Answer:
Eukaryotic transcription refers to an elaborate procedure, which is used by the eukaryotic cells to duplicate the genetic information present in DNA into the units of RNA replica. The transcription of the gene takes place in both prokaryotic and eukaryotic cells. The expression of eukaryotic genes is mainly monitored at the level of the beginning of transcription, though in certain cases, transcription may be monitored and attenuated at subsequent steps.
Like in bacteria, the transcription in eukaryotic cells is monitored by the proteins, which combines with the unique regulatory sequences and modulate the activity of RNA polymerase. The intricate task of monitoring the expression of gene in various kinds of cells of multicellular species is achieved mainly by the combined activities of various different transcriptional regulatory proteins.
In supplementation, the packaging of DNA into chromatin and its variation by methylation produce further levels of complexity to the control of eukaryotic gene expression. The transcription in bacteria is monitored by the combination of proteins to the cis-acting sequences, which regulate the transcription of adjacent genes.
The same kind of cis-acting sequence monitors the expression of eukaryotic genes. These sequences have been determined in mammalian cells mainly by the application of gene transfer assays to examine the activity of suspected regulatory regions of cloned genes.
A mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with this plant by manually opening and pollinating the flowers with pollen from another plant. When closed x open crosses were done, all the F1 progeny were open. The F2 plants were 145 open and 59 closed. A cross of closed x F1 gave 81 open and 77 closed. How is the closed trait inherited? What evidence led you to your conclusion?
Answer:
This trait is inherited according to Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.
Explanation:
First, there are closed-flowers plants and open-flowers plants. A cross between them produces only open flowers at F1. With this information, we can say that open condition is dominant over closed, and probably, parental plants were homozygous, like this:
- Open flowers: FF
- Closed flowers: ff
- F1 open flowers: Ff
When F1 was crossed, the progeny was:
- 145 open
- 59 closed
When we cross two heterozygous, we obtain a phenotypical radio of 3:1, like this:
- (1) FF Open
- (2) Ff Open
- (1) ff Closed
This ratio is very similar to the F2 above reported (145 open and 59 closed).
When F1 (Ff) plants were crossed with closed plants the progeny was:
- 81 open
- 77 closed
This is alike to a ratio 1:1.
In the Punnett Square, it is possible to notice that in this cross it is expected a ratio of 1:1 closed/open.
In conclusion, this trait is inherited according to Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.
Describe the structure and function of haemoglobin (Hb). You should include a description of both tertiary and quaternary structures of the protein and the location and function of non-protein components.
Answer:
Heamoglobin is the structure present in red blood cells which is involved in carrying of oxygen from the blood to all parts of the body.
Explanation:
Tertiary Structure : A molecule of heamoglobin consists of four groups of heme and a group of globulin.A molecule of heamoglobin contains four iron atoms to which oxygen molecules get attached. As there are four iron atoms in a molecule of heamoglobin so a single heamoglobin molecule can carry four oxygen atoms.
Quaternary structure: Multi sub unit globular proteins combine to make the Quaternary structure of heamoglobin. Alpha helices are made by the amino acids present in heamoglobin. Non helical segments, short in length, connect the alpha helices. The helical segments are stabilized by hydrogen bonds which results in folding of the polypeptide into the specific Quaternary shape.
Function:
1.The basic function of heamoglobin is that it transfers oxygen from the lungs to different parts of the body.
2. CO2 ions and hydrogen ions are also transported by the heamoglobin.
A researcher wanted to study the effect of low light environments on the growth of alfalfa plants, so she grew six plants and measured their heights after two weeks. She calculated a sample mean of 5.35 cm and a sample standard deviation of 0.90 cm. If the heights of the first five plants in her sample were 6.0 cm, 5.9 cm, 4.0 cm, 5.5 cm, and 6.2 cm, what is the height of the other plant?
Answer:
Height of the other plant is [tex]4.5[/tex] cm
Explanation:
- Given -
Mean of the sample set of plants [tex]= 5.35[/tex]cm
Mean of a given set of data is equal to the sum of integral values of each data divided by the number of such data points
[tex]M = \frac{x}{n}[/tex]
Where M signifies the Mean values
X signifies the sum of all unit data with in a sample
and n signifies total number of data units with in a sample
On substituting the given values in above equation, we get -
[tex]\frac{6.0+5.9+4.0+5.5+6.2+A}{6} = 5.35\\5.35 * 6= 27.6 + A\\A = 32.1-27.6\\A=4.5[/tex]
Hence, height of the other plant is [tex]4.5[/tex] cm
The height of the sixth alfalfa plant is 4.5 cm, calculated by subtracting the total of the first five plant heights from the overall sum required to achieve the sample mean.
The student requires the calculation of the height of the sixth alfalfa plant given the sample mean, standard deviation, and the heights of the first five plants. To find the height of the sixth plant, we will use the formula for the sample mean: mean = (sum of all values) / number of values. Given that the sample mean is 5.35 cm and there are six plants, the sum of all the plant heights is 5.35 cm * 6 = 32.1 cm. The sum of the heights of the first five plants is 6.0 cm + 5.9 cm + 4.0 cm + 5.5 cm + 6.2 cm = 27.6 cm. Subtracting this from the total sum we get the height of the sixth plant: 32.1 cm - 27.6 cm = 4.5 cm.
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.
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.The mitochondrial inner membrane form a series of infoldings known as cristae to:
a. Increase surface area
b. Decrease surface area
c. Increase production of H+
d. Decrease production of H+
Answer: Increase the surface area.
Explanation:
The mitochondria is an important cell organelle that is found in both the plant cell and animal cell.
It basically aids in the production of ATP, hence also known as the power house of the cell.
The mitochondria is a double membrane system in which the inner membrane of mitochondria helps in increasing the surface area for integral proteins.
It helps in embedding the proteins in the folding known as cristae.
In humans, a. How many sperm develop from 100 primary sper matocytes? b. How many sperm develop from 100 secondary spermatocytes? c. How many sperm develop from 100 spermatids? d. How many ova develop from 100 primary oocytes? e. How many ova develop from 100 secondary oocytes? f. How many ova develop from 100 polar bodies?
Answer:
In males primary spermatocytes undergo two meiotic divisions. First meiotic division give rise to two secondary spermatocytes and secondary meiotic division give rise to four spermatids from two spermatocytes.
Therefore a. from 100 primary spermatocytes 400 sperms will develop (100*4= 400)
.
b. From 100 secondary spermatocyteS 200 sperms will develop(100*2= 200).
c. Spermatids do not undergo any further division rather they undergo differentiation to form mature sperms. Therefore from 100 spermatids, 100 mature sperms will develop.
d. One primary oocyte gives rise to one secondary oocyte and one secondary oocyte give rise to one mature ova so only one mature ova will develop from 100 primary oocytes.
e. 100 ova will develop from 100 secondary oocytes.
f. Polar body do not develop into ova and degenerates after some time so no ova will develop from 100 polar bodies.
Approximately 200 sperm develop from 100 primary spermatocytes.
b. Around 400 sperm develop from 100 secondary spermatocytes.
c. Roughly 100 sperm develop from 100 spermatids.
d. Only one ovum develops from 100 primary oocytes.
e. Only one ovum develops from 100 secondary oocytes.
f. No ova develop from 100 polar bodies.
What is the numbers of spermIn people, spermatogenesis makes sperm and oogenesis makes ova, but they make different amounts of each. Here are the estimated numbers of sperm and eggs that develop from certain cell stages:
When 100 cells called primary spermatocytes go through a process called meiosis, they turn into 200 cells called secondary spermatocytes. Each cell called primary spermatocyte breaks into two smaller cells named secondary spermatocytes.
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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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One oak tree cell with 14 chromosomes undergoes mitosis. How many daughter cells are formed, and what is the chromosome number in each cell?
Answer:
The correct answer will be option- 2 daughter cells with 14 chromosomes.
Explanation:
Mitosis is a type of cell division which is used by the organism to divide the somatic cells. This method is therefore considered an asexual mode of reproduction.
Mitosis proceeds in two stages: karyokinesis (a division of the nucleus) and cytokinesis (a division of the cytoplasm). Mitosis results in two daughter cells with the same amount of DNA content that is clone.
In the given question, since the tree cell contains 14 chromosomes therefore on the mitotic division will produce two daughter cells with the same amount of chromosomes that are 14 chromosomes.
Thus, 2 daughter cells with 14 chromosomes are the correct answer.
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 type of lipid would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water?
a. glycerophospholipid
b. cholesterol
c. galactolipid
d. shingolipid
e. free fatty acids
Answer:
e. free fatty acids
Explanation:
Free fatty acids would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water.
Free fatty acids produced by the activity of lipase enzyme on stored triglycerides in the cytoplasm.
Micelles are the molecules of lipid which are arranged in a spherical manner which have a polar head and a hydrophobic tail. Micelles transport the fatty acids and insoluble monosaccharides.
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 innervates the superior, medial and inferior rectus muscles of the eye and is a constrictor of the pupil?
a. Trochlear
b. Abducens
c. Oculomotor
d. Trigeminal
e. Facial
Answer:
The correct option is: c. Oculomotor
Explanation:
Oculomotor nerve is one of the twelve pairs of cranial nerves, emerging from the brain. Oculomotor is the third cranial nerve that originates from the third nerve nucleus in the midbrain.
The oculomotor cranial nerve (CN III), innervates the extrinsic eye muscles including the superior rectus, inferior rectus, and the medial rectus; that are responsible for raising the eyelid and movements of the eye.
The fibers of this nerve innervates the intrinsic eye muscles that are responsible for the constriction of the pupil.
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.
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.
The evolution of archosaurs was heavily impacted by the climate and environmental changes that occurred in the Early Triassic. Which of these best describes the environment in the Triassic?
a. The vegetation was dominated by grasses and angiosperms which were new food sources
b. The equatorial climate became much more arid compared to that seen in the Permian
c. The polar regions were dominated by permanent polar ice caps that were linked to lower sea levels.
d. Oxygen levels were extremely high due an increase in worldwide vegetation
Answer:
Option (B)
Explanation:
The Triassic period belongs to the Mesozoic era in which the dominant species was the dinosaurs. During this time, the environment and the type of climate was relatively dry and there was no precipitation to support the vegetation. During, this Triassic period, the super-continent Pangaea was present that eventually got split up into different smaller continents by Jurassic and Cretaceous. The climate in the equatorial region during that particular time on earth was comparatively much more arid than that was in the Permian.
There were no polar ice caps during that time and tectonic activities were rapid. The earth experienced a temperature of about 10°C more than the present day temperature and oxygen level was also relatively low then. The plants that existed during this period were mostly gymnosperms.
Hence, the correct answer is option (B).
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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Briefly describe the functional recovery of brain after brain injury or stroke.
Answer:
The brain is made up of billions of neural connections that transit and gather signals that are processed into information. Brain injury or stroke may cause damage to some of these connections. Neurons have a limited ability to repair themselves, unlike normal cells hence making brain injury difficult to recover fully from. The brain recovery after such as injury is are mainly driven by the brain finding new neural pathways around the damaged ones.
However difficult, there is a chance that a neuron can be repaired especially if the cell body is not damaged in the injury. If the axon is regenerated, it needs to make the right connections to re-establish signaling. The problem is that glial cells, that are integral to neurons structure and function, reduce the chances of recovery because they clean up damaged neurons and produce molecules that inhibit recovery
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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Atoms of which three elements could bond together to form an organic compound?
(a) carbon, iron, and nitrogen
(b carbon, hydrogen, and oxygen
(c) oxygen, hydrogen, and nitrogen
(d) oxygen, iron, and carbon
Answer:
b
Explanation:
carbon, hydrogen and oxygen
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HEY hun!
your answer is B) Carbon, Hydrogen, and Oxygen
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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.
Base-pair substitutions involving the replacement of a purine with a pyrimidine and vice versa are called:
a. isomers
b. transitions
c. transversions
d. inversions
e. translations
Answer:
The answer is: c. tranversions
Explanation:
A mutation is the result of changes in the structure of a gen, these variations happen in the nucleotide sequence of a genome, and can be as a consequence of DNA damage.
One type of DNA mutation is the substitution of one base pair for another. Two type of substitutions can happen. One of them are the transversions, which are interchanges of a purine (A or G) for a pyrimidine (C or T) and vice versa.
The other type of substitution are the transitions that are interchanges of one purine (A) for another purine (G) or a pyrimidine (C) to another pyrimidine (T).
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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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.
What element was pointed out for special attention in the structure of a DNA base?
a. oxygen
b. carbon
c. phosphorus
d. nitrogen
e. calcium
Answer:
The correct answer will be option-D.
Explanation:
Deoxyribose nucleic acid or DNA is the genetic material of the organisms which is composed of nucleotide monomers. Each nucleotide monomer is made up of a fiver carbon sugar, four different nitrogenous bases and a phosphate group
The DNA bases can be easily distinguished on the basis of nitrogen element as the structure of purine which is adenine and guanine contains four nitrogen molecule whereas cytosine and thymine contain two nitrogen molecules in their structure.
Thus, option-D is the correct answer.
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.
tRNAs:
a. have a "cloven hoof" secondary structure
b. serve as adapters between amino acids and DNA substitutions
c. have their corresponding amino acid covalently attached to their 5' end
d. are postranscriptionally modified to include abnormal bases such as dihydrosphinctosine
e. have an acceptor stem
Answer:
Have an acceptor stem.
Explanation:
RNA is present as genetic material in some viruses only. RNA is the single stranded molecules that are formed from the DNA template by the process of translation.
Different types of RNA are rRNA, mRNA and tRNA. Transfer RNA (tRNA) contains the anti codons that are complementary to the codon present on mRNA. The tRNA consists of the acceptor stem that acts as the amino acid attachment site during the process of translation.
Thus, the correct answer is option (e).
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.
Which of the following is not considered evidence supporting the evolutionary relationship among all plant taxa?
a. Plants share a significant number of genes
b. Plants and plant ancestors have chlorophyll
c. Fossils of plants show us transition series and evolutionary patterns of plants
d. Structures/characteristics among most plants are very similar
Answer:
The correct answer is option d. "Structures/characteristics among most plants are very similar".
Explanation:
There are extensive evidence to support the evolutionary relationship among all plant taxa, however, similarity of structures and characteristics among most plants is not one of them. Plants are one of the most diverse kingdoms having very different species such as flowering plants, conifers, lichens, cycads, algae, among others. The structures and characteristics of these plants are very different among them, and this diversity makes difficult to classify and clarify the evolutionary relationship that they share.