Since "the four bases in DNA could not code for the 20 amino acids," scientists assume that DNA does not directly code for proteins. Thus, Option E is correct.
In the early days of molecular biology, scientists were puzzled by how the genetic information encoded in DNA could be used to produce the vast array of proteins found in living organisms. The discovery of the genetic code, which showed how groups of three bases in DNA (codons) correspond to specific amino acids, provided a key insight into this process.
However, it also raised a new question: with only four different bases (A, C, G, and T), how could DNA possibly code for the 20 different amino acids that make up proteins? Scientists soon realized that the genetic code must be more complex than a simple one-to-one correspondence between bases and amino acids, and that additional factors, such as RNA intermediaries and post-translational modifications, are necessary to account for the diversity of proteins in living organisms.
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8. Many promising drugs remain limited by low stability, toxicity, inefficient administration, and the need for multiple doses. For example, while chemotherapy is effective for the treatment of malignancies, it is widely known for harmful side-effects on healthy tissues and reduce patient compliance. Targeted drug delivery, if successfully implemented, can reduce some of the toxic effects of BIOM 6615: DESIGN AND APPLICATION OF BIOMATERIALS Mid-term Exam March 17, 2021. drugs as minimum effective drug concentrations could be delivered. State what drug delivery strategy each of the following represents; passive or active. a. Cationic microparticles for enhanced delivery to phagocytic cells b. Surface modification with polyethylene glycol (PEG) to provide "stealth" behavior from immune recognition c. Antigen-antibody recognition d. Ligand-receptor recognition
The different drug delivery strategies mentioned in the question can be classified as passive or active based on their mechanism of action.
Here is the classification of the mentioned drug delivery strategies:
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Why did dr. Erwin focus on tropical trees to estimate the total numbers of animals in the world?
Dr. Erwin's focus on tropical trees was based on his observation that these trees support a particularly high diversity of insect species
What do you mean by tropical trees?
Tropical trees are trees that grow in the tropical regions of the world, which are typically located between the Tropic of Cancer and the Tropic of Capricorn. These regions are characterized by warm temperatures and high levels of precipitation, which provide ideal growing conditions for many types of trees.
Therefore, Dr. Erwin's focus on tropical trees was based on his observation that these trees support a particularly high diversity of insect species
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Photosynthesis evolved in bacteria long ago (before plants did), but across all living things what is it’s general purpose (what does it do for cells)? What are 2 ways in which photosynthesis in cyanobacteria is a) similar to photosynthesis in plants and b) different from photosynthesis in green or purple sulfur bacteria?
Photosynthesis provides energy for cellular processes. Cyanobacteria and plants both use chlorophyll a as their primary pigment and have a similar electron transport chain. Green/purple sulfur bacteria use different pigments and electron transport chains.
Its general purpose is to provide organisms with the energy they need to live and reproduce.
Photosynthesis in cyanobacteria is similar to photosynthesis in plants in that it uses the sun's energy to convert carbon dioxide and water into sugars and oxygen.
However, it is different from photosynthesis in green and purple sulfur bacteria in that it does not use sulfide as an electron donor, but rather water.
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why the core and rind differentiation not found in the flat
Rhizomorphs (Fungus)?
The core and rind differentiation is not found in the flat rhizomorphs of fungi because these structures are not present in this type of fungi. The core and rind differentiation is a characteristic of the Agaricomycetes, which are a class of fungi that includes mushrooms, bracket fungi, and puffballs. The flat rhizomorphs, on the other hand, are a type of structure found in the Basidiomycota, which is a different class of fungi that includes rusts and smuts.
The flat rhizomorphs are composed of parallel hyphae that are tightly packed together and are used for the transport of nutrients and water. They do not have the same structure as the Agaricomycetes, which have a core of densely packed hyphae surrounded by a rind of looser hyphae. This is why the core and rind differentiation is not found in the flat rhizomorphs.
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A saltwater aquarium can be used to model the ocean. What is one limitation
of this model?
OA. It is shallower than the ocean.
OB. Its temperature can be varied.
C. It can show how light affects ocean plants.
OD. It can show how certain fish swim in the ocean.
one limitation of this model of salt water aquarium is It is shallower than the ocean.
Saltwater aquarium explained
A saltwater aquarium is a tank or container filled with seawater and designed to simulate a marine ecosystem. It typically includes live plants, corals, and various species of fish, invertebrates, and other marine organisms. The water in a saltwater aquarium must be carefully maintained to ensure proper salinity, temperature, pH, and nutrient levels for the health and survival of the inhabitants. Saltwater aquariums can be used for scientific research, education, and recreation, and require significant knowledge and effort to maintain.
A saltwater aquarium can only hold a limited amount of water and is generally much shallower than the ocean. This means that the aquarium may not accurately represent the complex and dynamic conditions of the ocean, such as the depth, currents, and waves.
Therefore, the aquarium environment may be more stable and controlled compared to the ocean, which can limit the accuracy of the model.
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- Tissue cell donar -> Cells from animal to be cloned are main- ained in he hey do not grow or dvide -> Nucleus is removed -> Nucleus fusses with empty egg after electric current applied -> The reconstructed embryo grows for 7 days -> Embryo's eimplanted into surrogate mother -> Cloned animal is born with exact DNA as the tissue cell donor. - Donor Supplies unfertilised eggs -> Egg cell -> Nucleus is removed -> Nucleus fusses with empty egg after electric current applied -> The reconstructed embryo grows for 7 days -> Embryo's eimplanted into surrogate mother -> Cloned animal is born with exact DNA as the tissue cell donor.
Dolly was the first successfully cloned mammal. Which of the 3 adult female sheep was she considered to be a clone of? A. A combination of the tissue donor and egg donor females B. The tissue cell donor. C. The egg donor D. The surrogate mother
- Tissue cell donar -> Cells from animal to be cloned are main- ained in he hey do not grow or dvide -> Nucleus is removed -> Nucleus fusses with empty egg after electric current applied -> The reconstructed embryo grows for 7 days -> Embryo's eimplanted into surrogate mother -> Cloned animal is born with exact DNA as the tissue cell donor. - Donor Supplies unfertilised eggs -> Egg cell -> Nucleus is removed -> Nucleus fusses with empty egg after electric current applied -> The reconstructed embryo grows for 7 days -> Embryo's eimplanted into surrogate mother -> Cloned animal is born with exact DNA as the tissue cell donor.
Dolly was the first successfully cloned mammal. The 3 adult female sheep was she considered to be a clone of B. The tissue cell donor.
Dolly was the first successfully cloned mammal. She was considered to be a clone of the tissue cell donor, as cells from the tissue cell donor were taken and fused with an empty egg after an electric current was applied. The reconstructed embryo was grown for 7 days before being implanted into a surrogate mother, and eventually a cloned animal was born with the exact DNA of the tissue cell donor.
Since the genetic material in the nucleus of the tissue cell donor was used to create Dolly, she is considered to be a clone of the tissue cell donor. Therefore, the correct answer is B. The tissue cell donor.
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.
1. Answer the following characteristics for glomeromycota
Fungi.
A. Color
B. Texture
C. Form
D. Size
E. Starch storage (where)
The characteristics of Glomeromycota fungi are as follows:
A. Color: colorless
B. Texture: typically slimy or powdery
C. Form: usually mycelial and branching
D. Size: microscopic
E. Starch storage (where): in the hyphae and spores
Glomeromycota is a group of fungi that form mutualistic associations with the roots of plants. It is characterized by the following features:
Color: They are colorless.
Texture: They are typically slimy or powdery.
Form: They are hyphae-forming fungi.
Size: They are smaller in size than other fungi.
Starch storage: They store carbohydrates in their hyphae.
Glomeromycota does not produce any sexual spores, so it is hard to distinguish it from other fungi. The asexual spores of Glomeromycota are produced inside the sporangium.
Glomeromycota are considered primitive fungi because they are lacking certain cell organelles and have a simple genome structure. Their major ecological importance is in the formation of arbuscular mycorrhiza, which is a mutualistic relationship between the plant roots and fungi.
The fungi provide the plants with essential nutrients such as phosphorus and nitrogen, while the plants supply them with carbohydrates. This mutualistic association enhances the growth and development of the plants while also benefiting the fungi by providing them with a stable source of energy.
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Name the Biome:Precipitation: 230-420 cm/yearTemperature: 20°-28°Chigh biodiversity leads to different specialized niches-some occupy a single habitat layer for their entire liveswith the abundance of different species, plants form layers such as the canopy, midstory, and the understorybeing converted to agriculture despite poor nutrient content of its soil and difficulty of rehabilitation after the conversion; chemicals produced by the diverse organisms have been turned into medicineslittle seasonality with lots of rain fall and warm temps year-round (no freezing months)
The biome described in the question is the Tropical Rainforest Biome. This biome is characterized by high precipitation levels, with an annual rainfall of 230-420 cm.
The temperature in this biome is consistently warm, with temperatures ranging from 20°-28°C. The Tropical Rainforest Biome also has a high level of biodiversity, with many different species occupying specialized niches and different habitat layers. The abundance of different species leads to the formation of layers, such as the canopy, midstory, and understory. Despite the poor nutrient content of its soil, this biome is often converted to agriculture, which can be difficult to rehabilitate after the conversion. The chemicals produced by the diverse organisms in the Tropical Rainforest Biome have also been turned into medicines. This biome has little seasonality, with lots of rainfall and warm temperatures year-round, with no freezing months.
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Which TWO of the following are key movements of laryngeal vestibular closure (LVC) achieved via the intrinsic laryngeal muscles, innervated by CN X RLN
The two key movements of laryngeal vestibular closure (LVC) achieved via the intrinsic laryngeal muscles, innervated by CN X RLN are Adduction and Elevation.
1. Adduction of the arytenoid cartilages: This movement is achieved through the contraction of the lateral cricoarytenoid muscles, which are innervated by the recurrent laryngeal nerve (RLN) of the tenth cranial nerve (CN X).
2. Elevation of the larynx: This movement is achieved through the contraction of the thyrohyoid muscles, which are innervated by the recurrent laryngeal nerve (RLN) of the tenth cranial nerve (CN X).
These two movements are crucial for LVC, as they help to close off the larynx during swallowing, preventing food or liquids from entering the airway. The adduction of the arytenoid cartilages brings the vocal folds together, while the elevation of the larynx moves the entire structure upward, further closing off the airway. Both of these movements are controlled by the intrinsic laryngeal muscles, which are innervated by the RLN of CN X.
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Correct form of question should be:
Which TWO of the following are key movements of laryngeal vestibular closure (LVC) achieved via the intrinsic laryngeal muscles, innervated by CN X RLN?
A. Elevation
B. Adduction
C. Depression
D. Abduction
Set R to 1 mmHg min ml-1
If the difference between P1-
P2was 40 mmHg, what would the rate of
blood flow (F) equal in ml min-1?
Rate of blood flow (ml min-1) =
The rate of blood flow (F) would equal 40 ml min⁻¹ if the difference between P1 and P2 was 40 mmHg.
According to the question, R is set to 1 mmHg min ml⁻¹ and the difference between P1 and P2 is 40 mmHg. We can use the formula for the rate of blood flow (F) to find the answer:
F = (P1 - P2)/R
Plugging in the given values:
F = (40 mmHg) / (1 mmHg min ml⁻¹)
F = 40 ml min-1
Therefore, the rate of blood flow (F) is 40 ml min⁻¹.
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Lab 5
Cell Fractionation
Extraction of Mitochondria & Illustration of Electron Transport
Activity
1. What is differential centrifugation?
2. What is the role of rho-phenylenediamine in this assay?
3. Why the assay was carried out at 37ºC?
4. Which compound(s) inhibited electron transport and how?
Differential centrifugation is a technique used to separate different types of cellular or organelle components based on their size, shape, and density.
This process involves spinning the lysate cells at different speeds, which causes the different components to form layers or pellets based on their physical properties. One of the indicators used to measure electron transport activity is rho-phenylenediamine.
When electron transport occurs, the oxygen consumed and the amount of rho-phenylenediamine that interacts with oxygen will decrease. A decrease in the color intensity of this test is an indication of reduced electron transport activity.
The 3 answer is:
Q1: Differential centrifugation is a method used to separate components in a mixture based on their size, shape, density, and other characteristics. It involves spinning a solution at high speeds in a centrifuge, which causes the different components to separate based on their mass.Q2: Rho-phenylenediamine is used in this assay as a reagent to detect electrons. When electrons are produced during the electron transport activity, the rho-phenylenediamine reacts with the electrons to produce a colored complex, which can be observed.Q3: The assay was carried out at 37ºC because the optimum temperature for mitochondria is 37ºC. Q4: Compounds such as malonate, succinate, and antimycin A inhibit electron transport by blocking certain pathways of the electron transport chain.Learn more about differential centrifugation at https://brainly.com/question/30433874
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The following data correspond to one marker taken from a genome-wide association study (GWAS) of genetic variants related to heart disease. The number of individuals are categorized by case-control status, as well as their genotype at this marker (AA, AG, or GG).
AA AG GG
Cases
(heart disease)
315 702 1410
Controls
(no heart disease)
85 723 1717
What would you conclude based on these data?
Group of answer choices
a. This marker is likely to be closely linked to a gene that affects heart disease risk, and the A allele is associated with higher disease risk.
b. This marker is likely to be closely linked to a gene that affects heart disease risk, and the G allele is associated with higher disease risk.
c. This marker is not likely to be closely linked to a gene that affects breast cancer risk.
Based on these data you could conclude is b. This marker is likely to be closely linked to a gene that affects heart disease risk, and the G allele is associated with higher disease risk.
The genome-wide association study (GWAS) is used to recognize the association between genetic variants and diseases. It is a type of observational study that compares the genomes of groups of people with and without disease. This study determines which genetic variations are related to the disease.
In this study, the G allele is associated with higher disease risk. So, this marker is likely to be closely linked to a gene that affects heart disease risk. Therefore, option B This marker is likely to be closely linked to a gene that affects heart disease risk, and the G allele is associated with higher disease risk is correct.
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True or False. Ants are more closely related to beetles than to
flies because there are only two nodes between ants and beetles but
three nodes between ants and flies.
False. Flies and beetles are more distantly related to ants than ants are to beetles. It is not possible to establish how closely related two species are based on the number of nodes that exist between them.
Instead, the degree of proximity between two people is determined by the number of similar features, often known as synapomorphies.
Beetles are classified under the order Coleoptera, whereas ants and flies are classified under the order Hymenoptera.
Because of this, ants are more closely related to beetles than they are to beetles themselves.
Therefore, the statement is False. Ants are more closely related to flies than to beetles.
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What is the methylene blue staining procedure? What is the
procedure of methyl blue staining (Process based on yeast and mold
staining)
The methylene blue staining procedure is a technique used to identify and visualize different types of microorganisms, such as bacteria, yeast, and mold. The process involves treating a sample with methylene blue dye, which binds to certain cellular structures and makes them more visible under a microscope.
The procedure of methylene blue staining for yeast and mold is as follows:1. Prepare a slide by placing a small amount of the sample on a glass microscope slide.
2. Add a drop of methylene blue solution to the sample.
3. Spread the solution evenly over the sample using a sterile loop or needle.
4. Allow the slide to sit for a few minutes so that the dye can bind to the cellular structures.
5. Rinse the slide gently with water to remove any excess dye.
6. Allow the slide to air dry or gently blot it with a clean paper towel.
7. Place the slide on the microscope stage and observe under the appropriate magnification.
By following these steps, you can visualize the yeast and mold cells in your sample and identify any structural features that may be present. This can help in the identification and classification of different types of microorganisms.
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Data from a series of genetics crosses give the. following x2 values. For each, determine the degrees of freedom and, by consulting Table 6-5, the proba
bility value for the x2 value. Indicate whether the deviation involved is
significant or nonsignificant at the 0.05 level of significance and whether
there is support for the hypothesis assumed for each cross.
a. Two classes of progeny, x2 = 3.020.
b. Four classes of progeny, x2 = 10.360.
c. Three classes of progeny, x2 = 1.555.
6-4. Refer back to the information given in problem 6-3. Assume the significance
level is changed from 0.05 to 0.01. How does this influence the interpretation
of the deviations involved in each set of data?
For problem 6-3, the degrees of freedom for the three classes of progeny are (a) 2, (b) 3, and (c) 2.
For each, the probability value for the x2 value is (a) 0.215, (b) 0.010, and (c) 0.456.
This means that the deviation involved in
(a) is nonsignificant at the 0.05 level of significance and there is no support for the hypothesis assumed for the cross;
(b) is significant at the 0.05 level of significance and there is support for the hypothesis assumed for the cross; and
(c) is nonsignificant at the 0.05 level of significance and there is no support for the hypothesis assumed for the cross.
If the significance level is changed from 0.05 to 0.01, the interpretation of the deviations involved in each set of data will be
(a) nonsignificant at the 0.01 level of significance and there is no support for the hypothesis assumed for the cross;
(b) significant at the 0.01 level of significance and there is support for the hypothesis assumed for the cross; and
(c) nonsignificant at the 0.01 level of significance and there is no support for the hypothesis assumed for the cross.
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which term is best described as the production of proteins based on the cell's genetic information?
transcription
Gene synthesis
Gene repression
gene expression
Transcription is best described as the production of proteins based on the cell's genetic information.
What is transcription?
Transcription is the process of turning a segment of DNA into RNA. DNA segments that have been translated into RNA molecules that can encode proteins are known as messenger RNA (mRNA). When extra DNA segments are translated into RNA molecules, non-coding RNAs are created (ncRNAs). Just 1% to 3% of all RNA samples contain mRNA. Human genome coding vs. non-coding DNA analysis reveals that while at least 80% of mammalian genomic DNA can be actively translated (in one or more types of cells), the majority of this 80% is non-coding RNA (ncRNA), while less than 2% of the mammalian genome can be actively translated into mRNA.
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Answer: Gene expression
Def: Chemical reaction by which the cells convert energy from one form to another and build and break down molecules
A chemical reaction that occurs in cells to convert energy from one form to another and to build and break down molecules is called metabolism.
Metabolism is the process by which cells convert nutrients into energy and use that energy to perform various cellular functions, such as building and breaking down molecules. There are two types of metabolism: catabolism, which breaks down molecules to release energy, and anabolism, which uses energy to build molecules. Both types of metabolism are necessary for cells to function properly and maintain homeostasis.
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compartment A has a concentration of 125 mosm/L and a volume of 13.5 L, compartment B has a concentration of 225 mosm/L and a volume of 6 L, and the compartments are only permeable to water. If the initial volume of compartment A were doubled, what would be the final concentration in compartment B at equilibrium?
A. 185
B.155.64
C. 170
D. 140
The final concentration in compartment B at equilibrium if the initial volume of compartment A were doubled is 155.64 mosm/L.
To find the finаl concentrаtion in compаrtment B аt equilibrium, we cаn use the formulа:
[tex]C_{1}V_{1}[/tex] = [tex]C_{2}V_{2}[/tex]
where [tex]C_{1}[/tex] is the initiаl concentrаtion of compаrtment А, [tex]V_{1}[/tex] is the initiаl volume of compаrtment А, [tex]C_{2}[/tex] is the finаl concentrаtion of compаrtment B, аnd [tex]V_{2}[/tex] is the finаl volume of compаrtment B.
Since the initiаl volume of compаrtment А is doubled, we cаn plug in the vаlues into the formulа:
125 mosm/L * 13.5 L * 2 = 225 mosm/L * 6 L * [tex]V_{2}[/tex]
Solving for [tex]V_{2}[/tex], we get:
[tex]V_{2}[/tex] = (125 * 13.5 * 2) / (225 * 6)
[tex]V_{2}[/tex] = 7.5 L
Now, we cаn plug in the vаlues for [tex]C_{1}[/tex], [tex]V_{1}[/tex], аnd [tex]V_{2}[/tex] into the formulа to find the finаl concentrаtion in compаrtment B:
[tex]C_{2}[/tex] = ([tex]C_{1}[/tex] * [tex]V_{1}[/tex]) / [tex]V_{1}[/tex]
[tex]C_{2}[/tex] = (125 * 13.5 * 2) / 7.5
[tex]C_{2}[/tex] = 155.64 mosm/L
Therefore, the finаl concentrаtion in compаrtment B аt equilibrium is 155.64 mosm/L.
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Which descriptions accurately characterize rectal temperature measurement? Check all that apply.
It most closely matches the core body temperature.
It can be used for uncooperative patients.
It can be affected by recent consumption of food or drink.
It gives a reading lower than the actual core body temperature.
It can be used for infants.
The descriptions that accurately characterize rectal temperature measurement are as follows:
It can be used for uncooperative patients.It can be affected by recent consumption of food or drink.It can be used for infants.Thus, the correct options for this question are B, C, and E.
What do you mean by Rectal temperature?The rectal temperature may be defined as a type of process that significantly deals with measuring a person's temperature by inserting a thermometer into the rectum via the anus.
The rectal temperature is one of the most accurate ways in order to determine whether a child has a fever or not. This is because this temperature is usually taken in the rectum and is the closest way to finding the body's true temperature. Rectal temperatures run higher than those taken in the mouth or armpit (axilla) because the rectum is warmer.
Therefore, the descriptions that accurately characterize rectal temperature measurement are well-mentioned above.
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You have an original cell density of 5.8 x 108 CFU/mL. What is this number in its non-scientific notation or "regular" format?
a. 0.000000058 CFU/mL
b. 0.0000000058 CFU/mL
c. 58,000,000 CFU/mL
d. 5.8 CFU/mL
e. 5800,000,000 CFU/mL
f. 580,000,000 CFU/mL
The number in its non-scientific notation or "regular" is option f. 580,000,000 CFU/mL.
To convert a number from scientific notation to regular format, you need to move the decimal point to the right the same number of places as the exponent. In this case, the exponent is 8, so you need to move the decimal point 8 places to the right.
5.8 x 10^8 = 58 x 10^7 = 580 x 10^6 = 5800 x 10^5 = 58000 x 10^4 = 580000 x 10^3 = 5800000 x 10^2 = 58000000 x 10^1 = 580000000 x 10^0 = 580,000,000 CFU/mL
Therefore, the original cell density of 5.8 x 10^8 CFU/mL is equivalent to 580,000,000 CFU/mL in regular format.
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Why is protein shape such an important thing? What is
denaturation, and how does it affect the protein's usefulness in
the living cell where it was produced?
Protein shape is important because it determines the protein's function. Each protein has a specific shape that allows it to interact with other molecules and perform its function. Denaturation is the process of a protein losing its specific shape.
This can happen due to changes in temperature, pH, or other environmental factors. When a protein is denatured, it can no longer perform its function.
In the living cell, denaturation can have serious consequences. If a protein that is essential for a cell's function is denatured, the cell may not be able to perform its normal functions and may die. This is why it is important for cells to maintain a stable environment to prevent denaturation of their proteins.
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3. (6pts) What would the translation of these mRNA transcripts produce? (mRNA codon
→
anticodon
→
protein) a. UAA CAA GGA GCA UCC b. UGA CCC GAU UUC AGC
The translation of UAA CAA GGA GCA UCC will not produce any protein
The translation of UGA CCC GAU UUC AGC will also not produce any protein.
Translation of mRNA transcriptTo translate the mRNA transcripts into protein, we need to use the genetic code to convert the mRNA codons into amino acids. Each mRNA codon corresponds to a specific amino acid, and the sequence of codons determines the sequence of amino acids in the protein.
UAA CAA GGA GCA UCCThe first codon, UAA, is a stop codon and does not code for an amino acid. Therefore, this mRNA transcript does not produce a protein.
UGA CCC GAU UUC AGCUGA is also a stop codon and does not code for an amino acid.
Therefore, the two mRNA transcripts do not produce proteins.
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Birth control works by providing _________ to the hypothalamus and anterior pituitary about the level of circulating estrogen and progesterone.
Question 8 options:
positive feedback
negative feedback
critical feedback
inverse feed-forward dynamics
Birth control works by providing B: negative feedback to the hypothalamus and anterior pituitary about the level of circulating estrogen and progesterone.
Birth control methods such as the pill, patch, and vaginal ring contain synthetic hormones, either estrogen alone or a combination of estrogen and progesterone. These hormones work by suppressing the production of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) in the hypothalamus and anterior pituitary gland. FSH and LH are the hormones responsible for stimulating ovulation.
By providing feedback to the hypothalamus and pituitary gland, the synthetic hormones in birth control mimic the natural hormonal changes that occur during the menstrual cycle. However, they do so at higher levels, which prevent the production of FSH and LH and stop ovulation from occurring.
This feedback is negative because it works to decrease the production of FSH and LH in response to the high levels of synthetic hormones in birth control, rather than increasing their production. In contrast, positive feedback would stimulate the production of FSH and LH, which would lead to ovulation, and critical feedback and inverse feed-forward dynamics are not applicable to this scenario.
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Eye color in fruit flies is a sex-linked trait. Red eye color is dominant over
white eye color. A homozygous, red-eyed female is crossed with a white-eyed
male.
In their offspring, what is the expected phenotypic ratio of red-eyed females
to white-eyed females to red-eyed males to white-eyed males?
OA. 1:2:1:0
OB. 1:1:1:1
OC. 2:0:1:1
OD. 2:0:2:0
Answer: D
Explanation: The RR female with the rY male would make 2 RY(red-eyed) males as only the female donates a allele and two Rr(red-eyed) females which are heterozygous
The expected phenotypic ratio of red-eyed females to white-eyed females to red-eyed males to white-eyed males is 1:1:1:1. The correct option is B.
It can be determined based on the principles of sex-linked inheritance.
In this case, since the eye color gene is sex-linked and located on the X chromosome, the genotype of the female parent would be [tex]\rm X^R X^R[/tex](homozygous for red eyes), and the genotype of the male parent would be [tex]X^{W }Y[/tex](white eyes).
The possible genotypes and corresponding phenotypes of the offspring are as follows:
Red-eyed females ([tex]X^R X^R[/tex]): All female offspring from the cross will inherit the red-eye color gene from the homozygous red-eyed mother.White-eyed females ([tex]X^R X^W[/tex]): Female offspring will inherit one red eye color gene from the mother ([tex]X^R[/tex]) and one white eye color gene from the father ([tex]X^W[/tex]).Red-eyed males ([tex]X^R Y[/tex]): Male offspring will inherit the red eye color gene from the mother ([tex]X^R[/tex]) and the Y chromosome from the father.White-eyed males ([tex]X^W Y[/tex]): No white-eyed males will be produced in this cross since the white-eye color gene is recessive and only located on the X chromosome.The expected phenotypic ratio of red-eyed females to white-eyed females to red-eyed males to white-eyed males is 1:1:1:1.Thus, the correct option is B.
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Which elements most easily give up electrons?
Responses
metalloids
nonmetals
metals
noble gases
Answer:Elements that give up electrons easily are called metals.
A dragon contains 6 pairs of chromosomes. How manysister chromosomes will it have at each of thefollowing:-Prophase of mitosis?-Prophase of meiosis 1?-Prophase of meiosis 2?
a. In prophase of mitosis will have 24 sister chromatids.
b. In prophase of meiosis 1 will have 12 sister chromatids.
c. In prophase of meiosis 2 will have 6 sister chromatids.
A dragon contains 6 pairs of chromosomes. At the prophase of mitosis, the dragon will have 12 sister chromatids since each chromosome has two identical sister chromatids.
At the prophase of meiosis 1, the dragon will have 24 sister chromatids. This is because it will have duplicated its chromosomes during the previous interphase. So, each chromosome is composed of two sister chromatids, which will separate during meiosis 1.
At the prophase of meiosis 2, the dragon will have 12 sister chromatids. After meiosis 1, the chromosome pairs separate, but the sister chromatids remain joined. This means that the chromosome number is halved, and there are now 3 chromosomes instead of 6.
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You have discovered several new antimicrobial compounds that inhibit bacterial growth and can be used an antibiotic. You have determined the specific cellular target for each. Based in your knowledge of replication, transcription and translation, indicate which process each compound will likely block and justify your answer. Compound A. Inhibit helicase.
Compound A inhibits helicase, an enzyme that is involved in the unwinding of DNA during replication. This means that it prevents the replication of the DNA template strand and therefore the production of the complementary strand, leading to an inhibition of the bacterial growth.
The inhibition of helicase also has an effect on transcription and translation, as it prevents the transcription of the complementary strand and prevents the translation of proteins from that strand. Therefore, Compound A can be used as an antibiotic, as it blocks the essential processes of replication, transcription and translation, thereby inhibiting bacterial growth.
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In eukaryotes, what is the first thing that binds to a gene's promoter for transcription to begin? TFIIF Sigma factor TFIIH TBP by itself. TFIIA TFIIB TFIID + TI
In eukaryotes, the first thing that binds to a gene's promoter for transcription to begin is TFIID.TFIID is the first transcription factor to bind to the promoter in eukaryotic cells to initiate transcription.
It specifically binds to the TATA box, a sequence of nucleotides in the promoter region of the gene.
TFIID recruits other transcription factors and binds to RNA polymerase II to initiate transcription.
Other transcription factors that bind to the promoter and RNA polymerase II to initiate transcription in eukaryotic cells include TFIIA, TFIIB, TFIIF, TFIIH, and TBP.
TBP stands for TATA-binding protein, which binds to the TATA box and causes DNA to bend, making it more accessible to other transcription factors.
TFIIH unwinds DNA and exposes the template strand for RNA polymerase, allowing it to synthesize RNA.
TFIIF stabilizes the RNA polymerase II complex and stimulates its activity, helping it to stay attached to the template strand and move forward to synthesize RNA.
TFIIB helps RNA polymerase II bind to the promoter region of the gene by binding to the BRE and recruiting RNA polymerase II to the promoter.
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Explain how teamwork was important for evolution of the modern-day eukaryotes that contains mitochondria and chloroplast.
The evolution of modern-day eukaryotes containing mitochondria and chloroplast was influenced by the importance of teamwork and explained by the endosymbiotic theory.
The endosymbiotic theory argues that mitochondria and chloroplasts, which are key components of eukaryotic cells, were once independent, free-living prokaryotic cells that evolved to form a partnership with a host cell in order to benefit from the cell's resources and protective environment. Both the host and the prokaryotic cell have benefited from this partnership, as the prokaryotic cell has become dependent on the host cell for the provision of a stable, protective environment, and the host cell has benefited from the prokaryotic cell's ability to carry out a variety of essential metabolic functions that are essential for life. F
or example, mitochondria are responsible for energy production in eukaryotic cells, while chloroplasts are responsible for photosynthesis. It is due to teamwork that these functions are carried out seamlessly.
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We are learning about the life cycle of viruses. You should be able to know the steps of viral replication from Entry to Exit. This assignment will help you understand this process. Please Pick one of
The steps of viral replication include attachment, penetration, uncoating, replication, transcription, translations and assembly.
Viral replication is the process of forming new viruses within a host organism. The following are the steps of viral replication from entry to exit:
Step 1: Attachment- The virus attaches to the host cell surface using surface proteins.
Step 2: Penetration- The virus enters the host cell by endocytosis or fusion with the plasma membrane.
Step 3: Uncoating- The viral capsid or envelope is degraded, releasing the viral genome into the host cell cytoplasm.
Step 4: Replication- Viral genome is replicated using host cell machinery.
Step 5: Transcription- The viral genome is transcribed into mRNA using host cell machinery.
Step 6: Translation- The viral mRNA is translated into viral proteins using host cell machinery.
Step 7: Assembly- New viral particles are assembled using the viral proteins and genome.Step 8: ReleaseThe new viral particles are released from the host cell, often causing host cell death.
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