Answer:
Temperate phages don't immediately kill their hosts.
Explanation:
There exist phages that have shown the capability to alternate lytic and lysogenic life cycles, where don't kill the host cell. This feature allows a phage to reproduce without killing the host cells, thereby having higher probabilities of replicating its genetic material
Which of the following statements about bacterial growth is false? Group of answer choices Agar is used as a solidifying agent in some types of media A turbid culture is indicative of bacterial growth. Each bacterium plated will represent a colony-forming unit. Bacteria growing in a liquid culture will generate colonies.
Answer:
Bacteria growing in a liquid culture will generate colonies.
Explanation:
The false statement about bacterial growth is that bacteria growing in a liquid culture will generate colonies. Bacteria do not form colonies in a liquid culture, but rather grow dispersedly. They only form colonies when grown on a solid medium like agar.
Explanation:The statement that is false about bacterial growth among the options provided is: 'Bacteria growing in a liquid culture will generate colonies'. This statement is incorrect because bacteria grow dispersedly in a liquid culture medium and do not form colonies. They only form colonies when grown on a solid medium such as agar. Agar is indeed used as a solidifying agent in some types of media, and a turbid, or clouded, culture is an indication of bacterial growth. Also, each bacterium plated in a solid medium will usually represent a colony-forming unit.
https://brainly.com/question/37562011
#SPJ3
GENETICS VOCABULARY QUIZ
The color of dog fur is inherited. Fur can be black (B) or white (b).
The father dog has black hair (BB), and the mother dog has white hair
(bb).
1. What is the trait that this story is talking about?
2. What are the alleles for this trait?
3. Which allele is dominant?
1. The trait that this story is talking about is the color of dog fur.
2. The alleles for this trait are B (for black fur) and b (for white fur).
3. The allele B is dominant.
1. The trait in question is directly stated in the scenario provided: it is the color of the dog's fur. This is the characteristic being inherited and displayed by the dogs in the story.
2. Alleles are different versions of the same gene. In this case, the gene responsible for fur color has two alleles: B, which codes for black fur, and b, which codes for white fur. These alleles are what the dogs' offspring can inherit from their parents.
3. Dominance refers to the ability of one allele to mask the presence of another allele in the phenotype of an organism. In the given scenario, the father dog has black hair and is homozygous for the black fur allele (BB), meaning he has two copies of the B allele. The mother dog has white hair and is homozygous for the white fur allele (bb), meaning she has two copies of the b allele. Since the offspring of these two dogs could have black fur (if they inherit at least one B allele), but not white fur unless they inherit two b alleles, we can conclude that the B allele is dominant over the b allele. This is because the dominant allele (B) will result in a black fur phenotype even if only one copy is present, while the recessive allele (b) will only result in a white fur phenotype if two copies are present.
In 1814, 15 British colonists founded a settlement on Tristan da Cunha, a group of small islands in the Atlantic Ocean, midway between Africa and South America. One of the early colonists apparently carried a recessive allele for retinitis pigmentosa, a progressive form of blindness that afflicts homozygous individuals. Of the founding colonists' 240 descendants on the island in the late 1960s, 4 had retinitis pigmentosa (rr). The frequency of the allele that causes this disease is ten times higher on Tristan da Cunha than in the populations from which the founders came. Calculate the frequency of the r allele in the original population of 15 colonists and in the 240 descendants. Please show calculations, Thanks!
Answer:
Frequency of the allele "r"causing the disease on Tristan da Cunha[tex]= 1.67[/tex] %
Frequency of the allele "r"causing the disease in the original population of 15 colonists [tex]= 16.7[/tex]%
Explanation:
Frequency of the allele "r"causing the disease on Tristan da Cunha
Given -
Out of 240 descendants on the island, 4 had retinitis pigmentosa (rr).
As per Hardy Weinberg's equllibrium equation
The frequecy of recessive individual in a given population is represented by [tex]q^2[/tex]
And q represents the frequency of allele r
So, in this case q is equal to
[tex]\frac{4}{240} * 100\\[/tex]
[tex]= 1.67[/tex] %
Frequency of the allele "r"causing the disease in the original population of 15 colonists
As it is given in the question statement , the frequency of allele "r"causing the disease in the original population of 15 colonists is ten times the frequency of the allele "r"causing the disease on Tristan da Cunha
i.e
[tex]1.67 * 10\\[/tex]
[tex]= 16.7[/tex]%
You argue with your boss, saying that your knowledge of biochemistry tells you that this mutant strain will not be viable. Your boss tells you that the mutant will grow aerobically (in the presence of oxygen) but that it will not be able to grow on glucose anaerobically (in the absence of oxygen). You reconsider, and decide that your boss is correct.
Answer:
As the given equation is correct,hence the boss is right,glucose utilization are therefore not able to grow ... and molecular biology to create new mutants.
The correct conclusion is that the mutant strain will not be viable under anaerobic conditions when glucose is the only energy source available.
In biochemistry, the viability of an organism, particularly a microorganism, is often linked to its ability to metabolize nutrients to produce energy. Aerobic respiration is a metabolic process by which cells use oxygen to convert glucose into energy. In contrast, anaerobic respiration is a process by which cells convert glucose into energy without the use of oxygen.
The information provided indicates that the mutant strain can grow aerobically, which means it can use oxygen to metabolize glucose or other substrates to produce energy. However, the mutant strain cannot grow on glucose anaerobically. This suggests that the mutant strain lacks the necessary biochemical pathways to metabolize glucose in the absence of oxygen.
In many organisms, particularly facultative anaerobes, the ability to switch between aerobic and anaerobic respiration is crucial for survival in environments where oxygen levels may fluctuate. For example, in the absence of oxygen, these organisms can ferment glucose to lactate or ethanol, or use alternative electron acceptors in anaerobic respiration.
If the mutant strain cannot grow anaerobically on glucose, it means that it cannot produce energy from glucose without oxygen. This would severely limit the mutant strain's viability in anaerobic environment where glucose is the primary or only available energy source. Therefore, the mutant strain would not be considered viable under those specific conditions, supporting the boss's assertion.
In summary, while the mutant strain can survive and grow in the presence of oxygen, its inability to metabolize glucose anaerobically would render it non-viable in anaerobic environments where glucose is the sole energy source. This conclusion is consistent with fundamental principles of microbial metabolism and the requirements for cellular energy production."
Label the parts in the diagrams (Plant Cell and Animal Cell) Please help!
Answer:
the ovals are mitochondria, the fat outer layer on the plant cell is a cell wall, the blank stuff is cytoplasm, the round things are nuclei, the squiglly stuff inside the nucleus is DNA, the thin outer later inside the cell wall on the plant but on the outside of the animal is the cell membrane
Explanation:
Can anyone help with this question for me
Answer:
Chameleon is the secondary consumer.
Different cloning vectors are able to accept inserts of different sizes. Small plasmid vectors, for example, can clone insert fragments up to about 20,000 bp long. The Human Genome Project scientists also used large plasmid vectors call Bacterial Artificial Chromosomes (BACs) that could clone fragments of the human genome 100,000 - 200,000 base pairs long. Which of the following statements is true about genomic libraries made in small plasmids vs. BAC vectors?
a. A complete genomic library made in small plasmid vectors will not contain overlapping recombinant clones: a complete genomic library made in BAC vectors will contain overlapping clones and this is necessary regardless of insert size.
b. A complete genomic library made in small plasmid vectors will contain more independent recombinant clones than a complete genomic library made In BAC vectors.
c. It is impossible to generate a complete genomic Runty in small plasnud vectors because it would require mote genomic DNA fragments than d Is possible to obtain
Answer:
A) A complete genomic library made in small plasmid vectors will contain more independent recombinant clones than a complete genomic library made In BAC vectors.
Explanation:
Different cloning vectors are able to accept inserts of different sizes. Small plasmid vectors, for example, can clone insert fragments up to about 20,000 bp long. The Human Genome Project scientists also used large plasmid vectors call Bacterial Artificial Chromosomes (BACs) that could clone fragments of the human genome 100,000 - 200,000 base pairs long. Which of the following statements is true about genomic libraries made in small plasmids vs. BAC vectors because A complete genomic library made in small plasmid vectors will contain more independent recombinant clones than a complete genomic library made In BAC vectors.
The refractory period is a:
a. postorgasmic period when females are relatively unresponsive to sexual stimulation.
b. postorgasmic period when males are relatively unresponsive to sexual stimulation.
c. preorgasmic period when males are relatively unresponsive to sexual stimulation.
d. prepubescent period when males and females are relatively unresponsive to sexual stimulation.
Final answer:
The correct answer is b. postorgasmic period when males are relatively unresponsive to sexual stimulation. The refractory period is a postorgasmic period when males are relatively unresponsive to sexual stimulation, occurring during the resolution phase of the sexual response cycle.
Explanation:
The refractory period is a time following an orgasm during which an individual is incapable of experiencing another orgasm. Specifically, in males, it is the time after ejaculation when they are unresponsive to sexual stimuli. According to Masters and Johnson's four phases of the sexual response cycle, which include excitement, plateau, orgasm, and resolution, the refractory period occurs during the resolution phase. The duration of the refractory period can vary, ranging from minutes to hours, and tends to increase with age. While both sexes experience orgasm, the refractory period is particularly noted in males, as they cannot maintain an erection or ejaculate during this time.
Which insect species is most distantly related to S. browderus?
Answer:
C. phillipensis
Explanation:
An enhancer may increase the frequency of transcription initiation for its associated gene when… (indicate true or false for each statement and explain your answer…) A. …it is located 1000 nucleotides upstream of the gene’s core promoter. B. ...it is located 1000 nucleotides downstream of the gene’s core promoter. C. …it is in the gene’s coding region.
Answer:
it is located 1000 nucleotides upstream of the gene’s core promoter - true
it is located 1000 nucleotides downstream of the gene’s core promoter- true
it is in the gene’s coding region - False
Explanation:
These enhancers are located 50 or more kilobases from the promoter they controlled upstream from a promoter, downstream from a promoter within an intron, or even downstream from the final exon of a gene which can be thousands of bp away from the gene's core promoter and can also occur thousands of nucleotides away from the gene's core promoter needing the activity of a DNA -bending protein that binds to the enhancer changing the shape of the DNA and allow interactions between the activators and transcription factors.
Question 19 of 20 The complete oxidation of one glucose molecule yields 30 or more ATP . Glucose catabolism includes glycolysis, pyruvate oxidation, and the citric acid cycle. The total yield of ATP includes ATP , GTP , and reduced cofactors that yield ATP from the electron transport chain and oxidative phosphorylation. Which processes yield the most ATP ? When determining the ATP yield for each process, include ATP derived from reduced cofactors. glycolysis citric acid cycle oxidation of pyruvate to acetyl‑ CoA
Answer:
Oxidative phosphorylation is a process which produces most of ATP that is produced during cellular respiration. Glycolysis produces 2 ATP molecules. Citric acid cycle produces 2 ATP molecules.
Explanation:
Oxidative phosphorylation process is also known as electron transport chain. This process produces ATP in large amount i. e. 28 ATP molecules. This process only occurs in aerobic respiration because this process requires oxygen while glycolysis and citric acid cycle produces only 2 ATP molecules.
a change in allele frequency due to random events is called?
Splenda is a fat replacer found in chips and pack foods.
True
O False
Answer:
false
Explanation:
it replaces sugar, not fat
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some populations along the coast of British Columbia. Kermit Ritland and his colleagues determined that white coat color in these bears results from a recessive mutation (G) caused by a single nucleotide replacement in which guanine substitutes for adenine at the melanocortin 1 receptor locus (mcr1), the same locus responsible for red hair in humans (K. Ritland, C. Newton, and H. D. Marshall. 2001. Current Biology 11:1468–1472). The wild-type allele at this locus (A) encodes black or brown color. Ritland and his colleagues collected samples from bears on three islands and determined their genotypes at the mcr1 locus. (Section 25.2) Genotype Number AA 42 AG 24 GG 21 a. What are the frequencies of the A and G alleles in these bears? b. Give the genotypic frequencies expected if the population is in Hardy–Weinberg equilibrium. c. Use a chi-square test to compare the number of observed genotypes with the number expected under Hardy–Weinberg equilibrium. Is this population in Hardy–Weinberg equilibrium? Note that DF =1. Show Chi Square value, DF, P value, and interpretation.
Answer and Explanation:
a) Frequencies of A and G lalleles are as follows:
f(A) =( 84+24)/174 = 0.62
f(B) = (42+24)/174 =0.38
b) Expected genotype frequencies:
f(AA) = (0.62) (0.62) = 0.384
f (AG)= 2(0.62) (0.38) =0.471
f(GG) = (0.38) (0.38) = 0.144
c) Genotype Observed Expected O-E (O-E)2 (O-E)2/E
AA 42 33 9 81 2.45
AG 24 41 17 289 7.05
GG 21 13 8 64 4.92
Chi squared = 14.42
The number of degrees of freedom is the number of genotypes minusthe number of alleles= 3-2 =1
The p value is much less than 0.05, therefore we reject the hypothesis that these genotype frequencies may be expected from HArdy Weinberg equilibrum
Club foot is one of the most common congenital skeletal abnormalities, with a worldwide incidence of about 1 in 1000 births. Both genetic and nongenetic factors are thought to be responsible for club foot. C. A. Gurnett et al. (2008. American Journal of Human Genetics 83:616–622) identified a family in which club foot was inherited as an autosomal dominant trait with reduced penetrance. They discovered a mutation in the PITXI gene that caused club foot in this family. Through DNA testing, they determined that 11 people in the family carried the PITXI mutation, but only 8 of these people had club foot. What is the penetrance of PITXI mutation in this family? Please round to two decimal places (e.g. 0.01).
Answer:
Penetrance might be characterized as the part of the predetermined genotypes populace that shows the normal phenotype.
Here, total number of genotypes (PITXI transformation observed) = 11
Genuine influenced phenotypes = 8
Penetrance = (Watched number of club foot occasions) / (all out number of people with PITXI transformation)
Penetrance = 8/11
Penetrance = 0.72
What are thought to have been present before vertebrates.
2. Cladograms are graphic representations of evolutionary history, which is called They
are sometimes referred to as phylogenetic trees.
3. Each node, or intersection, on a cladogram represents a/n_ _between two species.
4. Traits, or characteristics, that organisms develop and are passed down to become new
species are called traits.
5. Traits or structures that likely developed from common ancestors are called
structures.
6. Traits or structures that have a similar function, or job, but are not shared due to common
ancestry are called structures
7. Primates are a group of animals that have developed many adaptations such as larger brains,
binocular vision and thumbs that support arboreal life.
8. New World monkeys differ from Old World monkeys because they have which act as
additional hands when living in the trees.
9. is an early australopithecine skeleton, found in 1974.
10. Homo is not thought to have evolved into Homo sapiens. The two are now thought to
have been present at the same time as sister species.
Answer:
1. Chordates
2. phylogeny
3. most recent common ancestor
4. derived
5. homologous
6. analogous
7. opposable
8. prehensile tails
9. Lucy
10. neanderthalensis
11. Using comparative anatomy, scientists identify similarities and differences in the anatomy of different species. Scientists would search for homologous structures, analogous structures, and vestigial structures to provide clues as they construct a cladogram.
Explanation:
From Penn Foster
The study of a living being is called biology.
The correct answer is as follows:-
Chordates are thought to have been present before vertebrates. Cladograms are graphic representations of evolutionary history, which is called They phylogeny are sometimes referred to as phylogenetic trees Each node, or intersection, on a cladogram, represents the most recent common ancestor between two species Traits, or characteristics, that organisms develop and are passed down to become new species are called derived traits. Traits or structures that likely developed from common ancestors are called similar structures. Traits or structures that have a similar function, or job, but are not shared due to common ancestry are called different structuresopposable Primates are a group of animals that have developed many adaptations such as larger brains, binocular vision and thumbs that support arboreal life.New World monkeys differ from Old World monkeys because they have prehensile which act as additional hands when living in the trees. lucy is an early australopithecine skeleton, found in 1974. neanderthalensis is not thought to have evolved into Homosapiens. The two are now thought to have been present at the same time as sister species.
Hence, these are the answer to the following question.
For more information, refer to the link:-
https://brainly.com/question/20293447
One form of chromatin modification is acetylation, which is known to occur on positively charged histone tail amino acids, thus neutralizing the charge. Based on your understanding of chromatin packaging, we would expect this to (INCREASE/DECREASE/NOT IMPACT) DNA compaction. Which of the following statements best explains your answer?
A. The tails do not interact with the DNA.
B. The tails would now be unable to link together two nucleosomes.
C. The tails would now more tightly interact.
Answer:
A. The tails do not interact with the DNA
Explanation:
The acetylation refers to the transfer of the acetyl group from Acetyl-CoA to the N-terminal of the histone protein.
Lysine residues (positively charged amino acid) are present at the end of the N-terminal of the histone protein which is neutralized by the acetyl group.
This loses the compaction between the positively charged histone and the negatively charged DNA and the DNA becomes more relaxed. This relaxed state allows the transcription factors to easily bind the DNA and therefore the DNA becomes transcriptionally active.
Thus, Option-A is correct
Can anyone help me with this please
Answer:
Explanation: the organism is good at the function it serves in it's habitat
In the countercurrent exchange system of fish gills, blood and water flow in opposite directions. blood flow in the gills reverses direction with every heartbeat. blood and water are separated by a thick polysaccharide barrier. blood and water flow in the same direction.
Answer:
Topic:
Explanation:
What:
Final answer:
In fish, blood and water flow in opposite directions in a countercurrent exchange system for efficient gas exchange. Blood flows through the gills picking up oxygen, then through the body, and the single circuit heart pumps deoxygenated blood back to the gills.
Explanation:
In the countercurrent exchange system of fish gills, blood and water flow in opposite directions to maximize gas exchange efficiency. The blood flows through the gills, passing over deoxygenated veins first, where it picks up oxygen from the water. This system, known as gill circulation, is unidirectional, with the oxygenated blood then flowing to the rest of the body. Unlike in mammals, fish have a single circuit for blood flow and a two-chambered heart, comprising only one atrium and one ventricle. The oxygenated blood flows past the organs and the rest of the body, delivering oxygen before returning to the heart, a process called systemic circulation. The large surface area of the gills, due to their folded structure, is paramount for this efficient gas exchange, where oxygen molecules diffuse from areas of high concentration in the water to low concentration in the blood.
In an insect,
a. nutrients are circulated through an open circulatory system.
b. the blood vessels do not form a continuous circuit
c. the fluid flowing through the heart is the same fluid that bathes the tissues.
d. the circulatory system is responsible for exchange of respiratory gases
Answer:
In an insect, A. nutrients are circulated through an open circulatory system.
Explanation:
Insect respiration is independent of its circulatory system, that's way the blood does not play a direct role in oxygen transport and the circulatory system is responsible for exchange of respiratory gases. Insects also have an open circulatory system as opposed to our closed circulatory system.
What is the definition of bioarchaeology? Group of answer choices the study of all animal skeletal remains from archaeological sites the application of skeletal analysis to assist in legal investigations the study of the cultural life of living people the study of human skeletal remains from archaeological sites
Answer: bio archaeology:
the study of bones and other biological materials found in archaeological remains in order to provide information about human life or the environment in the past:
Explanation:
How do changes in the organism in a food chain affect the other organisms?
Answer:
Explanation: The orgaism can be an invasive species it would effect the amount of large or small organisms.things like elk that eat a large amount of certain plants can affect the other organisms source of food.Thats why their hunted by wolves.All the pieces matter.
Changes in a food chain can disrupt energy flow and nutrient cycles, with effects that can ripple through an ecosystem, potentially leading to biodiversity loss or collapse.
If a primary consumer decreases in population, this can lead to overgrowth of plants and the undernourishment of secondary consumers. Conversely, if a keystone species is removed or significantly altered, it can cause a ripple effect, disrupting the predatory and competitive relationships in an ecosystem, potentially leading to biodiversity loss or ecosystem collapse. The delicate balance of energy flow and nutrient cycles in an ecosystem makes it sensitive to changes in any single organism or group within a food chain, demonstrating the interconnectedness of all living things.
Determine if the limiting factors listed below are density-dependent or density-independent.
industrial pollution
habitat
food
a hurricane
number of mates
hunting by humans
industrial pollution independent
habitat dependent
food dependent
a hurricane independent
number of mates dependent
hunting by humans independent
The limiting factors industrial pollution and a hurricane are density-independent as they affect populations regardless of density; while habitat, food, and number of mates are density-dependent due to their reliance on population size. Hunting by humans can be considered both but is typically classified as density-independent.
Explanation:To determine if the limiting factors listed below are density-dependent or density-independent, we have to understand how they affect a population's mortality rate based on the population's density. Density-dependent factors usually increase in intensity as the population density increases and are mostly biotic such as predation, competition, and diseases. On the other hand, density-independent factors affect mortality at all densities and are typically abiotic like natural disasters or pollution.
Industrial pollution - Density-independent. It affects populations regardless of their density.Habitat - Density-dependent. It influences population based on how many individuals share the space.Food - Density-dependent. Food scarcity affects populations more when densities are higher due to competition.A hurricane - Density-independent. It can impact populations irrespective of their size.Number of mates - Density-dependent. Availability of mates affects reproductive success and varies with population density.Hunting by humans - This can be both, but generally considered density-independent as hunting can occur regardless of the population size, although at high densities, there may be more targets.Learn more about Density-Dependent vs. Density-Independent Factors here:https://brainly.com/question/29521423
#SPJ3
What morphology is represented in the picture? A) spirilla B) rod shaped C) filamentous D) cocci
Hi there!
This bacteria's morphology is rod-shaped. The flagella can be confusing and cause you to choose filamentous, but filamentous bacteria are simple strands. Spirilla bacteria are spiral shaped, and cocci bacteria are spherical.
I really hope this helps!!
The morphology of bacteria represented in the picture is rod shaped. Thus, option B is correct.
What is morphology?Morphology is the distinguishing characteristic of bacterial cell, which describes about the shape of the cell. Particular species of bacteria has particular morphology, hence it is the characteristic feature to distinguish between different bacterial species.
These morphologies are examined with the help of microscope. The basic morphologies are:
coccusbacillifilamentousspirochetevibrioCocci are spherical shaped cells which can be coccus (singular), diplococcus, sarcina, tetrad, stephylococcus, streptococcus based on number of cells and their arrangements.
Bacilli are rod shaped cells which can be bacillus (singular), cocobacillus, diplobacillus, streptobacillus, palisade. Therefore in the given picture, the morphology of cell is rod shaped. hence option B is correct.
Learn more about morphology, here:
https://brainly.com/question/8282896
#SPJ3
what is a 1:2:1 phenotypic ratio in the F2 generation of a mono hybrid cross is a sign of?
Incomplete dominance
1. If the ability to taste PTC were controlled by only two alleles: one dominant (T) and one recessive (t), would there be any way to distinguish between the heterozygous (Tt) individuals and the homozygous dominant (TT) individuals without mating or performing DNA analysis? Explain your answer.
Answer:No
Explanation: there would not be a way to distinguish between Tt and TT without mating or DNA analysis because T is dominant in Tt, therefore has the same physical characteristics as TT.
Summarize the development of a calf fetus during gestation.
Final answer:
The development of a calf fetus during gestation involves the growth from a zygote to a fetus, organ and structure development, and rapid growth in the third trimester with critical development in the brain, liver, and ossification process.
Explanation:
Fetal Development During Gestation
The development of a calf fetus during gestation is a complex process that starts with a zygote and progresses through the embryonic stage to form a fully developed fetus. During the first trimester, major organs begin to form, the backbone, muscles, and bone tissue start to develop, and the fetus will be able to move its small limbs. By the end of the second trimester, the fetus grows to about 30 cm (12 inches) and becomes active, with the mother typically feeling movements. The placenta now fully takes over nutrition and waste management and hormone production. In the third trimester, rapid growth occurs, and the fetus reaches 3 to 4 kg (6½ -8½ lbs.) and 50 cm (19-20 inches) in length. Organ development continues up to and after birth, with significant growth of the nervous system and liver.
Key Developments:
The brain continues to expand.The ossification process replaces cartilage with bone.The liver begins to secrete bile, and bone marrow starts erythrocyte production.The development of the amniotic sac and umbilical cord.Rapid growth and development of organs in the last trimester.The interaction between DNA and histone proteins (forming nucleosomes) plays a key role in the regulation of gene expression in eukaryotes and is a potential target mechanism in drug discovery. You are testing a drug which blocks the activity of histone acetyltransferases (HATS) in cancer cells. In general, what do you expect to happen with regards to chromatin structure and gene expression in cells treated with the drug? Please select the most correct answer.
A. The chromatin near cis-regulatory sequences will be more closed and there will be less transcription.
B. The chromatin near cis-regulatory sequences will be more open and there will be more transcription.
C. The chromatin near cis-regulatory sequences will be more open and there will be less transcription.
D.The chromatin near cis-regulatory sequences will be more closed and there will be more transcription.
E. There should be no effect on chromatin structure or gene transcription. HATs only function in the packaging of DNA prior to mitosis.
Answer:
A. The chromatin near cis-regulatory sequences will be more closed and there will be less transcription.
Explanation:
In the presence of histones, the cis-regulatory sequences of DNA like promoter, enhancers etc. are not exposed. The function of the histone acetyltransferases (HATS) is to cause chromosome decondensation i.e. removal of histones from the DNA so that transcription of the DNA could occur. Histone acetyltransferases (HATS) cause acetylation of lysine amino acid of the histone proteins. Acetyl group is negatively charged so the acetylation of histone proteins leads to the removal of their positive charge which ultimately leads to the decrease in the interaction between N terminal of histones and negatively charged phosphate group of the DNA molecule. As soon as histones are removed from the DNA where cis-regulatory sequences are located, the DNA becomes accessible for transcription.
But here a drug has been added which blocks the activity of histone acetyltransferases (HATS) in cancer cells. So it is quite evident that in these cells, histones will not get removed from the cis-regulatory sequences of DNA so the DNA will be more closer or tightly packed as a result of which less transcription will occur.
True or false 100% of what you look like or behave like come directly from both of your parents?
Answer:
False.
Explanation:
The way you behave does not usually come from either of your parents, you chose how you act. And how you look can could from your dad, mom, both or sometimes other ancestors. So it definitely False.
Steps in protein synthesis Step order : initiation complex formed with addition of large ribosomal subunit translocation small ribosomal subunit binds to mRNA ribosomal subunits dissociate ribosome reads a stop codon codon recognition (non-initiating site) transcription of mRNA from DNA polypeptide chain is released from the P site peptide bond formation
Answer:
1. transcription of mRNA from DNA
2. small ribosomal subunit binds to mRNA
3.initiation complex formed with addition of large ribosomal subunit
4.codon recognition (non-initiating site)
5.peptide bond formation
6.translocation
7. ribosome reads a stop codon
8.polypeptide chain is released from the P site
9. ribosomal subunits dissociate
Explanation:
The translation is a process which translates the nitrogenous bases or codons in the proteins.
The process of translation requires the mRNA, tRNA and ribosome and proceeds in three stages: the initiation, elongation and the termination.
The process begins with the binding of the small subunit of the ribosome to the mRNA. The charged tRNA with a first amino acid called methionine binds the mRNA and scans the mRNA until it finds the start codon.
After it finds the start codon, the large subunit complex binds the mRNA and form initiation complex. After this, the amino acid enters the P-site of the ribosome where elongation of peptide takes place.
The peptide then exits from the E-site and the ribosome dissociates.
Protein synthesis begins with the formation of an initiation complex involving the small ribosomal subunit, mRNA template, initiation factors, and initiator tRNA. The initiator tRNA has anticodon UAG and interacts with the start codon AUG, carrying the amino acid methionine.
Explanation:Protein synthesis begins with the formation of an initiation complex. This complex involves the small ribosomal subunit, the mRNA template, initiation factors, and a special initiator tRNA, called tRNAMet. The initiator tRNA has anticodon UAG, which interacts with the start codon AUG and is charged with the amino acid methionine. Methionine is therefore the first amino acid of every polypeptide chain.
Learn more about Protein synthesis initiation here:https://brainly.com/question/31033742
#SPJ3