Answer:
The post transcription modification include three ways i.e., capping, polyadenylation and splicing.
1. 5' Capping- mRNA capping is catalyzed by capping enzyme which is associated with RNA polymerase II. This enzyme adds the guanosin residue to 5' end of mRNA then methyl transferase adds methyl group to N⁷ position this guanine. Capping protects mRNA from degradation by enzymes presents in the cell.
2. Polyadenylation: In polyadenylation many adenosine residue is added at 3'end of mRNA by poly(A) polymerase. This poly A tail helps in exporting mRNA out of nucleus and binding of translational factors.
3. Splicing: In splicing process the non coding introns are removed from RNA so that only coding region i.e., exons are left in mRNA. It is important to remove unwanted sequence like introns which hinders in protein formation.
Tc cells are involved in the following type of immune response
Select one:
a. humoral immune response
b. allergic response
c. cell-mediated immune response
d. complement
Answer:
The correct answer will be option-C.
Explanation:
The lymphocytes are a type of immune cells which provides immunity to the organism. The lymphocytes differentiate into two types: B cell and t cell.
T cell lymphocytes are programmed to recognize and respond to kill the antigen containing cell. The T cells provide immunity wither by directly killing cell containing antigen or by producing lymphokines which induce other cells to kill the affected cell. This type of immune response is known as cell-mediated immune response.
Thus, Option-C is the correct answer.
Two short-haired cats mate and produce six short- haired and two long-haired kittens. What does this information suggest about how hair length is inherited?
Answer:
In the progeny, six kittens were short haired and two were long haired. The ratio of short to long haired kittens is 3:1 which is a classic monohybrid cross ratio. Here, short hair is inherited dominantly and long hair is inherited recessively. Both the parents are heterozygous for short hair trait (Ss). Cross between them:
Ss X Ss :
S s
S SS Ss
s Ss ss
Genotypically: 1 : 2 : 1
SS Ss ss
Phenotypically: 3 : 1 = 6 short and 2 long haired kittens
short hair long hair
Hence short hair is dominant over long hair in these cats.
Final answer:
Hair length in cats is inherited according to Mendelian genetics, with short hair being dominant and the observed offspring ratio suggesting both parent cats are heterozygous for the trait.
Explanation:
The mating of two short-haired cats resulting in both short-haired and long-haired kittens suggests that the trait for hair length in cats is inherited in a Mendelian fashion, with the allele for short hair (S) being dominant over the allele for long hair (s). Since both parents are short-haired but have produced long-haired offspring, it implies that each parent is heterozygous (Ss), carrying one allele for short hair and one for long hair. The distribution of short-haired to long-haired kittens (a roughly 3:1 ratio) fits the expected outcome of a Mendelian monohybrid cross, where both parents are heterozygous for a trait.
The difference between gross primary productivity and net primary productiviry is determined by the respiration of primary producers.
a. True
b. False
Answer:
True
Explanation:
Gross primary productivity (GPP) of a system is defined as the rate by which the primary producers trap the solar radiation during the process of photosynthesis. The gross primary productivity is represented as total photosynthesis per unit area in a given time.
The net primary productivity (NPP) refers to the energy available to the plants for their growth after cellular respiration and is represented as plant growth per unit area in a given time.
Plants use part of the gross primary productivity in cellular respiration to produce the energy required to perform other vital functions. Hence, the energy consumed in cellular respiration differentiates the gross primary productivity and the net primary productivity.
NPP = GPP - Plant respiration
Net biomass productivity is the difference between gross productivity and respiration. It is influenced by factors like climate and soil nutrients. Forests tend to have high net biomass productivity while deserts have low productivity.
Explanation:Net biomass productivity is the difference between gross productivity (production of plant material by photosynthesis) and respiration. Net productivity is closely related to a number of environmental factors like climate, soils, and available nutrients. Net biomass production will be highest where there is an ample supply of moisture to meet the needs of plants. Biomass productivity is also high where soils are rich in nutrients and have a positive soil moisture balance. Forest, especially tropical forest have high net biomass productivity while deserts have low productivity.
The statement that the difference between gross primary productivity and net primary productivity is determined by the respiration of primary producers is true. Gross primary productivity (GPP) is the rate at which photosynthetic primary producers, such as plants, incorporate energy from the sun. However, not all of this energy remains within the plants because they must use a portion of it for their own metabolic processes, primarily through cellular respiration. The energy that remains after accounting for this respiration is known as the net primary productivity (NPP). It is the energy that is actually available to primary consumers in an ecosystem. The NPP can be influenced by various environmental factors, such as climate, availability of nutrients, and moisture levels, leading to variations in the productivity of different ecosystems, like tropical forests having high NPP while deserts have relatively low NPP.
Which group is incorrectly paired with its description?
a. diatoms—important producers in aquatic communities
b. red algae—eukaryotes that acquired plastids by secondary endosymbiosis
c. apicomplexans—unicellular parasites with intricate life cycles
d. diplomonads—unicellular eukaryotes with modified mitochondria
Answer:
b. red algae—eukaryotes that acquired plastids by secondary endosymbiosis
Explanation:
Red algae did not acquire plastids by secondary endosymbiosis and thus, option b is an incorrect pair. Red algae are eukaryotes that are red in the color not because they acquired plastids y endosymbiosis but because they produce the enzyme phycoerythrin and phycocyanin which masks the other pigments present in it.Secondary endosymbiosis is a process by which the product of primary endosymbiosis is engulfed by another organism. Such type of process never occurred in case of red algae.What is ATP's importance in the cell? ATP stores energy in carbonyl groups. When a carbonyl group is removed, energy is released to be used in cellular processes. ATP contains a long hydrocarbon tail and is important in storing energy. ATP is an important component of cell membranes because it is nonpolar and hydrophobic. ATP can add phosphate groups, thereby releasing energy that can be used in cellular processes. ATP stores the potential to react with water, thereby removi
ATP is essential in a cell as it stores and transfers energy required for various cellular activities. Its energy is stored in high-energy phosphate bonds, which are broken to facilitate energy release. It does not contribute to the cell membrane structure due to its polar, hydrophilic characteristics.
Explanation:ATP, or Adenosine Triphosphate, plays a crucial role in the cell as it acts as a medium for energy storage and transfer. It is the main source of energy for most cellular processes. ATP doesn't store energy in carbonyl groups, rather it stores energy in the high-energy phosphate bonds. These bonds are broken to release the energy needed for various cellular activities.
ATP is not an important component of cell membranes, nor does it have a long hydrocarbon tail. It doesn't contribute to the structure of the cell membrane because it is polar and hydrophilic, unlike the major constituents of the cell membrane - phospholipids, which are both hydrophobic and hydrophilic.
The process of ATP releasing energy involves hydrolysis where a phosphate group is removed, allowing the stored energy to be used for various metabolic and physiological processes within the cell. It doesn't add phosphate groups but rather loses them in the process of energy release.
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ATP is the primary energy currency of the cell, storing energy in high-energy phosphate bonds and releasing it to power cellular processes. Cells produce ATP through cellular respiration, and it is used in various essential activities like metabolism and muscle contraction. ATP is continually recycled within the cell to maintain energy balance.
The Importance of ATP in Cellular Processes
Adenosine triphosphate (ATP) is often described as the energy currency of the cell. It is a molecule that stores and transports chemical energy within cells. ATP's energy is stored in its phosphate ester bonds, particularly the two high-energy phosphate anhydride linkages. When one of these bonds is broken, usually through the removal of a terminal phosphate group, energy is released. This energy is utilized in a variety of cellular processes, including metabolism, active transport, muscle contraction, and cell division.
Cells produce ATP during cellular respiration, a process which involves the oxidation of carbohydrates, proteins, and fats. The oxidative reactions that produce ATP occur in the cytoplasm and mitochondria of the cell. Once produced, ATP can be used immediately or stored for short periods until the energy is required for cellular reactions or for performing mechanical work, such as muscle contraction.
Phosphorylation is the process by which ATP transfers a phosphate group to another molecule. This process requires energy and is a key mechanism by which cells convert ATP's stored energy into a form that can do biological work, powering a vast array of endergonic reactions in the cell. ATP is continually recycled within the cell; it is constantly being made from ADP and inorganic phosphate and used up in cellular activities, maintaining the energy homeostasis of the cell.
Amino acids differ from one another by the ________________ that is present
a. amino group
b. carboxyl group
c. R group
d. R plasmid
e. core group
Amino acids, fundamental in biological processes, possess an amino group, a carboxyl group, a hydrogen atom, and a side chain or R group. Amino acids differentiate from each other based on their unique R group. This distinctive R group gives each amino acid its singular characteristics.
Explanation:Amino acids are organic compounds that play a crucial role in biological processes. They are composed of an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a side chain (R group). Each amino acid differs from another by its unique R group. So, the answer to your question is c. R group.
The R group can vary among different amino acids and this variation is what gives each amino acid its unique properties. For example, the R group can be as simple as a single hydrogen atom (as in the case of the amino acid glycine) or a more complex arrangement of atoms (as in the case of tryptophan).
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Amino acids differ from each other by their respective R groups, which represent unique side chains. These side chains could range from simple hydrogen atoms to complex multi-ring structures.
Explanation:Amino acids differ from one another by the R group that is present. Every amino acid has three main components: a central carbon atom, an amino group (-NH2), and a carboxyl group (-COOH). However, each also has a unique side chain, or R group, that sets it apart from the others.
This R group could be as simple as a single hydrogen atom or as complex as a multi-ring structure. It's due to these distinct R groups that we have 20 different common amino acids in proteins.
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What is the purpose of stem cell research?
Answer:
Stem cells may be defined as the potipotent cells that has the ability to differentiate into different cells. The stem cells are now used in research as well.
The main purpose of the stem cells are as follows:
The damaged tissue or organ can be replaced by growing stem cells in the laboratory.
The genetic defects can be corrected by the stem cells discovery.
The new drugs can be tested on the stem cell.
The reason of the development of specific cancer cells.
Hypothesize how over-washing of hands can affect the population of ""good"" bacteria that resides on human skin.
Answer:
The human body inhabits various microbes, which coexist and does not cause any kind of harm, these bacteria are considered as good bacteria and are known as microbiota. Even the skin comprises various categories of bacteria, that is, more than 500 species. These microbes demonstrate skin microbiota, that is, an effective system of safeguarding, which takes place in immune defenses.
The beauty of the skin is nearly associated with the balance of the microbes, which populate it, these are considered as the first line of defense from outside threats. When the bacterial ecosystem is differentiated and balance, the skin stays healthy.
It has been hypothesized that over-washing of hands diminishes the population of good bacteria from the skin. As it is considered that over-washing extricates the majority of the beneficial bacterial cells from the human skin. Hence, may result in the loss of natural microflora from the section of the skin.
Bacteria are single-celled prokaryotes that can be beneficial and harmful to organisms. Overwashing hands can harm the good bacteria on the skin surface.
What are good bacteria?The body of the organism is full of bacteria and microflora that can be beneficial to the organism. Good bacteria are helpful to the organism as they help in maintaining health.
The skin surface of the organism is home to various microbiota that are good and helpful. They help in safeguarding and protecting the skin cells and tissues. They are part of the first line of defence as they help to regulate the immune check.
When the skin surface is overwashed then the population of the good bacteria gets affected and their habitation gets destroyed. Overwashing can result in the loss of the natural and good microflora of the skin surface.
Therefore, overwashing can affect the population of "good" bacteria.
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Photosystem II in photosynthesis, obtains its electrons from
a. Water
b. Oxygen
c. Carbon Dioxide
d. NADPH
Answer:
a. Water
Explanation:
An arrangement of chlorophyll and other pigments in the chloroplast is known as photosystems. In the photosynthetic eukaryotes, there are two photosystems, Photosystem II and Photosystem I. Photosystem I uses chlorophyll a, with reaction centre as P700. Photosystem II uses chlorophyll a with reaction centre P680.
During light reaction or photochemical reaction of photosynthesis, the light energy causes the removal of an electron from chlorophyll a reaction centre of Photosystem II P680. After losing an electron the P680 is deficient of an electron. P680 takes an electron from photolysis of a water molecule into eletron, H⁺ ions and O⁻² ions.
Which of the following is not true of RNA processing?
a. Exons are cut out before mRNA leaves the nucleus.
b. Nucleotides may be added at both ends of the RNA.
c. Ribozymes may function in RNA splicing.
d. RNA splicing can be catalyzed by spliceosomes.
The correct answer is option d, as RNA splicing is indeed catalyzed by spliceosomes.
Explanation:d. RNA splicing can be catalyzed by spliceosomes.
In RNA processing, the process of removing introns and joining exons together is called RNA splicing. This process is facilitated by the presence of ribozymes, which are RNA molecules that catalyze chemical reactions. One type of ribozyme involved in RNA splicing is the spliceosome, a large and complex assembly of proteins and RNA.
Therefore, the correct answer is option d, as RNA splicing is indeed catalyzed by spliceosomes.
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The incorrect statement about RNA processing is that 'Exons are cut out before mRNA leaves the nucleus.', as it is actually the introns that are removed, not the exons. The correct answer is a.
The student has asked which statement is not true about RNA processing. The correct answer to this question is 'a. Exons are cut out before mRNA leaves the nucleus.' In fact, during RNA processing, it is the introns that are cut out while the exons are joined together. This process, known as splicing, is essential for converting the pre-mRNA transcript into a mature mRNA molecule that can be translated into protein.
The modifications that occur during RNA processing in eukaryotic cells include the addition of a 5' cap and a poly-A tail to the pre-mRNA, and the removal of introns by splicing which can be catalyzed by spliceosomes or ribozymes. Different splicing patterns can lead to the production of various mRNA transcripts from the same DNA sequence, a phenomenon known as alternative splicing.
Therefore, the statement that 'Exons are cut out...' is not correct since exons are the sections that remain in mRNA after the introns have been removed. RNA processing includes the removal of introns, addition of nucleotides at both ends, and splicing by spliceosomes or ribozymes.
If you came into contact with Vibrio, what defense do you have to keep it from even getting to your intestines?
A. Your skin is an effective barrier against bacteria you come in contact with.
B. Antimicrobial enzymes in saliva
C. Acid in the stomach
D. Mucous membranes in the digestive and respiratory tract
E. All of the above
Answer:
The correct answer is E. All of the above
Explanation:
The body contains different physical and chemical barriers to protect itself from the invasion of any pathogen. These barriers act as the first line of defense to the body and include skin, mucous membrane, acids, and antimicrobial enzymes.
Skin acts as a mechanical barrier to microbes and does not allow any pathogen to invade through it until it gets punctured.
Saliva contains antimicrobial enzymes like lysozyme which kills bacteria by destroying their cell wall. Hydrochloric acid released in the stomach is highly efficient to kill pathogens coming with food.
Mucous membranes in the digestive and respiratory tract contain WBCs, antimicrobial enzymes and proteins which kills the invaded pathogen like Vibrio.
In guinea pigs coat color is determined by a single gene with two alleles. A guinea pig from a true-breeding black strain is mated with a guinea pig from a true-breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated with each other. What proportion of the black F2 progeny is expected to be homozygous?
Answer:
1/3 of black F2 progeny will be homozygous.
Explanation:
Here, BB = black coat color = true breeding dominant black strain
bb = white coat color = true breeding recessive white strain
When true breeding black and white guinea pigs are mated to give F1 progeny:
BB X bb = Bb ( all black guinea pigs )
When two of the F1 guinea pigs are mated to give F2 progeny:
Bb X Bb = BB, Bb, Bb, bb
F2 progeny has 3/4 progeny as black (BB and Bb) . Out of them 1/3 are homozygous and 2/3 are heterozygous.
Hence, 1/3 of black F2 progeny will be homozygous.
In the F2 generation of guinea pigs, where the first generation was all black from a true-breeding black and white strain, 25% of the black progeny is expected to be homozygous.
Explanation:In guinea pigs, coat color is determined by a single gene with two alleles. If a guinea pig from a true-breeding black strain is mated with a guinea pig from a true-breeding white strain, the F1 progeny would be all black, indicating that black is the dominant color. If two of these F1 progeny are mated together, their offspring, or the F2 generation, would exhibit a phenotypic ratio of 3 dominant (black):1 recessive (white) according to Mendelian genetics. Therefore, in this case, about 3 out of 4, or 75%, of the black F2 progeny would be expected to be heterozygous (Bb), and 1 out of 4, or 25%, would be homozygous (BB). Thus, the proportion of the black F2 progeny expected to be homozygous is 25%.
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Construct a table that organizes the following terms, and label the columns and rows
Monosaccharides Polypeptides Phosphodiester linkages
Fatty acids Triacylglycerols Pepptide bonds
Amino acids Polynucleotides Glycosidic linkages
Nucleotides Polysaccharides Ester linkages
Answer:
Let's organize this with the four biomolechules:
-NUCLEIC ACID (Nucleotides, Polynucleotides, Phosphodiester linkages)
-LIPIDS (Fatty acids, Triacylglycerols, Ester linkages)
-PROTEINS (Polypeptides, Pepptide bonds, Aminoacids)
-CARBOHYDRATES (Monosaccharides, Polysaccharides, Glycosidic linkages)
Explanation:
Answer: Answers are detail in the attachment
Explanation:
Which of the following mutations would be most likely to have a harmful effect on an organism?
a. a deletion of three nucleotides near the middle of a gene
b. a single nucleotide deletion in the middle of an intron
c. a single nucleotide deletion near the end of the coding sequence
d. a single nucleotide insertion downstream of, and close to, the start of the coding sequence
Answer:
d. a single nucleotide insertion downstream of, and close to, the start of the coding sequence.
Explanation:
A single nucleotide insertion downstream and close to the start of coding sequence will produce the most harmful effect among all these given options because it will cause the entire reading frame to shift in the beginning itself leading to insertion of wrong amino acids in the entire polypeptide chain, hence the entire protein will have extremely different amino acids than the original protein was supposed to have and such protein will be dysfunctional.
In option (a), it is given that the three nucleotides near the middle of the gene will be deleted. This scenario will produce comparatively less harmful effect because in this case only one amino acid which is coded by this deleted triplet codon will be absent in the polypeptide.
In option (b), a single nucleotide deletion has occurred that too in an intron. It will produce no harmful effect in case of eukaryotes because introns are ultimately removed before translation so the amino acids in the polypeptide will remain the same. In case of prokaryotes such deletion may cause some harmful effect in case of mRNA only because introns in prokaryotic mRNA are not removed. But, if the mRNA is for prokaryotic tRNA and rRNA then there will be no harmful effect because they undergo processing in which introns are removed.
In option (c) also there will be very less harmful effect because deletion is occurring near the end of coding sequence which will lead to insertion of wrong amino acid in the polypeptide at the end of polypeptide chain that means very less abnormal amino acids will be present so effect will be less harmful.
Answer:
D. a single nucleotide insertion downstream of, and close to, the start of the coding sequence.
Explanation:
Just wanted to confirm. The explanation above me is correct.
Dominance hierarchies
a. are found only in monogamous and solitary primates
b. likely give dominant males and females a reproductive advantage
c. are permanent and inflexible
d. do not result in dominant females having priority access to food
Answer:
The correct answer is option B. likely give dominant males and females a reproductive advantage .
Explanation:
Dominance hierarchy is a social hierarchy in which members of a group of animals interact with each other to form a ranking system or superiority and inferiority of the organism. In such social groups, members have to compete for access for the food and resources which are limited.
Mating is also decided on the basis of the dominance of males and females in such social groups. Dominant male-female have always advantage for mating in comparison to lower members in the ranking.
For instance, female baboon mated to a dominant male or high ranked male so she can get high rank as well for the time of period.
Thus, the correct answer is option B. likely give dominant males and females a reproductive advantage .
Removal of one of a pair of endocrine glands often results in atrophy of the other gland.
a. True
b. False
Answer:
False
Explanation:
When one of the glands is removed, the other continues to perform its function because they are independent entities that can continue to secrete hormones. This happens with the adrenal glands for example.
The structural level of a protein least affected by a disruption in hydrogen bonding is the
a. primary level.
b. secondary level.
c. starch d. carbohydrate
c. tertiary level.
d. quaternary level.
Answer: a. Primary level.
Explanation:
The primary level is characterized by the sequence of aminoacids of the protein, aminoacids have a carboxyl and an amino group bonded to a carbon atom, and are held together by peptide bonds between the carboxyl end of an aminoacid and the amino end of the next one, making an aminoacid chain.
Secondary, tertiary and quaternary levels determine the shape and thus the function of the protein according to different forces, hydrogen bonds take part in it bending the chain in different ways.
The primary structure of a protein is least affected by a disruption in hydrogen bonding because it is maintained by covalent peptide bonds, which are not broken under conditions that disrupt weaker hydrogen bonds.
Explanation:The structural level of a protein least affected by a disruption in hydrogen bonding is the primary structure. This level of structure refers to the sequence of amino acids in the polypeptide chain, held together by peptide bonds, which are much stronger than hydrogen bonds. Despite the breaking of hydrogen bonds during denaturation, which will disrupt secondary, tertiary, and quaternary structures, the primary structure remains intact because the peptide bonds are not broken under conditions that disrupt hydrogen bonds. Therefore, the answer is a. the primary level.
The secondary structure is defined by the spatial arrangement of the polypeptide chain's backbone, primarily through hydrogen bonding, forming structures like α-helices and β-pleated sheets. Interruption of hydrogen bonding in this structure affects the correct folding and thus its stability and function.
Tertiary structure is the three-dimensional shape of a single polypeptide chain, stabilized by various interactions, including hydrogen bonds. Quaternary structure involves the interaction and assembly of multiple polypeptide subunits, which is also dependent on hydrogen bonds among other interactions for stability.
While all secondary, tertiary, and quaternary structures can be disrupted by the loss of hydrogen bonding, it is specifically the primary structure that will persist because its stability is owed to covalent peptide bonds.
Type I alveolar cells:
a. secrete surfactant.
b. are squamous epithelial cells.
c. are macrophages.
d. function mainly in the diffusion of gases.
e. Both B and D.
Answer:
The correct option is: e. Both B and D
Explanation:
Type I alveolar cells are the thin squamous epithelial cells that lines the alveolar surface of the pulmonary alveoli.
These cells are involved in the gas exchange process. The thin lining of the type I alveolar cells, allows the rapid diffusion of gas and thus enabling the exchange of gases between the blood present in the capillaries and the air present in the pulmonary alveoli.
Which of the following statements about DNA isolation from E. coli is not correct?
a. chemical extraction using phenol removes proteins from the DNA
b. RNA is removed from the sample by RNase treatment
c. detergent is used to break apart plant cells to extract DNA
d. lysozyme digest peptidoglycan in the bacterial cell wall
e. centrifugation separates cellular components based on size
Answer:
The correct answer is c. detergent is used to break apart plant cells to extract DNA.
Explanation:
Here we are talking about DNA isolation in E.coli and detergent is used to break apart the bacterial cell membrane by destabilizing the cell membrane. Detergent disintegrates the proteins present in the cell membrane which causes the rupture of membrane releasing the cell extract outside the cell.
Detergent is not able to break plant cell because it is made up of cellulose and require blending or mashing. Detergent can only be used to disintegrate the cell membrane because only cell membranes contain lipids and proteins.
Hence, the correct answer is c. detergent is used to break apart plant cells to extract DNA.
In organic soil with plenty of oxygen, organic notrogen will be converted to NH4+ by decomposens the NH4+ will be converted to NO3- by bacteria.
a. True
b. False
Answer:
False NH₄+ is transformed first to NO₂ and then to NO₃
Explanation:
Organic nitrogen which appears in different forms like urine, dead animals or plants, Nitrogen will be converted to ammonium (NH₄+) by either fungi or bacteria, this process is called ammonification.Nitrification is the process through which ammonium (NH₄+) is transformed to nitrites (NO₂) and then into nitrates (NO₃). Each transformation is performed by different types of bacteria species.Glucose travels in the plasma or liquid part of
your blood butcholesterol (triglycerides) travel bound to protein
carriers in theblood. Based on their respective chemical
properties, explain thereason for this difference.
Answer:
Explanation:
Glucose is a sugar, highly soluble in water because it has lots of polar hydroxyl groups (-OH) which can form hydrogen-bonds with water molecules. These types of bonds are intermolecular forces which are present in other macromolecules like the DNA or proteins.
Cholesterol and triglycerids are lipids. These have long chains of hydrocarbons, which are non polar and therefore insoluble in water.
Blood is made of a solid and a liquid part. The solid part contains cells. The liquid part (plasma), is made of water, salts, and proteins. Glucose and cholesterol/triglycerids travel in the liquid part of the blood, in an aqueous environment. Because the lipids are not soluble, they need to travel with a carrier protein that keeps them protected from the surrounding water.
If patient suffered from stroke just an hour ago, what testing can be done to assess brain function? What categories of stroke are recognized? What treatment approaches can be used to minimize the functional loss of brain?
Answer:
In the given case, the FAST or face, arms, speech, and time test can be done to evaluate the function of the brain. In supplementation, imaging tests, and blood tests can also be performed. In the given case, ischemic stroke is determined.
For the treatment of ischemic strokes, a tissue plasminogen activator also known as tPA is administered intravenously in the arm. The tPA functions by dissolving the clot and bettering the flow of blood to the section of the brain, which is being deprived of blood flow.
Apart from this, the antiplatelet agents like aspirin and anticoagulants like warfarin can be provided to the patient. In addition, the antihypertensives can also be administered on the basis of patient's condition.
Imagine you have two pure-breeding lines of canaries, one with yellow feathers and the other with brown feathers. In crosses between these two strains, yellow female x brown male gives only brown sons and daughters, while brown female x yellow male gives only brown sons and yellow daughters. Propose a hypothesis to explain these results.
Answer:
The results can be explained if the gene that determines color of feathers is located in the Z chromosome.
Explanation:
Unlike mammals that have X and Y sex chromosomes, birds have Z and W sex chromosomes. Females are ZW and males are ZZ.
The gene that determines color of feathers has the alleles:
B_ = brown color
bb = yellow color.
1st crossyellow female x brown male
♀️ [tex]Z^bW[/tex] x ♂️ [tex]Z^BZ^B[/tex]Gametes
Female: [tex]Z^b[/tex] , [tex]W[/tex]
Male: [tex]Z^B[/tex]
F1 [tex]Z^BZ^b[/tex] = brown males[tex]Z^BW[/tex] = brown females2nd crossbrown female x yellow male
♀️ [tex]Z^BW[/tex] x ♂️ [tex]Z^bZ^b[/tex]Gametes
Female: [tex]Z^B[/tex] , [tex]W[/tex]
Male: [tex]Z^b[/tex]
F1 [tex]Z^BZ^b[/tex] = brown males[tex]Z^bW[/tex] = yellow femalesThe results of crossing pure-breeding yellow and brown canaries suggest a sex-linked inheritance of feather color, with brown being the dominant allele. A hypothesis could be that the gene for feather color is on the X chromosome, resulting in sex-specific phenotypes among the offspring. This follows Mendelian inheritance principles, as demonstrated in monohybrid crosses with peas resulting in predictable phenotypic ratios.
To explain the results of crossing the two pure-breeding lines of canaries, one could hypothesize that the feather color is sex-linked and that brown is a dominant allele over yellow. The fact that all crosses result in brown offspring suggests that the allele for brown feathers is dominant. Since the crosses from yellow female canaries to brown male canaries results solely in brown offspring, but the cross from brown female canaries to yellow males results in brown sons and yellow daughters, it can be hypothesized that the gene controlling feather color is located on the sex chromosomes, possibly the X chromosome considering the sex-specific inheritance pattern. In this case, the brown allele is dominant, which is denoted by 'B', whereas the yellow allele is recessive, denoted by 'b'.
Monohybrid cross analyses using a Punnett square help determine probable offspring genotypes and phenotypes. When two true-breeding parents are crossed, such as YY and yy in pea plants, all F1 offspring are Yy showing the dominant phenotype. Similarly, in canaries, the same principles apply where the feather color gene could follow dominant or recessive patterns.
Using Mendelian genetics and large numbers of crosses, genotypic and phenotypic ratios can be calculated to predict other genetic crossing outcomes. For example, when F1 offspring with genotypes Yy are crossed among themselves, there is a 3 in 4 probability of offspring with a yellow phenotype due to the dominant yellow allele. This explains the common 3:1 phenotypic ratio seen in Mendel's F2 generation of pea plants. Drawing parallels, similar principles could be attributed to the canary crossing results mentioned in the question, suggesting that the inheritance of feather color follows a Mendelian pattern of inheritance assuming complete dominance and sex linkage.
Briefly describe Wernicke and Broca aphasia. What is a main difference of these two classic types of aphasia and can be useful in the process of clinical differentiation. What symptoms can help to identify the localization of damage?
Answer:
Broca's area refers to a motor speech region, which assists in movements needed at the time of the production of speech. When an injury takes place in the frontal parts of the left hemisphere, it can give rise to various kinds of language issues. This section of the brain plays an essential role in putting words together to produce complete sentences. The injury to the left hemisphere is known as Broca's aphasia, also known as expressive aphasia.
Wernicke's aphasia also called receptive aphasia, posterior aphasia, or sensory aphasia is a kind of aphasia in which individual encounters with difficulty in understanding spoken and written language. Thus, the two kinds of aphasia are expressive aphasia in which one faces difficulty in conveying thoughts via writing and speech. The other is receptive aphasia, in which one finds difficulty in understanding the written or spoken language.
By analyzing the symptoms that whether the patient exhibits difficulty in understanding speech and using accurate words to express thoughts or the movements that are needed to generate speech, one can find the site of destruction.
The functioning of enhancers is an example of
a. a eukaryotic equivalent of prokaryotic promoter functioning.
b. transcriptional control of gene expression.
c. the stimulation of translation by initiation factors.
d. post-translational control that activates certain proteins.
Answer:
b. transcriptional control of gene expression.
Explanation:
The binding sites of transcription factors (enhancers and silencers) can activate or deactivate, respectively, a gene in specific parts of the body.
Some genes must be expressed in a part of the body. For example, if you need to activate a gene in the spine, skull and toes, but not in the rest of the body, how can transcription factors do this task?
As a gene of this type of pattern it can have several enhancers or silencers, where each one can activate or repress the gene of a specific type of cell or part of the body, joining transcription factors that occur in that specific part of the body.
The functioning of enhancers exemplifies transcriptional control of gene expression in eukaryotes, where they augment the transcription rate by enabling interactions between transcription factors and the transcription machinery at the promoter.
Explanation:The functioning of enhancers is an example of transcriptional control of gene expression. Enhancers are regions in eukaryotic DNA that can be located quite far from the genes they regulate. They contain specific DNA sequences where transcription factors can bind, thus influencing the transcription of a gene. These enhancers can increase the transcription rate of associated genes by facilitating the interaction between transcription factors and the general transcription machinery, including RNA polymerase at the promoter site. This process involves the folding of DNA to bring enhancers into proximity with promoters, making it possible for the transcription factors and the transcription initiation complex to interact effectively. Unlike prokaryotes which rely heavily on the interaction of activators and repressors with promoters, enhancers provide an additional layer of transcriptional regulation in eukaryotes. The concept of enhancers expands the understanding of intricate gene regulation mechanisms present in eukaryotic cells beyond what is seen in prokaryotic systems.
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Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of ions on opposite sides of the membrane?
a. intracellular receptor
b. G protein-coupled receptor
c. phosphorylated receptor tyrosine kinase dimer
d. ligand-gated ion channel
Answer:
d. Ligand-gated ion channels
Explanation:
a.Intracellular receptor: it's activated through second messengers since its activation and actions happen only inside the cell there's no distribution of ions on opposite sides of the membrane.
b.G-protein-coupled receptor: these receptors bind specific ligands at the cell surface (hormones and neurotransmitters) to relay the signal across the membrane. This means they act as a messager, there's no distribution of ions.
c.Phosphorylated receptor tyrosine kinase dimer: works as a messenger as well, it propagates a signal through the plasma membrane when it allows tyrosine to be trans phosphorylated.
d. Ligand-gated ion channels open to allow ions such as Na+, K+, Ca2+ or Cl- in response to the binding of a chemical messenger, one example of this kind of channel in action is when a neurotransmitter binds to a receptor located on the postsynaptic neuron, opening ion channels, this lead to a flow of ions across the cell membrane that turns into a depolarization or a hyperpolarization by changing the distribution of ions on opposite sides of the membrane.
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Final answer:
The answer is d) ligand-gated ion channel. This receptor, upon binding to a signaling molecule, allows the passage of ions through a membrane pore, altering ion distribution.
Explanation:
The binding of a signaling molecule to a d) ligand-gated ion channel leads directly to a change in the distribution of ions on opposite sides of the membrane. This type of cell-surface receptor works by having an extensive membrane-spanning region, and upon binding to a ligand, it undergoes a conformational change that allows specific ions to pass through the pore. Ligand-gated ion channels facilitate the almost instantaneous passage of millions of ions, which drastically change the interior environment of the cell.
What is a biogeographic realm? List eight of them. How do biogeographic realms indicate relatedness in terms of biodiversity at the global level?
Answer:
Biogeographic realms refer to the huge spaces or regions on the surface of the Earth within which the development of ecosystems takes place. The animal and plant species witnessed in a biogeographic realm share identical features all the way through.
The eight biogeographic realms are the afrotropical realm, Indomalayan realm, Antarctica realm, Neotropical realm, Nearctic realm, Palaearctic realm, Oceanian realm, and Australian realm.
The biodiversity signifies the variability between the living species in every possible manner. Each of the biogeographic realm exhibiting identical environmental conditions or identical mediators of modification has been found to comprise an identical kind of biota.
However, diversity can be found within it too. For example, the tropical moist rainforest found in any section of the globe will share identical procedures and vegetation, however, its specific structure will vary on the basis of the biogeographic realm in which it is witnessed.
Analogous adaptation to the similar kind of environmental conditions between the distinct species of the distinct biogeographic realm can be witnessed at a global level.
A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Remember that extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?
Answer:
1/2 or 50%
Explanation:
Given that the extra digit is a dominant trait, the man with an extra digit may be homozygous or heterozygous dominant for the trait.
However, having a daughter with his normal wife makes him heterozygous dominant. Let's assume that the genotype of the heterozygous dominant male is "Dd" and that of his wife is "dd".
The heterozygous dominant male and homozygous recessive female would have progeny in the following ratio= 1/2 Dd (child with extra digit): 1/2 dd (child with no extra digit).
The fraction of this couple's children expected to have extra digits is 1 (all of the children) out of 1 (total number of children), which is equivalent to 1/1 or simply 1.
The man in this scenario has six fingers on each hand and six toes on each foot, a condition caused by a dominant gene. His wife and their daughter have the normal number of digits. Since having extra digits is a dominant trait, the man must be homozygous dominant (DD) for the gene responsible for the extra digits.
When the couple has children, all of their offspring will inherit one dominant allele from the father, resulting in the expression of the trait. Therefore, the fraction of their children expected to have extra digits is 1, meaning that all of their children will have the additional digits.
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Where is sperm first produced? Where do the sperm cells mature?
Sperm are first produced in the seminiferous tubules inside the testes. They mature in the epididymis where they acquire motility and the ability to fertilize. They are released during ejaculation, travelling through the ductus deferens and exiting via the urethra.
Explanation:Sperm are first produced in the seminiferous tubules that are coiled inside the testes. The least mature cells, the spermatogonia, dwell along the basement membrane inside the tubule and undergo division to produce primary and secondary spermatocytes, and then spermatids. These spermatids finally mature into spermatozoa, or formed sperm, in a process known as spermatogenesis.
The sperm cells mature further in the epididymis, where they gain the ability to move and fertilize an oocyte. The mature sperm are stored in the epididymis until ejaculation. During ejaculation, sperm exit the epididymis through the ductus deferens, where secretions from the prostate and bulbourethral glands help form semen, assisting the sperm in their journey through the urethra and into the female reproductive tract.
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Describe the steps in translation that require complementary base pairing.
Answer:
Translation is the synthesis of protein synthesis from RNA where genetic information encoded in mRNA, which is then translated into a sequence of amino acids present in a polypeptide chain. These are the steps in translocation:
1) Ribosomes bind to mRNA present in the cytoplasm and move with the molecule in a 5'-3' direction until it reaches a start codon AUG.
2) Anticodon present on tRNA molecule binds with appropriate opposite codon according to complementary base pairing. (translocation depends on complementary base pairing, e.g. AUG=UAC between codons present on mRNA and anticodons on tRNA)
3) Each tRNA molecule has specific amino acids, according to the genetic code.
4) Ribosomes catalyze the formation of peptides bonds with the help of condensation reaction between adjacent amino acids.
5) Ribosomes move with the mRNA molecule which is synthesizing a polypeptide chain until it reaches a stop codon.
6) At this stage, translation comes to stop and the polypeptide chain is released.