = 0.4\muM/s (30 μM)
0.3 \muM/s(Km) + 9 \muM2/sec = 12 \muM2/s
0.3 \muM/s(Km) = 3 \muM2/s
Km = 10 \muM
Another way to do this would be to first rearrange the Michaelis-Menten equation to:
V0/Vmax = [S] / Km+[S]
(300 nM/s) / (400 nM/s) = 3/4 = [S] / (Km + [S])
4 [S] = 3 Km + 3[S]
Km = [S]/3 = 30 \muM / 3 = 10 \muM
c. After removal of ANGER, the Vmax increased to 4.8 \muM/s and the Km became 15 \muM. It is a mixed because it is affecting both Vmax and Km.
Because Vmax increased by a factor of 3, \alpha'=3.
Similarly, Km varies as a function of \alphaKm/\alpha'. Given that Km increased by a factor of 1.5 when ANGER was removed (that is, the inhibitor decreased the observed Km by 2/3 and \alpha'=3, then \alpha=2.
d. Because both \alpha and \alpha' are affected, ANGER is a mixed inhibitor.
Explanation:
16. Most aquatic animals excrete ammonia, while land animals excrete urea or uric acid. What is the most likely explanation for the difference?
A. land animals cannot afford the energy needed to make ammonia
B. they have different diets.
C. ammonia is very toxic and it takes lots of water to dilute it.
D. fish need to get rid of ammonia, but land animals need it to live
E. land animals can get the extra energy needed to make urea or uric acid.
Answer:
It would be C
Explanation:
The most likely explanation for aquatic animals excreting ammonia, and land animals excreting urea or uric acid, is due to ammonia's high toxicity which requires significant amounts of water to dilute, something that is readily available in aquatic environments but not on land.
Explanation:The most likely explanation for the difference in nitrogen waste products excreted by aquatic animals (ammonia) versus land animals (urea or uric acid) is that ammonia is very toxic and requires large amounts of water to dilute it. Aquatic animals are ammonotelic; they excrete ammonia directly into the water, which is a less energy-intensive process and utilizable in an environment where water is abundant. Terrestrial animals, on the other hand, are either ureotelic or uricotelic. Mammals excrete urea, and birds, reptiles, and some terrestrial invertebrates excrete uric acid, primarily because these compounds are less toxic than ammonia and can be safely transported in the body until they can be eliminated. The conversion of ammonia to urea or uric acid requires more energy, but this trade-off is essential for survival on land where water is not as readily available for dilution purposes.
A letter to the editor of the Austin American-Statesman, published on December 23, 2009, asks this question: "The trillion-dollar question that Copenhagen has not answered [is this]: Because carbon dioxide molecules arc all identical, why is it that carbon dioxide from carbonated beverages, pets, cattle, farm animals, and humans, yeast, dry ice, fireplaces, charcoal grills, campfires, wildfires, alcohol and ethanol is good, and carbon dioxide from fossil fuel is bad? Can anyone in the United States answer this question?" What is your answer? Your aunt asks you how we know that volcanoes arc not responsible for the observed increase in carbon dioxide. What do you tell her?
Answer:
It's very simple. We all know that the property of a compound never change no matter what the source of that compound is. In case of carbon dioxide it's also the same. Yet, there is one huge difference in emissions of carbon dioxide. The amount of carbon dioxide emitted from all the other sources other than that of fossil fuels is several hundred times less as compared to that of condensed fossil fuels. That's the reason fossil fuels contribute the most in the overall emissions of carbon dioxide all over the world.
The answer is same from volcanic arcs because no volcano is able to emit carbon dioxide amount that is near compare to that of fossil fuels. Furthermore, fossil fuels are burned regularly throughout the world while volcanoes may erupt just a few times in a century.
what color would the ecoli cell appear under the microscope following a gram stain? explain why with a clear explanation of the gram staining process and the peritnent cellular componests that cause cells to stain
Answer:E. Coli will appear pink in color under the microscope following gram staining
Explanation:
The Gram stain is a differential technique that is commonly used for the purposes of classifying bacteria. The staining technique distinguishes between two main types of bacteria (gram positive and gram negative) by imparting color on the cells.
Being Gram-negative bacteria, E. coli have an additional outer membrane that is composed of phospholipids and lipopolysaccharides. The presence lipopolysaccharides on the outer membrane of bacteria gives it an overall negative charge to the cell wall. Because of these properties, E. coli does not retain crystal violet during the Gram staining process.
The diagram shows the effect of a point mutation on a section have a gene. The nitrogenous base adenine (A) is inserted into the sequence. ( NOTE: The diagram shows the best sequence of the union the top row in the transcribe mRNA a in the bottom row) What best describes this mutation and it’s effects on the protein that the gene produces? PLEASEEEE HELPPP HURRRYYY LLEASEE
Answer: Option C) It is a frameshift mutation that changes many amino acids.
Explanation:
Due to the triplet nature of gene expression by codons, the insertion of the single Adenine nucleotide will disrupt the reading frame of the gene (frameshift mutation), such that it is no longer evenly divisible by three.
Thus, resulting in a completely different translated protein from the original.
Answer:
option c
Explanation:
because the addition of that base creates a shift in all of the rest of the bases, not just one
Epidemiologic studies that examine the role of a suspected factor in the etiology of a disease may be observational or experimental. The most important difference between experimental and observational studies is that in experimental studies: ___________
a. The study and control groups are equal in size.
b. The study is prospective.
c. The study and control groups are always comparable with respect to all factors other than the exposure.
d. The investigator determines who shall be exposed to the suspected factor and who shall not.
e. Controls are used.
Answer: the correct option is D (The investigator determines who shall be exposed to the suspected factor and who shall not.)
Explanation:
Epidemiology is the branch of medicine that deals with the study of diseases in a defined population to obtain evidence that will aid in the prevention, treatment and control. It is divided into 2 categories:
- experimental studies and
- Observational studies.
In Observational studies, the epidemiologist collects data of interest based on what is seen or observed from the exposure or disease status of each study participants. While in experimental studies, the investigator determines through a controlled process the exposure for each individual and then track over time to detect the effects of the exposure. Therefore in experimental studies,the investigator determines who shall be exposed to the suspected factor and who shall not. I hope this helps, thanks!
14. Valuable molecules like glucose, amino acids, and vitamins are reabsorbed into the blood at which location in the nephron?
A. Loop of Henle
B. Bowman's capsule
C. proximal convoluted tubule
D. collecting duct
E. glomerulus
Answer: Option C) proximal convoluted tubule
Explanation:
The nephron has the proximal tubule where some water, amino acids and salts which are useful to the body are reabsorbed into the blood capillaries against concentration gradient by active transport.
Thus, the process of selective reabsorbtion takes place in the proximal convoluted tubule
Answer: Option C.
Proximal convoluted tubule.
Explanation:
Renal reabsorption is the process molecules like glucose, amino acids, vitamins are reabsorbed from urine preventing them from being loss from the body) (urine)into the circulating blood . Once they are in the convoluted tubule, the molecules diffuse into the blood capillaries. This take place in the nephron.
Which of the following processes directly require ATP? Choose all that apply.
a.Diffusion of calcium into the motor neuron at the neuromuscular junction.
b.Release of cross-bridge (interaction) between actin and myosin.
c.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.
d.Conformation change of troponin resulting in the movement of tropomyosin off the actin active sites.
Answer:the following processes directly require ATP includes:
B (Release of cross-bridge (interaction) between actin and myosin.)
C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction.)
Explanation:
Muscle contraction occurs in various parts of the body to ensure proper body functioning. This process requires the release of calcium ion and the use of ATP ( Adenosine Triphosphate) as source of energy at various levels for the process to take place. The distinct role of ATP in muscle contraction includes:
-ATP is directly required as it causes detachment from actin after power stroke when it binds at one of the reactive sites of myosin. This explains option B (Release of cross-bridge (interaction) between actin and myosin.)
-it powers the pump that transports calcium ions back into the sarcoplasmic reticulum after contraction. This explains option C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction)
- it activates the myosin head so it can bind to actin and rotate by the action of ATpase enzyme.
The following processes which directly require ATP include
Release of cross-bridge (interaction) between actin and myosin.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.The myosin head has to move in order for the release of cross-bridge
between them to be achieved. The myosin head movement brings about
movement of the actin also.
The movement of calcium also requires the use of ATP as it helps in the
pumping of the calcium ions into the sarcoplasmic reticulum.
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Write out the form of the partial fraction decomposition of the function (See Example). Do not determine the numerical values of the coefficients. (If the partial fraction decomposition does not exist, enter DNE.) (a) x4 + 7 x5 + 5x3 A x+ B x2+ C x3+ Dx+E x2+5 (b) 2 (x2 − 9)2
Answer:
a.[tex]\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+5}[/tex]
b.[tex]\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{(x-3)^2}+\frac{D}{(x+3)^2}[/tex]
Explanation:
a.We are given that
[tex]\frac{x^4+7}{x^5+5x^3}[/tex]
[tex]\frac{x^4+7}{x^5+5x^3}=\frac{x^4+7}{x^3(x^2+5)}[/tex]
Using partial fraction decomposition of the given function
[tex]\frac{x^4+7}{x^3(x^2+5)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+5}[/tex]
Using the formula
[tex]\frac{1}{x^3(x^2+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+a}[/tex]
b.[tex]\frac{2}{(x^2-9)^2}[/tex]
[tex]\frac{2}{(x^2-3^2)^2}=\frac{2}{(x+3)^2(x-3)^2}[/tex]
Using property [tex]a^2-b^2=(a+b)(a-b)[/tex]
[tex]\frac{2}{(x+3)^2(x-3)^2}=\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{(x-3)^2}+\frac{D}{(x+3)^2}[/tex]
Using the property
[tex]\frac{1}{x^2}=\frac{A}{x}+\frac{B}{x^2}[/tex]
An individual suffers a blood clot in an artery that delivers blood to his leg. The leg begins to take on a blue hue, becomes colder than the rest of his body and he experiences numbness in the leg.
He is most likely experiencing:
a) anemic hypoxia
b) ischemic hypoxia
c) hypoxic hypoxia
d) histotoxic hypoxia
e) allergic hypoxia
Answer: Anemic hypoxia.
Explanation:
Anemic hypoxia is a medical condition that result when few haemoglobin are present in the blood leading to decreased ability of the blood to carry oxygen.
Oxygen is essential for human proper body functioning.
This is normally cause by lack of red blood cells ( haemoglobin) that carry oxygen.
The symptoms of anemic hypoxia are;
Skin color change to blue or cherry red.
Numbness
Cold hands and feet.
Headache
Weakness.
From the questions, the symptoms are similar to anemic hypoxia.
Compare the movement of nematodes (vinegar eel) with that of annelid worms (earthworm) relating these different kinds of movement to the arrangement of muscle layers in these animals. Why might peristaltic motion be considered an advancement relative to sinusiodal movement?
Answer:
they both use the muscles along side their body which moves them through the ground
Explanation:
1. Annelid worms have two layers of muscles: a layer of longitudinal muscles and a layer of circular muscles. Nematodes have a single layer of longitudinal muscles that runs the length of their body. 2. Peristaltic movement is more forceful than sinusoidal movement, allowing annelid worms to move more easily through thick or viscous materials.
1. Nematodes move through their longitudinal muscles, causing pressure throughout their bodies, and it flexes rather than flattens, moving forward with the thrust.
Circular muscles are one of annelids' three muscular layers. They are found in the outer layer of the body wall and are in charge of restricting the body. Circular muscles are necessary for locomotion, breathing, and excretion.
2. Peristaltic motion:
Furthermore, peristaltic movement permits annelid worms to travel more precisely. This is due to the ability of the longitudinal and circular muscles to be coordinated in order to produce a smooth, wave-like motion. Sinusoidal movement, on the other hand, is jer-kier.
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Blood flow to the skin ________. A) is controlled mainly by decreasing pH B) increases when environmental temperature rises C) increases when body temperature drops so that the skin does not freeze D) is not an important source of nutrients and oxygen for skin cells
Answer:
Blood flow to the skin increases when environmental temperature rises
Explanation:
The blood flow to the skin adapts to both internal and external temperatures, helping the body with thermoregulation while also delivering crucial nutrients and oxygen to skin cells.
Explanation:The blood flow to the skin mainly fluctuates due to variations in body and environmental temperatures. When the environmental temperature rises, the body responds by increasing blood flow to the skin known as vasodilation. This process permits the body to release excess heat. On the other end of the scale, when the body temperature drops, blood flow to the skin decreases, preventing the skin from freezing, a process known as vasoconstriction. Besides its role in thermal regulation, blood flow to the skin is an important source of nutrients and oxygen for skin cells.
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The main purpose of the chloride channel proteins on the apical surface of the intestinal epithelial cells is to create an osmotic gradient that ultimately causes __________ to move through the intestinal epithelium and into the intestinal lumen to assist in the formation of mucus. These are the same chloride channel proteins, adversely affected in cystic fibrosis.
Answer:
Water.
Explanation:
Cystic fibrosis may be defined as the medical condition in which the excess mucus is produced by the body that might affects the lungs and pancreas as well.
This occurs due to the abnormal functioning of chloride channel proteins. The main function of this protein is the transport of the water molecule that helps in the mucus formation. The water moves through the epithelial cells and maintain the role in epithelial fluid and fluid transport.
Thus, the answer is water.
Brain imaging studies have implicated an area of the cortex called the TPO, lying at the junction of the temporal, parietal, and occipital lobes, which may be the cause of this condition:
a. Anesthesia
b. Synesthesia
c. Senesthesia
d. Synosthesia
e. Sensorithesia
Answer:
B) Synesthesia
Explanation:
Synesthesia is a sensation produced in one modality when a stimulus is applied to another modality, as when the hearing of a certain sound induces the visualization of a certain color. It may be caused by unusual cases of cross-talk between normally separated brain regions. Synesthesia alters connectivity in temporo-occipital and parietal areas. A brain imaging studies that implicated an area of the cortex called the TPO, lying at the junction of the temporal, parietal, and occipital lobes is likely caused by a condition known a synesthesia.
Answer:
Synesthesia
Explanation:
Brain imaging is a relatively new discipline and a breakthrough technology within medicine, neuroscience and psychology used for cognitive neuroscience, behavioral conditioning, and brain science. It involves different techniques(PET,MRI,CT, MRS,MRSI, MEG etc) for image the structure, function, or pharmacology of the nervous system and brain.
Brain imaging studies have implicated an area of the cortex called the TPO, lying at the junction of the temporal, parietal, and occipital lobes, which may be the cause of a condition called Synesthesia
Wikipedia describe Synesthesia as a perceptual phenomenon in which stimulation of one sensory or cognitive pathway leads to involuntary experiences in a second sensory or cognitive pathway i.e crossing of senses.For instance, people with synesthesia may see sounds, tastes word or probably feel a sensation on their skin when they smell certain scents.
Synesthesia forms as either grapheme-color (where letters or numbers are perceived as inherently colred) or number form synethesia
Little is known about how synesthesia develop but through brain imaging studies, it has been implicated that an area of the cortex called the TPO which lies at the junction of the temporal,parietal, and occipital lobes causes synesthesia.
Imagine a genomic researcher who is analyzing the genome of different types of cats. She finds that a particular sequence in the North American Bobcat genome is exactly homologous to a sequence found in the common house cat, while all other sequences in those two genomes differ at many nucleotides. (The most recent common ancestor between bobcats and house cats is estimated to be about 6.8 million years, plenty of time for mutation to generate DNA sequence variation.)
Which of the following could explain the identical sequence in these otherwise differing genomes?
01. The sequence encodes a ribosomal protein that is critical for life and can not be easily mutated while retaining function.
02. The sequence is contained in a virus that has infected germline cells in both species.
03. The sequence is from an intron of a gene that encodes a muscle protein.
Provide a rationale for why each of the three statements above are correct or incorrect.
Answer:
Is 1. The sequence encodes a ribosomal protein that is critical for life and can not be easily mutated while retaining function.
Explanation:
Along with a balanced diet, weight-bearing exercise promotes bone mineral density. Ruben's body-builder diet would be unhealthy for his bones over the long term because it is __________.
Answer: it is lacking adequate levels of Vitamin D and Calcium.
Explanation:Vitamin D deficiency can lead to loss of bone density, which is a contributory factor to osteoporosis and broken bones.
Severe vitamin D deficiency can also lead to other diseases. Vitamin D deficiency leads to osteomalacia. Osteomalacia which causes weak bones, bone pain, and muscle weakness. Because Reuben's body builder diet lacks Vitamin D, it can cause fractures and even osteoporosis over when used for a long time especially without trying to make amends for the deficient Vitamin D.
Answer:inadequate in vitamin D and excessive in calcium
Explanation:
Vitamin D plays a significant role in the regulation of calcium and maintenance of phosphorus levels in the blood. These factors are vital for maintaining healthy bones. People need vitamin D to allow the intestines to stimulate and absorb calcium and reclaim calcium that the kidneys would otherwise excrete.
Foods that provide vitamin D include:fatty fish, like tuna, mackerel, and salmon.
Foods fortified with vitamin D, like some dairy products, orange juice, soy milk, and cereals.
Beef liver.
Cheese.
Egg yolks.
You collect the following data on genotypes for a sunflower population:
AA: 15, AB: 70, BB: 15 (so there is an A and a B allele).
Based on Hardy-Weinberg predictions you expected the following numbers:
AA: 25, AB: 50, BB: 25.
Which of the following is a plausible explanation for the deviation?
a) Balancing selection
b) Negative frequency-dependent selection
c) Inbreeding
d) Genetic drift
Answer:
C. InbreedingExplanation:
Breeding of organisms which are closely related genetically (ancestry).
Inbreeding is useful in the retention of desirable characteristic ( which we want to retain) or the elimination of undesirable one (which we do not want to retain).
Indreeding often results in decreased vigour, size, and fertility of the offspring.
in general terms, what two general factors in a medium affect small speed of sound?
Please explain! I will give brainliest!
Answer:
Density of the medium, temperature of the medium, and stiffness of the medium.
Explanation:
Medium
Medium has a huge effect of the speed of sound. When most people discuss the “speed of sound” they are talking about the propagation of sound waves through the medium of “Air”. For anyone who has gone underwater and listen to people talking above it is likely that one would notice the muted an “odd” way that voices sound underwater. This is because the “medium” of water greatly bends, distorts and changes the speed of sound wave.
There is a whole aspect of science that measure and defines the effect of different mediums (gaseous and liquid) on the speed of sound. This is called Fluid Dynamics. Underwater communication is possible if you understand how this wave propagation as well as another important factor (pressure).
Because of elasticity of materials sound will, as a rule of thumb, generally travel faster in solids than in liquids and faster in liquids than in gases.
♦ Temperature
Temperature has a large effect on the speed of sound. Not as much as the “Medium” does, but far more than anything else. Temperature affects the speed of sound because temperature can affect the “elastic” qualities of different mediums. At the very basics lower temperatures will decrease the speed of sound while higher temperatures will increase the speed of sound, all other factors being equal.
♦ Pressure
Pressure is the final factor that has a significant impact on the speed of sound. The effect of pressure on the speed of sound is due to the materials inertial properties. In short, the more pressure that is applied to the material or medium the denser it becomes and the greater the “inertia” becomes. This makes any interactions between particles slower. Therefore the speed of sound throughout the medium is slowed due to the greater pressure.
Answer:
Elasticity and density
Explanation:
The speed of sound depends on elasticity and density of the medium through which it is travelling. The sound travels faster in liquids than gases. The sound travels faster in solids than in liquids. The lesser the elasticity and higher the density , the sound travels slower in a medium.
Speed = elasticity / density
"What would Avery, Macleod, and McCarty have concluded if their results had been that only RNAse treatment of the heat-killed bacteria prevented transformation of genetic virulence?
Answer:
RNA was the genetic material
Explanation:
If avary, Macleod, and McCarty would have seen that only RNAse treatment of the heat-killed bacteria prevented the transformation of genetic virulence than might have concluded that RNA is the genetic material as transformation does not occur because the RNAase degraded the RNA.
But if the genetic material is DNA then RNAase will not work because RNAase can not degrade DNA and then DNA will pass from virulent heat killed strain to nonvirulent strain and will cause transformation.
The blood of a normal healthy adult rat contains 5 million red blood cells (RBCs) per microliter of blood. Each deciliter of blood contains 15 grams of hemoglobin. Each RBC contains 250 million molecules of hemoglobin (a fairly close approximation for a healthy adult retired breeder female rat). Suppose you have 8.56 mL of blood drawn from a healthy female retired breeder rat, then how many RBCs are in this blood sample?
Answer:
There are 4.28×10^10 RBCs in the blood sample.
Explanation:
Concentration of blood = 5×10^6 RBC/10^-6 L = 5×10^12 RBC/L of blood
Volume of blood sample = 8.56 mL = 8.56/1000 = 0.00856 L
Number of RBCs = concentration of blood × volume of blood sample = 5×10^12 RBC/L × 0.00856 L = 4.28×10^10 RBCs
Some renewable energy resources, such as solar energy and hydropower, have several important advantages over nonrenewable resources. Which is an advantage of renewable energy resources?
Answer:
A,B,F
Explanation:
I just took the test and got it correct.
Renewable energy resources such as solar and hydropower are sustainable, enable decentralized power generation, and help in combating climate change by reducing greenhouse gas emissions.
One significant advantage of renewable energy resources such as solar energy and hydropower over nonrenewable resources is their ability to be replenished within a short timeframe. This makes them sustainable for long-term use. Solar energy, for example, is essentially inexhaustible, as it comes from the sun, which is expected to continue shining for another 4-5 billion years.
Renewable energy also facilitates local, decentralized control over power, allowing homes, businesses, and isolated communities to generate electricity with technologies such as solar panels without being dependent on a grid or centralized power plant. This is particularly beneficial for remote locations where connecting to the power grid is impractical or too expensive.
. The Vmax of a glucose transport into a certain preparation of red blood cells is determined to be 1206nmol glucose/s without ATP present in the buffer. The Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s when ATP is present in the buffer. However, when cells that do not express any glucose transporter are probed for rate of glucose transport, Vmax is infinite. Explain these data.
Answer:
During a process of facilitated diffusion, presence ATP increase the rate of glucose transport through a protein membrane of a red blood cells. Vmax will then be on increased.
Explanation:
The rate of glucose transport in to the certain sample of blood is a facilitated diffusion process. It is understandable to be on increase through a protein membrane with the presence of enzyme to speed up the rate of the biological reaction. This is reflected when it was initially observed that with the enzyme ATP presence in the buffer, the Vmax of a glucose transport into the certain preparation of red blood cells is determined to be 1206nmol glucose/s(considerably low).
When ATP is added, the Vmax is on increase. That is, the Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s. Then Vmax rises to infinity, when cells that do not express any glucose transporter are probed for rate of glucose transport.
It can be further explained that glucose moves into the blood through the permease in the membrane between the cell and the blood.Thus, ATP is used as an energy source to drive Na+ out of the cell, resulting in glucose transport from the intestine to the blood.
Membrane proteins must have an asynchronous distribution on the cell membrane for the system to function. This is an example of the membrane synthetic apparatus determining where in a membrane a protein should be localized. The Na+K+ ATPase must be localized to the membrane between the cell and our blood.
The data reflects that glucose transport via GLUT1 is highly efficient and typically passive, but slightly inhibited by ATP. Without GLUT1, glucose is minimally transported across the membrane, emphasizing the necessity of transport proteins. The minor reduction in Vmax with ATP indicates possible regulatory mechanisms.
The data indicates different behaviors for glucose transport under various conditions:
Vmax of glucose transport without ATP is 1206 nmol/s, showing a high rate of passive, protein-mediated transport via GLUT1.When ATP is present, Vmax decreases slightly to 1158 nmol/s, possibly due to competition or inhibition of the transport process.Without glucose transporters, Vmax is infinite, suggesting that glucose does not pass through the membrane efficiently without these transporters, highlighting the necessity of proteins like GLUT1 for effective glucose transportation.The data helps illustrate passive transport via GLUT1, a glucose transporter protein, and how ATP impacts the transport rate.
Two strains of true-breeding maize that both produce ears of corn with white kernels are crossed and found to produce F1 plants that all make ears of corn with red kernels. If these F1 plants are backcrossed to one of the parents, what proportion of the offspring should have white kernels
Answer:
50% or 1/2
Explanation:
Since the two parental strains are true-breeding, that is, homozygous; it means that the red kerneled ears of corn offspring produced are heterozygous and one of the parental strains must have been homozygous recessive and the other homozygous dominant.
Assuming the ears of corn's colour is coded for by the allele A, it means that one of the parents has the genotype AA while the other has aa with the offspring having genotype Aa.
AA and aa genotypes produce white ears of corn while Aa genotype produce red ears of corn.
Thus, if Aa is backrossed to either of AA and aa:
Aa x AA = AA (white), AA (white), Aa (red) and Aa (red)
Aa x aa = Aa (red), Aa (red), aa (white) and aa (white)
Hence, the proportion of the offspring with white kernels will be 50% in each case.
In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the F1 generation consists of all solid white individuals. After many F1×F1 matings, the following ratio was observed in the F2 generation: 12/16 solid white 3/16 black-and-white spotted 1/16 solid black Part A How many gene pairs are involved in the inheritance of cattle coat color?
Answer:
Two locus
Explanation:
Let assume the gene for white (1st locus) be W i.e. ww = recessive
here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bb
Black&White: wwB_
Black: wwbb
This is dominant epistasis. In dominant epistasis, where the dominant allele of the 1st locus (W) masks or hide the expression of the 2nd locus. When the two alleles in the 1st locus are recessive (ww), the alleles in the 2nd locus can be expressed(B and b).
Two locus gene pairs are involved in the inheritance of cattle coat color
Assumption of Generation:Let us assume the gene for white (1st locus) be W i.e. ww = recessive here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bbBlack&
White: wwB_Black: wwbb
This represents dominant epistasis. In this, the dominant allele of the 1st locus (W) should mask or hide the expression of the 2nd locus. At the time when the two alleles in the 1st locus are recessive (ww), so the alleles in the 2nd locus could be expressed(B and b).
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