Explanation of true and false statements regarding cytokinesis in animal cells. A FTFT is the correct option.
FTFT
The correct answer sequence is FTFT:
True: The force for cytokinesis in animal cells is generated by kinesin motors on microtubule bundles that form the contractile ring.
False: As the contractile ring constricts, its thickness does not increase to keep a constant volume.
True: The midbody forms from bundles of actin and myosin II.
False: Local activation of Ran GTPase does not trigger the assembly and contraction of the contractile ring.
The answer is option D: FFFF, as explained in the detailed response regarding the statements related to cytokinesis in animal cells.
The correct answer is option D: FFFF.
Let's break down each statement:
(F) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring. This statement is false because the force is mainly generated by myosin II motors on actin filaments.(F) As the contractile ring constricts, its thickness increases to keep a constant volume. This statement is false; the thickness of the ring decreases as it contracts.(F) The midbody forms from bundles of actin and myosin II. This statement is false; the midbody contains microtubules and other proteins but not actin and myosin II.(F) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring. This statement is false; Rho GTPase, not Ran GTPase, plays a crucial role in activating the contractile ring.Complete question is-
Indicate true (T) and false (F) statements below regarding cytokinesis in animal cells. Choose the correct answer represented by a four-letter string composed of letters T and F only, e.g. TFFF.
( ) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring.
( ) As the contractile ring constricts, its thickness increases to keep a constant volume.
( ) The midbody forms from bundles of actin and myosin II.
( ) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring.
A. FTFT
B. FFTF
C. FTFF
D. FFFF
In some circumstances, when two different carbon sources are available, growth will occur first using one carbon source, then after a short lag period, growth will resume using the second carbon growth source.
a. This process is called _____ growth.
Answer: Diauxic growth
Explanation:
The diauxic growth or diphasic growth is a bacterial growth which is characterized by the growth in two phases depending upon the source of carbon used.
The preferred carbon source is consumed first, this leads to enhance the growth in the bacteria, followed by the lag phase. During the process of lag phase the bacteria start to metabolize the second carbon source and the growth resumes.
For example, a colony of E.coli bacteria was cultured in a medium containing the glucose, and lactose sugars. At the initial level the bacteria was capable of using the glucose sugar and growth became rapid this is followed by a lag phase. In the lag phase the bacteria started to utilize the lactose sugar and again the growth resumes.
Final answer:
Diauxic growth is when a cell uses one carbon source and then switches to another after the first is depleted. In the given E. coli growth curve example, glucose is used first, leading to rapid growth, followed by a lag phase and then slower growth on xylose once glucose is exhausted.
Explanation:
The process described in the student's question, where a cell uses one carbon source for growth and then switches to another after the first is depleted, is known as diauxic growth. Tackling Monod's research, we can clarify what is occurring at different points (A-D) in the growth curve. At point A, the E. coli is primarily using glucose as its carbon source for growth, causing a rapid increase in population. The xylose-use operon is not being expressed at this point because the presence of glucose typically represses the expression of enzymes for the metabolism of other sugars.
After glucose is exhausted, demonstrated by a leveling off of the growth curve at point B, the cells enter a short lag phase as they begin to express the necessary enzymes for the uptake and metabolism of xylose. At point C, we observe the resumption of growth as the bacteria start to use xylose. As these enzymes are less efficient or the substrates are utilized less favorably, the second phase of growth is slower compared to the first phase. Finally, at point D, the growth rate declines again as the bacterial population exhausts the xylose.
The dynamics are different when the advantageous allele is rare, compared to when it is near 100%. Explain what happens when a recessive advantageous allele is rare versus near fixation
Answer:
An advantageous allele might be dominant or recessive in population. In case of dominant advantageous allele the dynamics is relatively simple because being dominant the allele is expressed both in homozygous and heterozygous condition. Combined with the advantage of natural selection, its frequency increases rapidly.
However, recessive advantageous allele does not increase rapidly because despite being advantageous it gets masked by the dominant allele. When it is rare, it is present in very less number of recessive homozygotes and in slightly more number as heterozygotes. Over the generations, natural selection selects the recessive allele so the number of heterozygotes start to increase slowly. Recessive homozygotes are still rare because they need both the copies of recessive allele. Eventually recessive homozygotes also start increasing in number which leads to the fixation of recessive allele in the population.
What would happen if someone stabbed your leg with a syringe full of calcium and injected the calcium directly into your muscle?
a.The actin active sites would stay covered by tropomyosin.
b.Cross-bridges would form in the absence of an action potential from a motor neuron.
c.Tropomyosin would bind the calcium and change the conformation of troponin.
d.Myosin would be unable to hydrolyze ATP.
Answer:
Cross-bridges would form in the absence of an action potential from a motor neuron.
Explanation:
The injected calcium ions would bind to troponin. Troponin would make tropomyosin move away from the myosin-binding sites on actin. The presence of free binding sites on the actin would be followed by the contraction cycle. This would include hydrolysis of ATP to energize myosin heads and binding of these heads to actin to form cross-bridges. Therefore, cross-bridge formation would occur without any action potential if calcium ions are injected directly into the muscle.
Which of the following membranes is correctly matched to its function? (A) allantois .. food absorption (B) yolk sac .. embryonic bladder (C) amnion .. gas exchange (D) dura mater .. brain protection (E) peritoneum .. heart protection
Answer:
The only correct answer is D) dura mater ..brain protection
Explanation:
Allantois helps the embryo exchange gases and handle liquid waste, it does not do food absorption, yolk sac is not embryonic bladder, chorion does the gas exchange not amnion, peritoneum is the abdominal protection not heart.
Scientists construct an experimental bacteriophage that is composed of the T2 phage protein coat and T3 phage DNA. If a bacterium is infected by this phage, the new phages produced would be expected to have:
Explanation:
A bacteriophage is an infection that attacks bacteria. At the point when the tail strands identify an objective host the bacteriophage to the cell, injected its DNA, and utilizations the microscopic organisms' apparatus to reproduce. T4 is a sort of bacteriophage that infects of E. coli. The bacteriophage T4 capsid is a prolonged icosahedron, 120 nm long and 86 nm wide, and is worked with three essential proteins such as gp23*, which shapes the hexagonal capsid cross section, gp24*, which structures pentamers at eleven of the twelve vertices. gp20, which frames the extraordinary dodecameric entry vertex through which DNA. T4 DNA Ligase is ligation catalyst which utilized the parts of DNA by the catalyzing between compared 5'phosphate and 3' hydroxyl ends and phosphodiester bonds in the double stranded DNA utilizing ATP as a coenzyme.gHow would an inhibitor of cAMP phosphodiesterase affect glucose mobilization in muscle? It would reduce cAMP levels and inhibit glucose mobilization. It would maintain high cAMP levels and elevate glucose mobilization. It would increase AMP concentration, thereby increasing glucose mobilization. It would increase cAMP levels, which would inhibit glucose mobilization.
Explanation:
It would maintain high cAMP level and elevate glucose mobilization
Phosphodiesterase is an effector enzyme which degrades secondary messenger cAMP(cyclic adenosine monophosphate)Here in this case an inhibitor is inhibiting the phosphodiesterase therefore cAMP level will increaseAs cAMP level rise it activates a protein called protein kinase A which phosphorylates phosphorylase kinase and activates it Phosphorylase kinase becomes active that phosphorylates glycogen phosphorylase and makes it active,glycogen phosphorylase catalyse breakdown of glycogen(in liver and muscle cells)In liver cells breakdown of glycogen occurs and glucose 1 phosphate gets converted into glucose and supplied to whole body through blood
Imagine you are cutting a bagel (one of the most common household injuries) and you get a cut. The cut heals. How do the new cells compare to the original (pre-cut) cells
Answer:
The new cells are the same as the previous ones, since they are the result of the mitosis process.
Explanation:
When we cut our skin, our brain sends information to millions of cells to take action and prevent this cut from putting us in danger. At that moment, the blood cells begin their work, supplying enough oxygen to stop possible bleeding and start the healing process. Then another group of cells swap out possible bacteria that may be trying to get into the wound. Last but not least, skin cells enter cell division and undergo mitosis, to generate new cells and create a new skin layer.
New cells are the same as old cells, as they are the result of mitosis. Mitosis is the process of cell division where one cell gives rise to two cells exactly the same as it.
The new cells should be the same as the previous ones, because they are the result of the mitosis process.
What is the mitosis process ?
Mitosis refers to the process where a eukaryotic cell nucleus divides in two, followed by the division of the parent cell into two daughter cells. Also, it is the process of cell division where one cell gives rise to two cells exactly the similar it is.
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HIV does not have any enzymes to ensure that the replication of its genome is error free. Use this information to propose an explanation as to why it has been difficult for scientists to create a vaccine to HIV?
Answer:
The human immunodeficiency viruses (HIV) belong to the category of retrovirus and leads to fatal condition to the affected individual due to its action on the immune system.
Explanation:
This virus is known to contain its own replication machinery that lacks the property of proof reading. But despite vaccine against its is not yet formulated. This is due to the ability of higher mutation rate of the associated virus. This genetic variability allows the spread of the virus all throughout the body but with specific or unique genetic composition.
As the genetic composition of the virus is different, it adds to the difficulty level in creation of the antiviral drug as the target for the drug will not be specified.
If your sample does not grow on the plate in the anaerobic chamber but does grow in the presence of oxygen, it produces bubbles with the addition of hydrogen peroxide, and the dextrose tube remains orange, you can conclude:
Answer:
it is an obligate (or strict) aerobe
Explanation:
By whether an organism requires oxygen or not for respiration. We can classify it as either aerobic or anaerobic.
Aerobic organisms require oxygen and anaerobic do not.
For aerobes, it can be facultative or obligate. Facaltative aerobes require oxygen but they can however switch to fermentation when oxygen is not available.
Obligate aerobes are aerobes require oxygen for cellular metabolism.
In the test the dextrose tube remained yellow because fermentation had not taken place.
Therefore we can conclude the sample contained an obligate aerobe which was catalase positive since it produces bubbles when Hydrogen peroxide was added.
When you scratch a mosquito bite, you damage some cells. Damaged cells release histamine, which causes localized swelling. The swelling can crush cells, causing them to release more histamine. This is an example of
Answer:
of a cycle where scratching will cause even more of an itchy sensation
Answer:
Positive feedback.
Explanation:
Positive feedback is a process in which the end product of an action cause more of that action to occur in a feedback loop.
Inflammation is thhe local reaction of bodily tissues to injury caused by physical damage , infection or due to any allergic reaction. Injured tissue mast cells release histamine, which causes surrounding blood vessels to dilate and increase permeability. This allows fluid and cells of immune system to leak from bloodstream through vessel walls and migrate to site of tissue injury or infection where they fight infection and heal injured tissues.
hich of the following occurs during bacterial conjugation? a. The genome is duplicated and then diverges. b. An insertion sequence finds a target site on another DNA strand then uses recombination to insert itself into that strand. c. An F+ plasmid transfers via rolling circle replication through a relaxosome to an F- cell. d. A bacteriophage packages its capsid with degraded host DNA then infects a new host, delivering the previous host’s genes.
Answer: Option D.
An F+ plasmid transfers via rolling circle replication through relaxosome to an F- cell.
Explanation:
Bacteria conjugation is the transfer of DNA between bacteria cell through cell to cell contact or bridge like connection between bacteria cells.
During bacteria conjugation, DNA is transferred from the bacterium donor of a mating pair to the recepient through pilus.
An F+ plasmid which is the donor transfers via rolling circle replication through a relaxosome to an F- cell.
Picornaviruses can avoid detection by synthesizing virally induced vesicles, or replication complexes, formed from the Choose one: A. Golgi apparatus. B. nuclear membrane. C. endoplasmic reticulum. D. lysosome.
Answer:
Option-C
Explanation:
Picornaviruses is the virion or naked particles which cause many animal and human infections.
The mechanism of their action is not explained in detail till now but it has been predicted on the basis of certain research that the virus escapes the immune response by enclosing themselves in the lipid membrane-enclosed vesicles formed by the host cells.
These vesicles are produced by the cells during certain physiological mechanisms from the endoplasmic reticulum of the cells.
Thus, Option-C is correct.
A rare recessive allele inherited in a Mendelian manner causes the disease cystic fibrosis. A phenotypically normal man whose father had cystic fibrosis marries a phenotypically normal woman from outside the family, and the couple consider having children. a.) draw the pedigree as far as described? b.) If the frequency in the population of heterozygotes for cystic fibrosis is 1 in 50, what is the chance the couples first child will have cystic fibrosis? c.) If the first child does have cystic fibrosis, what is the probability that the second child will be normal?
Answers:
a.) draw the pedigree as far as described?
Pedigree:
C/– c/c
C/c C/–
?
b.) If the frequency in the population of heterozygotes for cystic fibrosis is 1 in 50, what is the chance the couples first child will have cystic fibrosis?
Man: has the disease
Wife: 1/50 chance to have the c allele
First child: 1.0 x 1/50 x 1/4 = 1/200 = 0.005
c.) If the first child does have cystic fibrosis, what is the probability that the second child will be normal?
If the first child has the disease, then the mother is a carrier of the
c allele. In consequence, the probability is 3/4
Variation in a trait is a required condition for natural selection to act on a population for that trait. Assuming a population of organisms started with only one form of a trait, what are two ways variation in the trait could be introduced into the population? Explain your answer.
Answer:
1. Mutation
2. Epigenetics
Explanation:
1. Mutation occurs when there is a change in an organism's DNA sequence as a result of mistakes in DNA replication or as a result of environmental factors like smoking. The mutation in a single organism can be passed on to other generations hence causing a genetic variation in the population, this obeys the Darwin's law that inherited traits (genetic) are passed on to other generations
2. Epigenetics are changes in gene expression that doesn't involve changes in the DNA sequences unlike mutation. This changes can be passed on to other generations and hence cause a variation in the population. This obeys the Lamarckian evolution that acquired traits are passed on to other generations.
A mutation is a change that occurs in our DNA sequence, either due to mistakes when the DNA is copied or as the result of environmental factors such as UV light and cigarette smoke.
Variation in a trait can be introduced into a population through mutations, which are random changes in DNA, and sexual reproduction, which shuffles alleles during gamete formation. These variations must be heritable for natural selection to act on them.
Variation in a trait is essential for natural selection to act on a population. Assuming a population starts with only one form of a trait, there are two primary ways that variation could be introduced:
Mutations: Random changes in DNA sequences can create new alleles of a gene, leading to new variations in traits. These mutations can occur due to errors in DNA replication or due to the influence of environmental factors like radiation.
Sexual Reproduction: During the formation of gametes, processes such as crossing over and independent assortment of chromosomes can reshuffle alleles to create new combinations of genes. When individuals with different genetic makeups mate, the offspring inherit a unique set of alleles, contributing to the genetic diversity of the population.
It is important to note that these variations must be heritable and have a genetic basis to contribute to the process of natural selection. Otherwise, natural selection cannot effectively lead to evolutionary change across generations.
4. Celery has very small flowers clustered into an inflorescence called an umbel. Based on what you noted about flower morphology in the sunflower and the iris, what do you predict would be the morphology of an individual celery flower? (1 pt)
Answer:
Compound Umbel.
Explanation:
Celery is more of an annual crop which is a herbaceous plant usually 60 to 120 cm high with white Flowers.
The Celery plant belongs to the Apiaceae and they are known to be mainly Annual.
Morphology or the shape of the Celery plant is that of a Compound Umbel, in which all Umbel inflorescences arises from a common point and appears to be at the same level.They change from
elongated axes (racemes and panicles) to flattened axes (corymbs and umbels) which results in inflorescences thereby making the flowers been arranged closely together. This close association encourages efficient pollination,and the extreme condensation of the inflorescences, as in the
head, gives rise to an inflorescence that appears
to be a single flower and example of such happen to be the sunflowers commonly found around us.
Answer:
I would predict that celery flower would have a petals of four(4) or five (5)
Explanation:
If you hiked in Pocahontas State Park 100 times last year and you saw a White-crowned Sparrow 43 times, what is the probability that you will observe a White-crowned Sparrow next time you go hiking at Pocahontas State Park?
Answer: 43/100
Explanation:
Total number of hikes (T) = 100
Number of times White-crowned Sparrow was seen (N) = 43
Then, probability of seeing White-crowned Sparrow again = N / T
= 43/100
Thus, the probability of seeing White-crowned Sparrow again in Pocahontas State Park is 43/100
Answer: 43/100 or 0.43
Explanation:
number of hikes in the period of a year (H) = 100
Number of times you observed White-crowned Sparrow (S) = 43
Therefore, the probability of seeing White-crowned Sparrow again =
S ÷ H
= 43/100 or 0.43
Which of the following pairs of microbe classification terms and optimal growth temperatures is mismatched? Which of the following pairs of microbe classification terms and optimal growth temperatures is mismatched? hyperthermophiles growth at 95°C psychrotroph growth at 22°C psychrophile growth at 37°C mesophile growth at 37°C
Psychrophile growth at 37°C is a mismatch.
Explanation:
Psychrophiles are also known as cryophyles.These organisms are extremophilic because they inhabit extremely cold places whose temperatures are at about 10 degree Celsius to -20 degree Celsius.These places include polar belts, high snow covered mountains, deep sea beds and other regions with permafrost etc.Examples include many bacterial genus like Polaromonas, Psychrobacter, Arthrobacter etc.The mismatched pair of microbe classification terms and optimal growth temperatures is psychrophile and growth at 37°C.
Explanation:The pair of microbe classification terms and optimal growth temperatures that is mismatched is psychrophile and growth at 37°C. A psychrophile is a type of microorganism that grows best at very low temperatures, typically around 0-20°C. On the other hand, mesophiles are microbes that grow best at moderate temperatures, between roughly 20-45°C. The other pairs in the question are correctly matched: hyperthermophiles grow at extreme high temperatures (around 80-113°C), psychrotrophs grow at cold temperatures but have an optimum growth temperature around 20-30°C, and mesophiles grow at moderate temperatures around 20-45°C.
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Match the # in the diagram with the correct structure/term.
Column A Column B
1. ____ 1 A. codon
2. ____ 2 B. amino acid
3._____ 3 C. tRNA (anit-codon)
4._____ 4 D. Polypeptide chain (protein)
Answer:
1 A. amino acid
2 B. Polypeptide chain (protein)
3 C. tRNA (anit-codon)
4 D. codon
Explanation:
The whole diagram explains protein synthesis from transcription to translation.
- The codon, GGU is a codon with a triplet nature coding for glycine, and it consists of three nucleotides
- The amino acid, phenylanine is encoded from the codon, UUU found on the mRNA molecule
- transfer RNA (tRNA) helps to combines covalently with a specific amino acid, threonine and transfers the amino acid to the ribosomes to join the polypeptide chain
- polypeptide chain is a pentapeptide (with five amino acids) and is formed in the final stage of protein synthesis.
MHC class II molecules expressed on the surface of thymic cortical epithelial cells normally have a wide repertoire of different peptides bound to them. By engineering a construct that fuses the MHC class II protein to a single peptide sequence, and expressing this construct in thymic cortical epithelial cells that have their endogenous MHC class II genes knocked out, it is possible to generate a mouse line where all MHC class II proteins expressed on all thymic cortical epithelial cells are bound to the same peptide. These mice are often referred to as ‘single-peptide’ mice. Examination of the T cell developing in these single peptide mice would likely show:
A. A significant reduction in the numbers of mature CD4 T cells
B. No change in the numbers of mature CD4 T cells
C. A block in T cell development at the CD4+CD8+double-positive stage
D. A repertoire of T-cell receptors on mature CD4 T cells restricted to a single Vbeta
E. A block in T cell development at the CD4-CD8-double-negative stage
Answer:
A. A significant reduction in the numbers of mature CD4 T cells
Explanation:
In the given problem, there is an engineering of the MHC class II protein with the sequence of a single peptide. In addition, it was expressed in the epithelial cells of the thymic cortical. Based on the result obtained from the engineering construction, it is obvious that there would be a large decreases in the CD4 T cells numbers.
Colored aleurone in the kernels of corn is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not the kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive yallele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained:Colored, green 88Colored, yellow 12Colorless, green 8Colorless, yellow 92Based on the data, what can you say about the genotype of the unknown plant?A. It was heterozygous for one gene, and homozygous for the other.B. It was homozygous for both genes.C. It was heterozygous for both genes.
Answer: C. It was heterozygous for both genes.
Explanation: To produce a generation that has individuals with trait for colored aleurone in the kernels, the plant being analised has to have a dominant allele R. In the same way, to have offspring with the recessive trait, it has to carry the recessive allele r. So, the unknown plant has to be heterozygous for colored aleurone in the kernels, Rr.
The same thought can be applied to plant color: Since there are green and yellow plants, the unknown plant has to be heterozygous for that trait, Yy.
In conclusion, the unknown plant is heterozygous for both genes.
The DNA sequence below is transcribed completely to generate anmRNA. The mRNA is then translated to synthesize a protein.On which strand is thelongestopen reading frame (ORF)?
Answer:
The ORF open reading frame is an RNA sequence that is located between the start codon and the termination codon, bounded by untranslated sequences or UTR.
Explanation:
For the translation of an mRNA into a protein, a reading frame is required. This framework allows the division of the mRNA sequence into different codons during translation. The start codon is the main signal, since the translation starts at the start codon, this position allows the mRNA to read in the appropriate frame.
The infants in the strange situation who waver as they move from mother to toys, are hesitant to explore, are cautious when meeting the stranger, are very upset when mother leaves, and push their mother away at the reunion are the category of
Answer:
Insecurely attached.
Explanation:
Infants that are insecurely attached have learned that adults are not to be trusted. Children who have had negative encounters with their caregivers tend to develop insecure attachment.
Children with insecure attachment refuse to associate with others, show fear and anger.
Unequal crossing over results in A. an exchange between nonhomologous chromosomes. B. a loss of genetic material. C. a repair of UV-induced damage. D. a production of eggs containing Y chromosomes. E. a creation of deletions and duplications.
Answer: OPTION E
Explanation:unequal crossover usually leads to duplication or deletion of chromosome. In this case,. A DNa strand is deleted and replace usually by another DNA strand which is mostly a duplicate from a sister chromatid and this process leads to Gene families been produced beause one is deleted and again and again duplicate is produced on the same place (2 product formation). It is a form of chromosomal crossing over that exists between homologous sequence which were initially not paired together. In Gene duplication and mutation in organism, unequal crossover is said to be the pioneer or chief cause of it with Gene conversion beside it.
The production of a continuous new strand of DNA using the many separate Okazaki fragments (in other words, the joining of the already made fragments) found on the lagging strand requires all of the following except which one?
A. nuclease
B. ATP
C. repair polymerase
D. DNA primase
E. DNA ligase
Answer:B
Explanation:
because if we strand dna we wouldnt have genes
Production of continuous new strand of DNA using Okazaki fragments found on lagging strand requires following components: ATP, repair polymerase, DNA primase, and DNA ligase,thus correct options are all except A.
1. ATP: ATP provides the energy necessary for the synthesis of DNA. It is required during the formation of phosphodiester bonds between nucleotides, which link them together to form the new DNA strand.
2. Repair polymerase: Repair polymerase, also known as DNA polymerase I, is responsible for replacing the RNA primers used in DNA replication with DNA nucleotides. It removes the RNA primers and fills in the gaps with complementary DNA nucleotides.
3. DNA primase: DNA primase synthesizes short RNA primers that provide a starting point for DNA synthesis. These primers are required for DNA polymerase to initiate DNA synthesis.
4. DNA ligase: DNA ligase is an enzyme that seals the gaps between the Okazaki fragments on the lagging strand. It catalyzes the formation of phosphodiester bonds between adjacent nucleotides, joining the fragments together to form a continuous DNA strand.
Based on this information, the correct answer is A. nuclease. Nuclease is not required for the production of a continuous new strand of DNA using the Okazaki fragments. Nucleases are enzymes that break down DNA or RNA molecules by hydrolyzing the phosphodiester bonds between nucleotides. In the context of DNA replication, nuclease activity is not involved in joining the Okazaki fragments.
Thus, production of continuous new strand of DNA using Okazaki fragments found on lagging strand requires following components: ATP, repair polymerase, DNA primase, and DNA ligase,thus correct options are all except A.
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Mismatch repair systems that maintain DNA replication fidelity: Require a protein that detects the damaged base as well as a protein that excises the base from the strand. Rely on DNA glycosylase for proper function. Require a protein that detects the mismatch as well as a protein that recruits an endonuclease to the site of the mismatch. Are responsible for repairing C T point mutations. None of the above
Answer:
Require a protein that detects the damaged base as well as a protein that excises the base from the strand.
Explanation:
The DNA can undergo the process process of mutation during the DNA replication process. The DNA repair process occurs in the body to repair the mismatched DNA.
The mismatch repair system identifies the insertion, deletion and mismatch DNA that can be corrected by mismatch repair system. The Mut S protein is required for the detection of mismatch DNA and mut H then acts as endonuclease to cut the protein and then DNA polymerase fills the gap.
Thus, the correct answer is option (1).
Please help with number 8,9and 10
If the killer whale obtained 26000 joules of energy by eating crab-eater seal, how much energy was available at each of the following trophic levels
Answer:
it is not very specific im sorry if i could see the paragraph i could help
Explanation:
One of your lab partners has followed the recommended procedure of running Gram-positive and Gram-negative control organisms on her Gram stain of an unknown species. Her choices of controls were Escherichia coli and Bacillus subtilis. She tries several times and each time concludes she is decolorizing too long because both controls have pink cells (one more than the other). What might you suggest she try and why?
Answer:
Reduced in holding time of decolrization step and also used less Alcohol because decolrization step is important in Gram's staining.The decolorization step must be performed carefully. Otherwise over-decolorization may occur. This step is critical and must be timed correctly otherwise the CV stain will be removed from the Gram-positive cells. If the decolorizing agent is applied on the cell for too long time, the Gram-positive organisms to appear Gram-negative..
Explanation:
Gram' staining is a technique used in microbiology labs to differentiate between Gram's positive and negative
Gram-positive bacteria :Stain dark purple due to retaining the primary dye called CV in the cell wall.
:Gram-negative bacteria Stain red or pink due to retaining the counter staining dye called Safranin or neutral red.
There are four basic step in Gram" staining
1) Application of the Primary Stain to a Heat-Fixed Smear of Bacterial Culture
2)Addition of Gram's Iodine
3)Decolorization with 95% Ethyl Alcohol:Alcohol or acetone dissolves the lipid outer membrane of Gram-negative bacteria, thus leaving the peptidoglycan layer exposed and increases the porosity of the cell wall. The CV-I complex is then washed away from the thin peptidoglycan layer, leaving Gram-negative bacteria colorless.
On the other hand, alcohol has a dehydrating effect on the cell walls of Gram-positive bacteria that causes the pores of the cell wall to shrink. The CV-I complex gets tightly bound into the multi-layered, highly cross-linked Gram-positive cell wall thus staining the cells purple.
The decolorization step must be performed carefully. Otherwise over-decolorization may occur. This step is critical and must be timed correctly otherwise the CV stain will be removed from the Gram-positive cells. If the decolorizing agent is applied on the cell for too long time, the Gram-positive organisms to appear Gram-negative. Under-decolorization occurs when the alcohol is not left on long enough to wash out the CV-I complex from the Gram-negative cells, resulting in Gram-negative bacteria to appear Gram-positive.
An enzyme that follows Michaelis-Menten kinetics has a KM value of 20.0 μM and a kcat value of 231 s−1. At an initial enzyme concentration of 0.0100 μM, the initial reaction velocity was found to be 1.07×10−6 μM/s. What was the initial concentration of the substrate, [S], used in the reaction ?
The initial substrate concentration was 0.52 μM.
We can apply the Michaelis-Menten equation to solve for the initial substrate concentration ([S]):
v = (Vmax * [S]) / (KM + [S])
where:
v is the initial reaction velocity (1.07 × 10^-6 μM/s)
Vmax is the maximum reaction velocity (determined by kcat and enzyme concentration)
KM is the Michaelis-Menten constant (20.0 μM)
[S] is the initial substrate concentration (unknown)
Step 1: Calculate Vmax based on kcat and enzyme concentration:
Vmax = kcat * [E] = 231 s^-1 * 0.0100 μM = 2.31 μM/s
Step 2: Rearrange the Michaelis-Menten equation to solve for [S]:
[S] = (v * KM) / (Vmax - v)
Step 3: Substitute known values and solve for [S]:
[S] = (1.07 × 10^-6 μM/s * 20.0 μM) / (2.31 μM/s - 1.07 × 10^-6 μM/s) ≈ 0.52 μM
Therefore, the initial concentration of the substrate ([S]) used in the reaction was approximately 0.52 μM.
Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. Indicate the probability of producing an AB gamete from an AaBb individual.
Explanation:
Step 1. The two quality loci, A and B, assort autonomously. The alleles A and B are predominant over the alleles an and b. In this way, when a cross happens between AaBb X AaBb, the subsequent gametes would be AB, Ab, aB, and ab.Step 2.The offsprings which have in any event one A and B allele, will show AB phenotype. Along these lines, AABB, AaBb, AABb, AaBB, will all have AB phenotype.If two gene loci, A and B, assort independently, the probability of producing an AB gamete from AaBb would be 1/4 or 25%
Since both A and B are independently inherited, Aa and Bb can be crossed just we would have it in a monohybrid cross following Mendelian pattern.
Thus:
Aa x Bb
AB Ab aB ab
The gametes and their respective probabilities of appearing would be:
AB - 1/4Ab - 1/4aB - 1/4ab - 1/4In other words, all 4 gametes have an equal chance of being produced during gametogenesis.
More on independent assortment of genes can be found here: https://brainly.com/question/13041062
One of the first diagnostic tools used at the hospital was an electrocardiogram (EKG or ECG), which reflects the electrical activity of the cardiac muscle. We know that the atria contract first (the P wave) and then, after a brief delay, the ventricles contract (the QRS complex). Given that the heart does not have any nerves to stimulate the cardiac muscle cells, how is the timing of contraction coordinated? How do action potentials get from muscle cell to muscle cell? If the EKG shows a long delay between the P wave and the QRS complex, which type of cardiac tissue might have been damaged?
Answer:
The heart has an intrinsic conduction system that causes electrical activity in the heart muscles causing them to contract. The intrinsic conduction system is made up specialized cells, that contain nerve and muscular characteristics.The muscle cells in the heart are linked together by gap junctions, allowing cardiac action potentials to travel from one muscle cell to another.Atrioventricular (AV) node. The damage to the AV node causes the electrical signals traveling from the upper chambers to the lower chambers to be impaired causing an AV block.Explanation: