The point at which the rate that a deleterious allele is being eliminated from a population by natural selection is in equilibrium with the rate at which the deleterious allele is being replaced by a new mutation is termed a. mutation-dependent selection b. frequency-dependent selection c. mutation-dependent equilibriunm d. mutation-selection balance
Answers
The point at which the rate that a deleterious allele is being eliminated from a population by natural selection is in equilibrium with the rate at which the deleterious allele is being replaced by a new mutation is termed d. mutation-selection balance.
This balance is important for maintaining genetic variation in a population and preventing the complete loss of a deleterious allele. If the rate of elimination of a deleterious allele is greater than the rate of replacement by mutation, the allele will eventually be lost from the population.
Conversely, if the rate of replacement by mutation is greater than the rate of elimination, the frequency of the deleterious allele will increase in the population.
The mutation-selection balance maintains a balance between these two processes, allowing for genetic variation to be maintained in the population.
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Related Questions
which nutrients often limit the distribution and abundance of photosynthetic organisms? select all that apply.
Answers
For nutrients like nitrogen or phosphorus are frequently a source of restriction for photosynthetic organisms (like kelp or phytoplankton).
While phosphorus (P) limitation typically occurs in tropical locations, nitrogen (N) is generally regarded as the primary limiting component in temperate regions. However, nutrient limitation in subtropical regions is little known.
Nitrogen (N), phosphorus (P), and potassium are the three main limiting nutrients in crop yield (K). These three nutrients are frequently added as parts of commercial fertiliser or manure to agricultural areas. You need to understand that the three limiting parameters for photosynthesis are light intensity, CO2 concentration, and temperature. Although a scarcity of water can slow down photosynthesis, other plant functions are typically affected before photosynthesis
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which of the following organ systems can be used as a control center in a homeostatic control system? group of answer choices digestive system nervous system urinary system cardiovascular system
Answers
The organ system that can be used as a control center in a homeostatic control system is the nervous system.
Thus, the correct answer is B.
Homeostаsis, in а generаl sense, refers to stаbility, bаlаnce, or equilibrium. Physiologicаlly, it is the body’s аttempt to mаintаin а constаnt аnd bаlаnced internаl environment, which requires persistent monitoring аnd аdjustments аs conditions chаnge. Аdjustment of physiologicаl systems within the body is cаlled homeostаtic regulаtion, which involves three pаrts or mechаnisms: the receptor, the control center, and the effector. Homeostаsis is controlled by the nervous аnd endocrine system of mаmmаls.
Your options aren't well arranged, but most probably your options were
A. digestive system
B. nervous system
C. urinary system
D. cardiovascular system
Thus, B is the correct answer.
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Examine a two-gene cross in which a mutation in herc2 is epistatic to the oca2 gene. In a cross between a blue-eyed oohh woman and a blue-eyed oohh man, what eye color will the progeny have?.
Answers
In a cross between a blue-eyed oohh woman and a blue-eyed oohh man, what eye color will the progeny have blue eye color.
According to research on the genetics of eye color, blue eyes are caused by a mutation in the neighboring, unrelated gene HERC2. The gene OCA2, which controls the production of the brown pigment melanin, is silenced by this mutation. Today, between 20 and 40 percent of Europeans have blue eyes. The quantity of functional P protein that is generated is decreased by a number of frequent changes (polymorphisms) in the OCA2 gene.People with a variation in this gene have blue eyes instead of brown because less P protein implies less melanin is present in the iris.A disturbance in the normal generation of melanin brought on by OCA2 gene mutations results in visual issues and a loss of hair, skin, and eye color. Oculocutaneous albinism type 2 refers to oculocutaneous albinism brought on by mutations in the OCA2 gene.Learn more about the Inheritance with the help of the given link:
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A biologist in Antarctica is trying to fudge some penguin data by moving the cute little guys across the ice. The evil biologist ties three, 4 kg penguins together such that there is a one meter length of rope between each penguin. The evil biologist then pulls the first penguin with a force of 24 N. What is the tension in the cord between the last two penguins?
Answers
A biologist in Antarctica is trying to fudge some penguin data by moving the cute little guys across the ice. The evil biologist ties three, 4 kg penguins together such that there is a one-meter length of rope between each penguin. The evil biologist then pulls the first penguin with a force of 24 N. The tension in the cord between the last two penguins is 102.4 N.
There are three penguins with a mass of 4 kg each. They are tied together such that there is a one-meter length of rope between each penguin. The biologist pulls the first penguin with a force of 24 N. We need to calculate the tension in the cord between the last two penguins.
Here, the distance between each penguin is 1 m, and the total distance between the 1st penguin and the 3rd penguin is 2 m. The penguins are tied together, so they will move together at the same acceleration. The weight of the penguins is given by:
w = mg
where,
w = weight of the penguins
m = mass of each penguin = 4 kg
g = acceleration due to gravity = 9.8 m/s²
w = 4 × 9.8 = 39.2 N
Now, let's find out the acceleration of the penguins:
a = F/m
where,
a = acceleration
F = force applied = 24 N
M = total mass of the system = 3 × 4 = 12 kg
a = 24/12 = 2 m/s²
The tension in the cord between the last two penguins can be calculated using Newton's second law of motion.
F = ma
Where,
F = force applied
m = mass
a = acceleration
The forces acting on the last two penguins are their weights and the tension in the cord. These two forces are acting in the opposite direction from the force applied by the evil biologist.
Therefore, the net force will be the difference between the tension and weight. Let's find the weight of the last two penguins:
W = mg
where,
W = weight
m = mass of each penguin = 4 kg
g = acceleration due to gravity = 9.8 m/s²
W = 4 × 9.8 = 39.2 N
Let the tension in the cord between the last two penguins be T. So, the net force will be:
Fnet = T - 2w
where,
Fnet = net force = ma = 2 × 12 = 24 NT = 2w + Fnet
T = 2 × 39.2 + 24T = 102.4 N
Therefore, the tension in the cord between the last two penguins is 102.4 N.
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Witch statement correctly compares the “analysis” and “conclusion” section of a lab report
Answers
Answer:
Analysis section of lab report comprises of making comparisons while Conclusion section is used to make further research about the experiment.
Explanation:
Analysis section of lab report comprises of making comparisons between specific data. After knowing the scope and objectives of the experiment, data is collected either by performing the experiment or adopted data from other organization such as hydrological data obtained from hydrological agency, analysis of such data comprises of making comparisons.
While
Conclusion section is used to make further research about the experiment. It is used to report the outcome of the result and also to determine other possibilities of results from the experiment.
The organ that monitors and adjusts the composition of tubular fluid, recycles damaged spermatozoa, and is the site of sperm maturation is the
a. epididymis.
b. prostate gland.
c. seminal gland (seminal vesicle).
d. ductus deferens.
e. rete testis.
Answers
The organ that monitors and adjusts the composition of tubular fluid, recycles damaged spermatozoa, and is the site of sperm maturation is the epididymis.
A is the correct answer.
The key reproductive organ known as the epididymis is in charge of sperm maturation, storage, protection, and acquisition of sperm motility. The human epididymis, like those of other mammalian species, is a lengthy, convoluted tubule that acts as a pathway for the transportation of spermatozoa from the testis to the vas deferens and is the site where spermatozoa mature and acquire their abilities of increasing motility and fertility.
The human epididymis, a six-meter-long tubule curled up against the testis, is where spermatozoa must pass in order to mature. The cells change from immobile, vibrating spermatozoa to smooth, swift-moving sperm over a period of several days.
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The complete question is:
The organ that monitors and adjusts the composition of tubular fluid, recycles damaged spermatozoa, and is the site of sperm maturation is the ______.
a. epididymis.
b. prostate gland.
c. seminal gland (seminal vesicle).
d. ductus deferens.
e. rete testis.
Explain ONE possibe advantage of vivipary when compared to
ovipary
Answers
Embryos are protected and physiologically maintained by the pregnant female.
Help with this please
Answers
Answer:
Cc
Explanation:
lagit just did this i think
Men who are kidnapped for human trafficking are usually held under debt bondage or _____.
soldiering
peonage
sex slavery
sex tourism
Answers
Answer:
It is Soldering not peonage
Explanation:
Peonage is what they do for Men not young boys.
Men who are kidnapped for human trafficking are usually held under debt bondage or peonage. Option B. This is further explained below.
What is peonage?Generally, Men who are kidnapped for human trafficking are frequently held hostage.
In conclusion, peonage defines Men who are kidnapped for human trafficking.
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Which buffers in GFP purification by HIC have the highest salt content and how can you tell
Answers
The buffers with the highest salt content in GFP purification by HIC are typically the elution buffers.
In GFP purification by HIC (hydrophobic interaction chromatography), the buffers with the highest salt content are usually the elution buffers. This is because higher salt concentrations are needed to disrupt the hydrophobic interactions between the GFP and the stationary phase of the HIC column. Typically, the salt concentration in the elution buffer is higher than that in the equilibration buffer and wash buffers. One way to determine the salt content is by looking at the ionic strength of the buffer, which is a measure of the concentration of all ions in solution. A higher ionic strength indicates a higher salt content. Another way to determine the salt content is by measuring the conductivity of the buffer. Conductivity is a measure of the ability of a solution to conduct electricity, which is influenced by the presence of ions. A higher conductivity indicates a higher salt content.
In summary, the buffers with the highest salt content in GFP purification by HIC are typically the elution buffers, which can be identified by their higher ionic strength and conductivity compared to the other buffers used in the process.
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When the body's digestive system breaks down starch, a complex carbohydrate, into simpler carbohydrates, the reaction is:________
Answers
The reaction is catabolic. Catabolism is a series of metabolic pathways that break down molecules into smaller units, either oxidized to release energy or used in other anabolic reactions.
Among complex carbohydrates, starch serves as the main glucose storage molecule in plants. A large number of different glucose monomers combine to form starch and are linked by glycosidic bonds. An example of a catabolic reaction that releases chemical energy is the breakdown of starch, a complex carbohydrate, by the body's digestive system into simpler carbohydrates. C. Catabolic, which causes the release of energy, is the correct response. The anabolic process is incorrect because an anabolic reaction involves the synthesis of less complex monomers into more complex monomers. Catabolic processes also lack the ability to store energy.Therefore, catabolic reaction is the correct answer.
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The molecules in a solid vibrate in place.
Answers
The answer is thanks for the extra points
Gravitational PE = _____ x ______
Answers
g.p.e = m × g × h
g.p.e = Gravitational potential energy (J)
m = Mass (kg)
g = gravity (ms-¹) [9.8 or 10]
h = height (m)
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A malignant lesion of the forehead measuring 1.0 cm was removed. The operative report states skin margins are 1.1 cm on all sides. Layered closure of 3.5 cm was performed. How is this coded
Answers
The code is 11644 and 12052-51.
What are lesions?An aberrant region of tissue inside or outside the body that could expand or alter in appearance, and which might or might not be malignant.
It is possible for lesions to have sources other than underlying diseases. Birthmarks, bug bites, and scars are a few examples.
Cancerous malignant tumors exist (ie, they invade other sites). They spread through the lymphatic or blood systems to distant locations. We refer to this spreading as metastasis. Although the metastatic disease can spread to any part of the body, it most frequently affects the liver, lungs, brain, and bones.
11644 (Excision, malignant lesion including margins, face, ears, eyelids, nose, lips; excised diameter 3.1 to 4.0 cm),
12052-51 (Intermediate repair; 2.5 cm or less; face, ears, eyes, nose, lips, and/or mucous membranes).
51 modifier is for numerous processes
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HELP!
1. Which best compares bacteria and viruses?
A. Bacteria are smaller than viruses.
B. Bacteria cause disease, but viruses do not.
C. Bacteria are living organisms, but viruses are not.
D. Bacteria have genetic material, but viruses do not.
Answers
Answer:
C
Explanation:
This is because viruses are bigger than bacteria, so it is not A. Bacteria and viruses both cause disease so it is not B. And viruses have genetic material, not bacteria so it is not D. Also viruses are considered nonliving because they don't meet all the characteristics of life, so it is C.
The tiny water flea, Daphnia, i placed in a petri dih under a diecting microcope fixed with a heat-producing lamp. Two minute after the light ha been turned on, the heartbeat i counted for 15 econd at two-minute interval. After ten minute, the Daphnia i removed from the dih and returned to a pecimen jar. Which quetion i thi experiment deigned to tet?
Group of anwer choice
What i the effect of increaing heat on the heart rate of Daphnia?
How i the heart rate of Daphnia affected by different amount of water?
What i the effect of light on the heart rate of Daphnia?
How i the circulation of Daphnia affected when placed in a petri dih?
Answers
The experiment is designed to test "What is the effect of increasing heat on the heart rate of Daphnia?" The microscopic water flea is placed in a petri dish and examined under a directed microscope that is fastened with a heat-producing lamp.
The heartbeat of the Daphnia is then counted for 15 seconds at two-minute intervals, allowing the researchers to observe and measure any changes in the heart rate of the Daphnia as a result of the increased heat. The fact that the Daphnia was placed under a heat-producing lamp and its heart rate was measured at various intervals further supports the conclusion that the experiment was designed to test the effect of heat on the heart rate of Daphnia.
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HELP FOR MIDTERM STUDY GUIDE
During photosynthesis, what best describes the role of the compound with the chemical formula C6H12O6?
a. product that is not useful because of its low-energy chemical bonds
b. a product that is useful because of its high-energy chemical bonds
c. a reactant that is useful because of its high-energy chemical bonds
d. either a product or a reactant, depending on whether the cell needs to store energy or
release energy
Answers
You are recruiting 25 space-travelers to populate a new planet many light-years away. The travelers will need to be cryogenically frozen and to fit in the cryogenic pods they will need to be between 165cm and 198cm tall. (Show your work). Amongst the species to which you belong height can be approximated by a normal distribution with a mean of 180cm and a standard deviation of 10cm. What is the probability that a randomly selected potential traveler will fall within the permitted height range? What is the probability that a randomly selected group of 25 travelers will all be within the permitted height range? The maximum total weight of the 25 travelers must not exceed 2000kg meaning that their average weight should not be above 80kg. If individuals' weight is described by a normal distribution with a mean of 78kg and a standard deviation of 4kg what is the distribution of the mean weight of a sample of 25? What is the probability that a randomly chosen group of 25 travelers is underneath the weight limit? A colleague of yours suggests that to get the overall probability of a group of 25 travelers meeting both the height and weight constraints you should multiply the probability in (b) with the probability in (d). Do you agree? Why or why not?
Answers
The probability that a randomly selected potential traveler will fall within the permitted height range is approximately 0.8186. The probability that a randomly selected group of 25 travelers will all be within the permitted height range is approximately 0.1544. The probability that a randomly chosen group of 25 travelers is underneath the weight limit is approximately 0.0014.
To find the probability that a randomly selected potential traveler falls within the permitted height range, we can calculate the area under the normal distribution curve between the given height limits. Using the Z-score formula, we can standardize the height range and then use the standard normal distribution table or a calculator to find the corresponding probabilities. The result is approximately 0.8186.
To calculate the probability that a group of 25 travelers will all be within the permitted height range, we need to consider the probability of each individual being within the range and then multiply those probabilities together. Since the heights of individuals are independent, we can use the multiplication rule for independent events. The probability that an individual is within the range is approximately 0.8186, so the probability that all 25 individuals are within the range is (0.8186)^25, which is approximately 0.1544.
Step 3: To determine the distribution of the mean weight of a sample of 25 travelers, we need to consider the central limit theorem. According to the theorem, if the sample size is large enough (typically n ≥ 30), the distribution of sample means will be approximately normal, regardless of the shape of the population distribution. The mean weight of the sample will still have a mean of 78kg, but the standard deviation will be the population standard deviation divided by the square root of the sample size, which in this case is 4kg/sqrt(25) = 0.8kg. Therefore, the distribution of the mean weight will be a normal distribution with a mean of 78kg and a standard deviation of 0.8kg.
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What is evaporation?
Answers
Answer:
When liquid turns into vapor
Explanation:
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A liquid turning into a gas is evaporation. It is how water gets from the surface of the Earth to the atmosphere.
Temperature, surface area, wind speed, and humidity affect a liquid's evaporation rate.
Latent heat is a form of energy held in the evaporated water when it evaporates and cools.
Most of the air's moisture evaporates from rivers, lakes, seas, and oceans. Water at the surface of the Earth evaporates into water vapor, which rises into the sky and forms clouds, which eventually release water back onto Earth as precipitation.
2/9/2021
Chapter 8 Test
13.
9. Which activity that increases the genetic variation within a population of earthworms?
Mark only one oval.
Binary fission
Sexual reproduction
001
Budding
Fragmentation
14.
10. In sexual reproduction
two arant
Answers
Answer:
sexual reproduction
Explanation:
It is secusl reproduction because it increases genetic variation
In animal cells, cytokinesis is accomplished by the formation of a Multiple choice question. cleavage furrow composed of tubulin filaments. cell plate composed of tubulin filaments. cell plate composed of actin filaments. cleavage furrow composed of actin filaments.
Answers
Cytokinesis is the process of cell division in which a single cell divides into two new daughter cells. In animal cells, cytokinesis is accomplished by the formation of a cleavage furrow composed of tubulin filaments.
Correct option is A.
Tubulin is a family of globular proteins whose primary function is to help organize and support the cells. During cytokinesis, tubulin filaments assemble themselves into a cleavage furrow which pinches the cell in half and separates it into two new cells. This cleavage furrow is capable of contracting and elongating just the same as any other cellular muscle.
It is an active process, meaning that energy must be expended in order to form the cleavage furrow. Tubulin filaments interact with other cellular organelles including vesicles, organelles, and proteins in order to successfully form the cleavage furrow and ultimately divide the cell in two.
Correct option is A.
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What is the name of the protein that helps fold other proteins into their final functional form? 0.5 pt a. Ubiquitin b. Collagen c. Proteasomes d. Spliceosomes e. Chaperone f. Flippase
Answers
The name of the protein that helps fold other proteins into their final functional form is e. Chaperone. They play a crucial role in maintaining proper protein conformation and preventing protein aggregation.
The name of the protein that helps fold other proteins into their final functional form is a Chaperone protein. Chaperone proteins assist in the folding of newly synthesized proteins and help in the refolding of misfolded or denatured proteins. Spliceosomes, on the other hand, are complexes of RNA and protein that remove introns from pre-mRNA, allowing the exons to be spliced together to form a functional mRNA molecule.
Proteasomes are large protein complexes that degrade unneeded or damaged proteins, while Ubiquitin is a small protein that can be attached to other proteins, marking them for degradation by the proteasome.
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Look at the table for energy used in activities. James goes walking for 1.5 hours a week and running for 0.5 hours. James has a mass of 65kg. How much extra energy will he need a week in order to go running and walking?
Answers
The extra energy needed by James in a week is 1950 joules.
Dimensionally speaking, the energy (\(E\)) , in joules, need by a person while doing an activity is defined by the following expression:
\(E = m\cdot t\cdot c\) (1)
Where:
\(m\) - Mass, in kilograms.\(t\) - Time, in hours.\(c\) - Energy index, in joules per kilogram-hour.The extra energy needed by James is the sum of the energy spent both by running and walking:
\(E = 65\,kg \times \left(1.5\,h \times 10\,\frac{J}{kg\cdot h} + 0.5\,h \times 30\,\frac{J}{kg\cdot h} \right)\)
\(E = 1950\,J\)
The extra energy needed by James in a week is 1950 joules.
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A species of lizards migrates to a group of islands in the ocean. Over time, they diversified and evolved into several different species. Each species has a unique jaw shape that allows it to eat its own specific diet. This is an example of
gradualism because one species evolved into several different species over a relatively short amount of time
adaptive radiation because one species evolved into several different forms that occupy different niches
coevolution because the evolving lizard species changed in order to interact with the changing environment
convergent evolution because one species evolved into multiple species that live alongside each other
Answers
Answer:
adaptive radiation because one species evolved into several different forms that occupy different niches
No, because its genes are still the same. To evolve would mean that the frog had one set of genes as a tadpole, then a different set as an adult. It just doesn't work that way. The genes are the same.
Within the food web provided, look at all the ABIOTIC and BIOTIC factorsHow do they work together to create an ecosystem?
Answers
Biotic factors in an ecosystem are the living organisms or the participants in the foob web. They count on each other for their survival. The biotic factors in the foob web provided are the producers, consumers and decomposers. Abiotic factors in an ecosystem are rocks, soil, and water that collaborate with biotic factors to provide them with the nutrients they needed to survive. Sunlight, air, soil, and minerals are also examples of abiotic factors. Sunlight gives energy so that plants may use them i order to grow. This ultimately holds up the whole community of an ecosystem. Air also connects with plants to help them grow by providing an origin of CO2 and other nutrients.
I need help this is for 20 points.
Answers
Answer:
Below
Explanation:
Square One:
Top Left: TT
Top Right: Tt
Bottom Left: TT
Bottom Right: Tt
Two:
TL: Tt
TR: Tt
BL: tt
BR: tt
Three:
TL: BB
TR: Rr
BL: Rr
BR: rr
Four:
TL:RR
TR: RR
BL: Rr
BR: Rr
2. First square: top is T then t, the side is t then T
Second: top is T then t, side is T then t
Third: top is R then r, side is r then r
Fourth: All R
the reason for a "blind spot" in the visual field is that
Answers
The reason for a "blind spot" in the visual field is the absence of photoreceptor cells, specifically rods and cones, at the location where the optic nerve exits the retina.
The retina, located at the back of the eye, contains specialized light-sensitive cells called photoreceptors, which are responsible for detecting and converting light into electrical signals. There are two types of photoreceptors: rods, which are responsible for vision in low-light conditions, and cones, which are responsible for color vision and visual acuity.
The optic nerve, which carries visual information from the retina to the brain, exits the eye at a specific point called the optic disc. At this location, there are no photoreceptor cells. As a result, light that falls on the optic disc cannot be detected or processed by the retina, leading to a "blind spot" in the visual field.
Despite the presence of the blind spot, most people are not aware of it in their everyday vision. This is because the brain compensates for the blind spot by filling in the missing information based on surrounding visual input. The brain uses information from the other eye and compares it with the corresponding visual information from the other eye to create a seamless perception of the visual field.
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1. When “cleaning” a cadaver, what is removed to better see defined muscles?
Answers
When cleaning a cadaver, a professional will remove the fascia in order to better see defined muscles.
The fascia is a thin layer of connective tissue that wraps muscles. It is membraneous and extends throughout the entire body. Among the many functions of the fascia are:
Protectionisolation CompartmentalizationOf its functions, the last it's perhaps the most relevant to this question. It divided the muscles into groups. This is one of the main reasons that in order to see and properly study the muscles of a cadaver, the professional must remove the fascia.
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Answer:
deep fascia
Explanation:
pressure at highest in arteries during ventricular contraction is called?
Answers
The pressure at its highest in arteries during ventricular contraction is called systolic blood pressure.
What is artery?An artery is a blood vessel that carries oxygenated blood away from the heart to the body's tissues and organs. Arteries are a vital component of the circulatory system, which is responsible for transporting blood throughout the body. Arteries are thick-walled and have a muscular and elastic structure, allowing them to withstand the high pressure of blood pumped by the heart. As they branch out from the heart, they become smaller and eventually divide into smaller vessels called arterioles, which then lead to even smaller capillaries where oxygen and nutrients are exchanged with the body's tissues.
Here,
Systolic blood pressure is the pressure in the arteries when the heart contracts and pumps blood into the circulatory system. It is measured as the first number in a blood pressure reading and is typically around 120 mmHg in healthy adults. The normal range for systolic blood pressure is considered to be less than 120 mmHg.
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Which types of bacteria does the Monera Kingdom consist of ?
Answers
Answer:
Prokaryotic Bacteria
Answer:
Monera is a kingdom in biology that comprises prokaryotes, which are single-celled organism that have no true nucleus.Apr 28, 2017
Explanation:
Kingdom Monera
You have suffered from a lot of diseases that bacteria cause. But, do you think all bacteria are bad? No! The bacteria have a huge kingdom of their own that consists of a number of varieties of them. Each bacteria has a different role to play. They belong to the kingdom Monera. Here, we will know all about the kingdom Monera. We will look at both their characteristics and divisions in greater detail.
Characteristics of Monera
Monera (Monos – single) includes prokaryotes and shows the following characters:
They are typically unicellular organisms (but one group is mycelial). The genetic material in these organisms is the naked circular DNA. A nuclear envelope is absent. Both, ribosomes and simple chromatophores, are the only subcellular organelles in the cytoplasm.
Sap vacuoles do not occur. Instead, gas vacuole may be present.
The predominant mode of nutrition is absorptive but some groups are photosynthetic (holophytic) and chemosynthetic.
The organisms are non-motile or move by the beating of simple flagella or by gliding.
Bacteria Shape
Cocci: They are oval or spherical in shape.
Bacilli: They are rod-shaped. They may or may not have flagella.
Vibrios: These are small and ‘comma or kidney’ like. They have a flagellum at one end and are also motile. Vibrio bacteria has a curve in its cell.
Spirillum: They are spiral or coiled like a corkscrew. The spiral forms are usually rigid and bear two or more flagella at one or both the ends e.g., Spirillum, Spirochaetes etc.
Filament: Just like fungal mycelia, the body of the bacterium is filamentous. The filaments are very minute in size. Examples include Beggiota, Thiothrix etc.
Stalked: The body of bacterium possesses a stalk e.g., Caulobacter.
Budded: The body of the bacterium is swollen at places e.g., Rhodomicrobiu
Structure of Bacteria
Capsule: In a large number of bacteria, a slimy capsule is present outside the cell wall. It is composed of polysaccharides and the nitrogenous substances (amino acids) are also present in addition. This slime layer becomes thick, called, capsule. The bacteria, which form a capsule, are’ called capsulated or virulent bacteria. The capsule ‘is usually found in parasitic forms e.g., Bacillus, anthracite, Diplococcus pneumoniae, Mycobacterium tuberculosis.
Cell wall: All bacterial cells .are covered by a strong, rigid cell wall. Therefore, we classify them under plants. Inner to the capsule, the cell wall is present. It is made up of polysaccharides, proteins and lipids. We also find D-glutamic acid and diaminopimelic acid.
Monera
Plasma membrane: Each bacterial cell has a plasma membrane. It is situated just internal to the cell wall. It is a thin, elastic and also differentially or selectively permeable membrane.
Cytoplasm: The cytoplasm refers to a complex and aqueous fluid or semifluid ground substance (matrix). This material consists of vitamins, salts, enzymes, carbohydrates, soluble proteins, co-enzymes, lipids, mineral and nucleic acids. The organic matter is present in the colloidal state.The cytoplasm is granular due to the presence of a large number of ribosomes.
Nucleoid: It has other common names like genophore, naked nucleus or incipient nucleus. There is nuclear material in these, DNA. It is double helical as well as circular. Some kind of typical protein surrounds it. However, these are not histone proteins.
Plasmids: In addition to the normal DNA chromosomes, many bacteria (e.g., E.coli) have extrachromosomal genetic elements or DNA. These elements are plasmids. They are small circular double-stranded molecules.
Flagella: These are fine, thread-like, protoplasmic appendages. These extend through the cell wall and the slime layer of the flagellated bacterial cells. These help in bacteria to swim about in the liquid medium.
Pili or Fimbriae: Besides flagella, some tiny or small hair-like outgrowths are present on the bacterial cell surface. These are pili. They comprise of pilin protein. They consequently measure about 0.5-2 mm in length and 3-5mm in diameter.
Nutrition in Bacteria
On the basis of mode of nutrition, we can group bacteria into two broad categories. First is autotrophic whereas second is heterotrophic bacteria.
Autotrophic bacteria: These bacteria are able to synthesize their own food from inorganic substances, as green plants do. They derive their carbon from carbon dioxide. The hydrogen needed to reduce carbon to organic form comes from sources such as atmospheric H2, H2S or NH3.
Heterotrophic bacteria: Most of the bacteria cannot synthesize their own organic food. They consequently depend on external organic materials. They require at least one organic compound as a source of carbon for their growth and energy. Such bacteria are heterotrophic bacteria. Heterotrophic bacteria are of three types: Parasites, Saprotrophs and Symbionts.