Answer:
Endosymbiont theory was proposed by Lynn Margulis in 1967 which suggested the origin of Chloroplast and mitochondria in eukaryotic cells.
The theory suggested that chloroplast and mitochondria organelle originated by engulfing the cyanobacteria and proteobacteria by other prokaryotic bacteria.
This shows that Prokaryotes exhibit the following characters:
1. These were the first organisms on Earth
2. Lack membrane-bounded
3. Aerobic bacteria are these types of cells
Eukaryotes exhibit:
1. Have mitochondria
2. Includes organisms that carry out aerobic respiration
3. Can be a multicellular organism
4. Includes organisms that are photosynthetic
5. Might have chloroplasts
Please match the Key Word to the correct definition.
Column A Column B
1. ___ chemical reaction a. element or compound produced by a chemical reaction
2. ___ endergonic reaction b. a chemical reaction that releases energy
3. ___ exergonic reaction c. element or compound that enters into a chemical reaction
4. ___ Photosynthesis d. a process that transforms two different sets of chemicals into a new set of chemicals when they come in contact with one another
5. ___ Product e. a chemical reaction that requires an input of energy
6. ___ Reactant f. process used by organisms to break down high energy molecules to capture the energy in a form that can be used to power metabolic reactions
7. ___ respiration g. process used by plants and other autotrophs to capture light energy and use it to power chemical reactions
Answer:
a. element or compound produced by a chemical reaction : 5.Product b. a chemical reaction that releases energy : 2.Endorgenic reaction c. element or compound that enters into a chemical reaction: 6.Reactantd. a process that transforms two different sets of chemicals into a new set of chemicals when they come in contact with one another : 1.Chemical reactione. a chemical reaction that requires an input of energy: 3.Exorgenic reaction f. process used by organisms to break down high energy molecules to capture the energy in a form that can be used to power metabolic reactions: 7.Respiration g. process used by plants and other autotrophs to capture light energy and use it to power chemical reactions: 4.PhotosynthesisThe matched pairs are: 1) d. a process that transforms two different sets of chemicals into a new set of chemicals when they come in contact with one another - chemical reaction, 2) e. a chemical reaction that requires an input of energy -endergonic reaction, 3) b. a chemical reaction that releases energy - exergonic reaction, 4) g. process used by plants and other autotrophs to capture light energy and use it to power chemical reactions - Photosynthesis, 5) a. element or compound produced by a chemical reaction - Product, 6) c. element or compound that enters into a chemical reaction - Reactant, 7) f. process used by organisms to break down high energy molecules to capture the energy in a form that can be used to power metabolic reactions - respiration.
Let's look at each term and its corresponding definition:
Chemical reaction: Matching: d. a process that transforms two different sets of chemicals into a new set of chemicals when they come in contact with one another
Endergonic reaction: Matching: e. a chemical reaction that requires an input of energy
Exergonic reaction: Matching: b. a chemical reaction that releases energy
Photosynthesis: Matching: g. process used by plants and other autotrophs to capture light energy and use it to power chemical reactions
Product: Matching: a. element or compound produced by a chemical reaction
Reactant: Matching: c. element or compound that enters into a chemical reaction
Respiration: Matching: f. process used by organisms to break down high energy molecules to capture the energy in a form that can be used to power metabolic reactions
A G:T base pair in DNA suggests that _____
(A) The T was originally a cytosine, which was deaminated to a T
(B) The T was originally a 5-methyl cytosine, which was deaminated to a T
(C) The T was originally a 5-methyl cytosine, which was oxidized to a T
(D) The G was originally an A, which was methylated to form an G
Answer:
(A) The T was originally a cytosine, which was deaminated to a T
Explanation:
Adenine and guanine are found in both DNA and RNA.
Cytosine is found in both DNA and RNA.
Thymine is normally found in DNA.
In a CpG site (5'—C—phosphate—G—3') , that is, Cytosine and Guanine separated by only one phosphate group; phosphate links any two nucleosides together in DNA.
In a G:T base pair in DNA, Adenine pair with Thymine and Guanine pair with Cytosine.
Here in the question, The G:T base pair in DNA suggests that the T was originally a cytosine, which was deaminated to a T.
How methylation of CpG sites followed by spontaneous deamination leads to a lack of CpG sites in methylated DNA.
A Cytosine base followed immediately by a Guanine base (a CpG) is rare in vertebrate DNA because the Cytosine in such an arrangement tend to be methylated. This methylation helps to distinguish the newly synthesized DNA strand from the parent strand, which aids in the final stages of DNA proofreading after duplication. However, over time methylated cytosines tend to turn into Thymine because of spontaneous deamination.
A G:T base pair in DNA suggests that the Thymine was originally a 5-methyl cytosine, which has been deaminated to a Thymine.
Explanation:The correct answer to the question is (B) 'The T was originally a 5-methyl cytosine, which was deaminated to a T'. In normative conditions, Guanine (G) in DNA usually pairs with Cytosine (C), but the presence of a G:T pair suggests a mutation. This mutation generally occurs when a 5-methyl cytosine (a modified form of cytosine) undergoes deamination, a process where the amino group (-NH2) is removed, and the cytosine is converted into Thymine (T), thus, breaking the typical G:C pairing and resulting in a G:T pairing.
Learn more about DNA base pair mutation here:https://brainly.com/question/30875580
#SPJ3
The age of oceanic bedrock on either side of a mid-ocean ridge is supporting evidence that at the ridges, tectonic plates are _______.
The age of oceanic bedrock on either side of a mid-ocean ridge is supporting evidence that at the ridges, tectonic plates are Diverging .
Explanation:
The underwater mountain range is the mid oceanic ridge, which were formed as a result of plate tectonics. when the convection current goes up in the mantle present beneath the oceanic crust results in the uplifting of the ocean floor.
Which leads to the formation of the tectonic plates through magma, that meet at divergent boundaries. The density and the age and the thickness of these oceanic crust increases due to the distance from the ridges. The mid oceanic ridges has been split on two sides as matching stripes ,
Answer:
Diverging
Explanation:
Recall Mendel's pea plants: the allele for purple flowers, P, is dominant to the allele for white flowers, p. A very large number of offspring from a cross of two plants are observed. If ALL of those offspring have purple flowers, what are the possible genotypes of the parents in the cross?
A. PP × PP only
B. PP×PP, PP×Pp, or PP x pp
C. PP×PP, PP×Pp, PP x pp, or Pp x Pp
D. pp×pp only
E. Not enough information is given
Answer:
B. PP×PP, PP×Pp, or PP x pp
Explanation:
The allele for the purple flower (P) is dominant over the one for the white flowers (p). To be a purple-flowered plant, the progeny must have homozygous dominant (PP) or heterozygous dominant (Pp). A cross between two homozygous dominant purple-flowered parents (PP x PP) would produce all the homozygous dominant progeny (PP) having purple flowers.
A cross between a homozygous dominant (PP) and a heterozygous dominant (Pp) parent plant would produce homozygous and heterozygous dominant progeny in a 1:1 ratio (PP x Pp = 1 PP: 1 Pp). Similarly, a cross between a homozygous dominant (PP) and homozygous recessive (pp) parent plant would produce all heterozygous dominant (Pp) progeny with purple flowers.
Answer:
pp
Explanation:
An individual is infected with Mycobacterium tuberculosis. The bacteria infect macrophages in the lungs, where it survives and reproduces intracellularly. What is the most likely mechanism the infected macrophages will use to combat the invading bacteria?
Answer:
Phagolysosomal fusion
Explanation:
Macrophages ingest the bacteria through a process called phagocytosis. The phagosome undergoes a series of acidification steps where it fuses with endosomal compartments and encounters bactericidal and antigen processing molecules. Ultimately, it will fuse with the lysosome (phagolysosomal fusion) where it encounters more acidification, hydrolases and bactericidal molecules which may kill the bacteria.
In humans, MITOSIS directly accomplishes all of the following EXCEPT
A. growth
B. production of 4 haploid gametes from a single diploid parent cell
C. repair of damaged tissues
D. development of organs
E. production of 2 diploid daughter cells from a single diploid parent cell
Answer:
The correct answer will be option-B
Explanation:
A cell divides its nuclear content and cytoplasmic content by one of the two ways that are mitosis and meiosis.
The process of mitosis takes place in the somatic cells of the organism where mitosis divides the parent cell into two genetically similar daughter cells. The type of division is involved in the growth, development and repair of the damaged tissues and organs.
Since mitosis does not form four haploid cells from parent cells but forms two daughter cells therefore, Option-B is the correct answer.
Mitosis aids in growth, tissue repair, organ development, and production of 2 diploid daughter cells from a single parent cell. However, it does not directly accomplish producing 4 haploid gametes from a single diploid parent cell, as it is achieved by meiosis.
Explanation:In the process of mitosis, the cells undergo various stages that lead to the creation of duplicate cells for specific purposes. It plays an essential role in the growth, repair of damaged tissues, development of organs, and the production of 2 diploid daughter cells from a single diploid parent cell. However, the production of 4 haploid gametes from a single diploid parent cell is not directly accomplished by mitosis. This is done through another cell division process known as meiosis, which specifically aims to create cells for sexual reproduction.
Learn more about Mitosis and Meiosis here:https://brainly.com/question/31658273
#SPJ6
Which of the following is not an example of adaptation? A) A plant population that is very drought-resistant due to the action of natural selection B) The process by which a plant population becomes drought-resistant C) The increased ability of an individual plant (above baseline) to withstand further drought after it has received a heat shock, which causes the expression of specific proteins that enable more efficient use of water D) A mammal species that has evolved an increased ability to store water through the operation of natural selection E) A mammal species that has evolved measures to use water more efficiently through the operation of natural selection
Answer:
C) The increased ability of an individual plant (above baseline) to withstand further drought after it has received a heat shock, which causes the expression of specific proteins that enable more efficient use of water
Explanation:
In biology, adaptations refer to the characteristics of organisms that allow them to survive and reproduce better in their environment than if they did not possess them. Many of these adaptations are very easy to recognize, such as bird beaks that are highly specific to the food they eat, for example. It is now accepted by science that only natural selection can consistently produce adaptations, although it is important to note that it is not the only evolutionary mechanism. Natural selection, whose idea is mainly attributed to Charles Darwin, acts directly on the phenotypic characteristics of individuals in a population, favoring those who are most likely to survive and reproduce in a given environment over those who are less adapted.
Based on this, we can conclude that adaptation is a biological process that occurs without human manipulation. Thus, we can conclude that among the options given in the above question, the letter C does not represent an example of adaptation. This is because the letter C, shows a plant that showed characteristics favorable to its survival in an inhospitable environment, after human intervention, which, through a thermal shock, caused the expression of specific proteins for that characteristic.
After graduating from college, you decide to put your biology skills to work at a local company that does genetic counseling. Your first case is working with a couple that is trying to decide if it would be wise to conceive a child given the family’s genetic history with Huntington’s disease. This is a very damaging neurological disorder that usually strikes individuals later in life. The Huntington’s disease allele is a dominant allele (represented as "H"). Individuals who will not develop Huntington’s disease carry two copies of the recessive allele (represented as "h"). After testing both the wife and the husband, you determine that the wife’s genotype is hh and the husband’s genotype is Hh.
a)Which of them will eventually develop Huntington’s disease?
b)What are the possible genotypes for their children? What is the chance that their child will inherit Huntington’s disease?
Answer:
The correct answers are a) "the husband", b) The possible genotypes for their children are Hh and hh, while the child have a 50% chance of inherit Huntington’s disease.
Explanation:
Huntington’s disease is a very damaging neurological disorder that follows an autosomal dominant inheritance. This means that a healthy person has two copies of the recessive allele "h", while a person with Huntington’s disease has at least a copy of the dominant "H" allele. Since the husband genotype is Hh, he is going to develop the Huntington’s disease eventually. Also, it should be informed to the couple that their children could either have the Hh or the hh genotype with a 50% chance of having either of the genotypes, which means that their child has a 50% chance of developing the Huntington’s disease.
Final answer:
The husband with genotype Hh will develop Huntington's disease. Their children could inherit genotype Hh or hh, with a 50% chance of inheriting the Hh genotype and thus the disease.
Explanation:
In the scenario where a wife's genotype is hh and a husband's genotype is Hh, we can deduce the following:
a) Which of them will eventually develop Huntington’s disease?:
The husband will eventually develop Huntington's disease as he has one copy of the dominant allele H. Since Huntington's disease is an autosomal dominant disorder, only one copy of the dominant allele is required for the disease to be expressed.
b) What are the possible genotypes for their children? What is the chance that their child will inherit Huntington’s disease?:
The possible genotypes for their children are Hh and hh. There is a 50% chance that each child will inherit Huntington's disease -- that is, the child inherits the Hh genotype. This can be determined through a simple Punnett square, where one axis represents the husband's alleles (H and h) and the other axis represents the wife's alleles (h and h), leading to two possible offspring genotypes: Hh and hh.
Which of the following statements about the central dogma is correct? Drag "True" or "False" to the end of each statement. ResetHelp The central dogma predicts that mRNAs are transcribed into DNA. False Transcription is the process of using a single strand of DNA as a template to produce a complementary sequence of RNA. True The central dogma predicts that a change in a DNA sequence will result in a change in an RNA sequence. False The central dogma summarizes how information is transferred from DNA to RNA to protein in cells. True The arrows connecting DNA, RNA, and protein in the central dogma model indicate a conversion of one type of molecule to another. True
Answer:
The central dogma predicts that mRNAs are transcribed into DNA. FalseTranscription is the process of using a single strand of DNA as a template to produce a complementary sequence of RNA. TrueThe central dogma predicts that a change in a DNA sequence will result in a change in an RNA sequence. TrueThe central dogma summarizes how information is transferred from DNA to RNA to protein in cells. TrueThe arrows connecting DNA, RNA, and protein in the central dogma model indicate a conversion of one type of molecule to another. TrueExplanation:
The Central Dogma of Molecular Biology was postulated by Francis Crick in 1958. It explains how the flow of information from the genetic code occurs. This model mainly shows that a sequence of a nucleic acid can form a protein, but the opposite is not possible. According to this dogma, the flow of genetic information goes along the following lines: DNA → RNA → PROTEINS.
Since one molecule originates from the configuration of the other, if an error in the DNA sequence occurs, a change in the RNA sequence occurs, which in turn results in a change in protein.
According to the central dogma in DNA, where genetic information is contained, which can be transcribed into RNA molecules. In the transcription process, a DNA molecule serves as a template for the creation of an RNA molecule. It is in this RNA molecule that the code used to organize the amino acid sequence and to form proteins in the translation process is found. This process consists in the union of amino acids, obeying the order of codons presented in a messenger RNA.
The central dogma of biology states that genetic information is transferred from DNA to RNA to proteins, with the process of transcription playing a key role.
Explanation:The central dogma of biology describes how the information in genes is transferred from DNA to RNA, and then from RNA to proteins. It does not predict that mRNAs are transcribed into DNA. Instead, transcription involves a single strand of DNA serving as a template to produce a complementary sequence of RNA. A change in a DNA sequence may consequently result in a change in an RNA sequence, and thus, potentially a change in a protein sequence as well. The arrows linking DNA, RNA, and protein in the central dogma model symbolize this conversion process, or flow of information, from one molecule type to another.
Learn more about central dogma here:https://brainly.com/question/34186556
#SPJ6
GTP hydrolysis and whether GTP or GDP is bound to tubulin is an important mechanism to control the dynamic instability of microtubules. Certain aspects of dynamic instability can be viewed using GFP-EB1. Which process(es) is it useful for visualizing and why?
Choose one:
A. growing and shrinking microtubules, because EB1 binds to the GTP-tubulin cap on microtubules
B. shrinking microtubules, because EB1 binds to the GTP-tubulin cap on microtubules
C. growing microtubules, because EB1 binds to the GTP-tubulin cap on microtubules
D. growing and shrinking microtubules, because EB1 binds to the GDP-tubulin cap on microtubules
Answer:
Growing microtubules, because EB1 binds to the GTP-tubulin cap on microtubules
Explanation:
Growing microtubules, because EB1 binds to the GTP-tubulin cap on microtubules. This process is useful for visualization because EB1 or it homolog recognize the GTP structural cap of growing microtubule ends.
Suppose that a woman had to have part of her thyroid gland surgically removed. She would most likely suffer from a condition known as hypothyroidism due to too little thyroid function.
Predict how this woman's hypothyroidism would affect prolactin levels in her body.
Which choice describes how surgical hypothyroidism would likely affect prolactin levels?
Which choice describes how surgical hypothyroidism would likely affect prolactin levels?
1.Thyroid hormone levels decrease, TRH levels decrease, and PRL levels decrease.
2. Thyroid hormone levels decrease, TRH levels decrease, and PRL levels increase.
3. Thyroid hormone levels decrease, TRH levels increase, and PRL levels decrease.
4.Thyroid hormones levels decrease, TRH levels increase, and PRL levels increase.
5.Thyroid hormone levels increase, TRH levels increase, and PRL levels increase.
6. Thyroid hormone levels increase, TRH levels increase, and PRL levels decrease.
7. Thyroid hormone levels increase, TRH levels decrease, and PRL levels increase.
Answer:
4. Thyroid hormone levels decrease, TRH level increase, PRL level increase
Explanation:
Surgical removal of Thyroid gland will lead to hypothyroidism.
Normally, the surgery is followed by maintenance dose of thyroxine to avoid side effects.
In the presence of hypothyroidism, however, the decreased thyroid hormone will lead to increase in thyrotropin releasing hormone (TRH). increased TRH increases production of prolactin. (but less than that in prolactinoma)
Answer: hypothyroidism causes an elevation of TRH which causes an elevation of prolactin.
4
Explanation: hormone decrease leads to rapid decrease in T3 and T4 which would allow the thyroid regulatory hormone(TRH) TO increase due to the non existence of thyroid gland necessary for T3 and T4, and without the presence of thyroid hormones to counter the production of prolactin hence, the level of prolactin would increase.
Which of the following statement(s) is/are true of the kidney? A. Each kidney is approximately the same size as an adult’s fist. B. Is located posterior to the abdominal peritoneum. C. Are well protected by ribs 10 – 12. D. Is very poorly vascularized. E. A, B, and C are correct
Answer: Option E.
A,B and C are correct.
Explanation:
Kidney is a bean shaped organ normally found in vertebrates. It is located posterior to the abdominal peritoneum. Kidney is well protected by ribs 10-12 , abdominal muscles, back muscles .kidney is about 11cm in length in adults. Kidney receives blood from the arteries and exist blood through the veins. Each kidney is attached to a ureter which harbour and carries urine to the bladder .Each kidney is approximately the same size as adult clenched fist(10cm). The main function of the kidney is excretion of wastes and urine.
Phytoplankton form the base of food webs, but are only present in the upper few hundred meters of the ocean. A number of factors here can limit their growth and hence limit the amount of biomass in higher trophic levels.
Which of the items below is not one of these limiting factors to phytoplankton?
a. Nitrogen
b. Light
c. Phosphate
d. Oxygen
Answer:
The correct answer is d. Oxygen
Explanation:
Phytoplanktons are responsible for the fixation of approximately half of the global carbon therefore seawater has high CO2 concentration which is required by phytoplanktons to make their food.
They are the primary producers of oceans and they are responsible to support the food chain of oceans. Factors that can limit their growth are mainly sunlight and nutrients like phosphorus, nitrogen, etc.
As phytoplanktons are photosynthetic they release oxygen itself as a byproduct therefore oxygen is not a limiting factor to phytoplanktons. So the right answer is d.
HELP ASAP!!!!
Question 1 (1 point)
How do glaciers form?
Question 1 options:
Snow accumulates on the ocean and freezes to become an iceberg.
Snow accumulates on water but does not melt.
Snow accumulates on land and does not melt.
Snow accumulates on land and then melts over the summer.
Question 2 (1 point)
Glaciers can form in all these areas except
Question 2 options:
on top of a mountain
near the equator
in the middle of a continent
over water
Question 3 (1 point)
Metal stakes were placed on the surface of the glacier in a straight line from position A to position B. Which diagram best shows the position of the metal stakes years later?
Question 3 options:
A
B
C
D
Question 4 (1 point)
Which force is primarily responsible for the movement of a glacier?
Question 4 options:
gravity
ground water
running water
wind
Question 5 (1 point)
What is happening to most glaciers all over the world?
Question 5 options:
They are advancing.
They are retreating as ice melts.
They are moving backwards.
They are staying the same size.
Answer:
1.) snow accumulates on land and does not melt
2.) over water
3.) C
4.) gravity
5.)They are retreating as ice melts.
Explanation:
You insert a gene for tetracycline resistance into one plasmid and a gene for ampicillin resistance into another plasmid. You successfully introduce both plasmids into a sample of E. coli cells, but fail to grow any of them in culture medium with both antibiotics present in it. What could best explain the problem?
A. Random mutation has inactivated the antibiotic resistance genes
B. Plasmid incompatibility will not allow both plasmids to persist
C. E. coli cannot maintain two plasmids
D. The plasmid(s) have integrated into the bacterial chromosome
E. A phage has neutralized one of the plasmids
F. None of the above is correct
Answer:
B. Plasmid incompatibility will not allow both plasmids to persist
Explanation:
The two possibilities of why the plasmids are incompatible are:
1) Both plasmids contain the same origin of replicon and compete for the same Rep proteins.
2) Tetracycline: inhibits protein synthesis by binding and inhibiting ribosomal proteins, thus ampicillin target proteins may also be inhibited.
Ampicillin: Causes cell lysis which may lead to inactivation of tetracycline activity since tetracycline needs to diffuse through membrane porin channels prior binding and inhibiting ribosomal proteins.
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the order here is arbitrary). The F1 females are test crossed back to the triple-mutant males and the F2 phenotypic ratios are as follows: a^+ b c = 18 a b^+ c = 112 a b c = 308 a^+ b^+ c = 66 a b c^+ = 59 a^+ b^+ c^+ = 320 a^+ b c^+ = 102 a b^+ c^+ = 15 total = 1000 Map these gene to a chromosome in a correct order and determine the map distance between them. Show all your work.
Answer:
a is the middle gene.
Distance [b-a]= 24.7 mu
Distance [a-c]= 15.8 mu
Distance [b-a} = 40.5 mu
Explanation:
A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). The order of the genes here is arbitrary.
The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).
The F2 shows the following phenotypic ratios:
320 a⁺b⁺c⁺308 a b c 102 a⁺ b c⁺112 a b⁺ c66 a⁺ b⁺ c59 a b c⁺18 a⁺ b c 15 a b⁺ c⁺Total = 1000
The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.
Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals: a⁺b⁺c⁺ and abc.
The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.
1st. Determine the gene orderCompare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.
2nd Determine the single crossover gametes
The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).
The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.3) Calculate the recombination frequencies between genesRecombination frequency (RF) = #Recombinants/Total progeny
RF [b-a]= (102+112+18+15)/1000= 0.247RF [f-br]= (66+59+18+15)/1000= 0.1584) Calculate the distance in map unitsDistance (mu) = RF x 100
Distance [b-a]= 0.247 × 100 = 24.7 mu
Distance [a-c]= 0.158 × 100 = 15.8 mu
Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu
The gene map therefore looks like:b------------24.7 mu--------------------------a---------15.8 mu-----------c
The DNA in a cell's nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell.
For example, consider these two proteins:
1. Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during
glycolysis.
2. Insulin, a protein that regulates blood sugar levels, is secreted from specialized
pancreatic cells.
Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations.
For each protein, identify its targeting pathway: the sequence of cellular locations in which the protein is found from when translation is complete until it reaches its final (functional) destination. (Note that if an organelle is listed in a pathway, the location implied is inside the organelle, not in the membrane that surrounds the organelle.)
Options:
Cytoplasm only, ER --> cytoplasm, ER --> Golgi --> outside cell, cytoplasm --> ER --> outside cell, Golgi --> ER --> outside cell, cytoplasm --> Golgi --> outside cell, nucleus --> cytoplasm, ER --> Golgi --> cytoplasm
Protein Targeting Pathway
PFK _______________
Insulin ____________
Answer:
PFK (protein): cytoplasm only
Insulin (protein): ER->Golgi->outside cell
Explanation:
The protein Phophofructokinase (PFK) has to function inside the cytoplasm of the cell where it is made. Hence, it does not have to go through the modification process needed for transport. Such kind of proteins which are made and function in the same cell are translated on free cytoplasmic ribosomes.
The proteins which have to be transported to other areas like insulin are translated in the rough endoplasmic reticulum. From there, they travel to the Golgi complex where they are modified and packaged so that they can travel outside the cell.
If a small founder population of 12 people ends up on a small island, and one member of the population had a mutation that creates a new allele, what is the likelihood that the allele will be eliminated from the population? Enter your answer as a percentage (not decimal), omit the percent sign however.
Answer:
96
Explanation:
This is genetic drift question, we can tell because the variable here is the population size. The probability of an allele of getting fixed is determined by its initial frequency, which is actually the probability of the allele of being fixed. So we need to calculate the initial allele frequency of this mutation:
Mutation allele frequency = # mutation alleles / total # of alleles = 1/24 = 0.04%
That is the probability of this allele to get fixed, so the probability of this allele getting eliminated is:
Elimination probability = 1 - fixing probability = 1- 0.04 = 0.96%
Answer: The probability of this allele getting eliminated is 96
Complete the sentences about DNA packaging. Some terms may be used more than once.
The less condensed form of chromatin is ______.
The inactive form of chromatin is ______.
A core composed of _____ proteins interacts with DNA through hydrogen bonding and ionic bonds.
If DNA structure is described as "beads-on-a-string," a "bead" is a _____.
The more darkly-staining form of chromatin is _______ .
If DNA structure is described as "beads-on-a-string," a "string" is the _____.
A ______ is a DNA–protein complex.
Answer:
the less condensed form of chromatin is... euchromatin
the inactive form of chromatin is... heterochromatin
a core composed of eight ___histone___ proteins interacts with DNA through hyrdogen bonding and ionic bonds.
if DNA structure is described as "beads on a string" a "bead" is a......nucleosome
The more darkly staining form of chromatin is.....heterchromatin
if DNA is described as "beads on a string" the "string" is the....
DNA molecule
A ___________ is a DNA protein complex.
nucleosome
Hope this help.. plz mark brainliest
The less condensed form of chromatin is euchromatin. The inactive form of chromatin is heterochromatin. A core composed of histone proteins interacts with DNA through hydrogen bonding and ionic bonds.
What is a chromosome?Chromosomes are thread-like structures that are found within the nucleus of animal and plant cells.
Protein and a single molecule of deoxyribonucleic acid make up each chromosome (DNA). Passed from parents to offspring, DNA contains the specific instructions that make each type of living creature unique.
The primary function of chromosomes is to transport DNA and genetic information from parents to offspring.
During cell division, chromosomes play an important role. They keep DNA from becoming tangled and damaged.
Euchromatin is the less condensed form of chromatin. Heterochromatin is the inactive form of chromatin. A core made up of histone proteins interacts with DNA via hydrogen and ionic bonds.
The string is made of DNA, and each bead is a "nucleosome core particle," which is made of DNA wound around a protein core made of histones.
Heterochromatin is the darker-staining form of chromatin. Chromatin is a DNA-protein complex with two main functions: tight DNA packaging and gene expression regulation.
Thus, this way one can complete the given scenario.
For more details regarding chromatin, visit:
https://brainly.com/question/11073325
#SPJ6
The Punnett square above shows a cross between two sweet pea plants in Mendel's greenhouse. Both parents have blue flowers (Bb).
Which statement describes the offspring expected from this cross?
Answer:
answer is 75% blue-flower and 25% white- flower pea plants
Explanation:
As Blue is dominant trait,
both parents are heterozygous dominant
Bb (father ) bb( mother )
So crossing between
Bb x bb
B b
B BB Bb
b Bb bb
Offspring will be BB, Bb, Bb, bb
so 75% blue-flower and 25% white- flower pea plants
If a cell is able to synthesize 30 ATP molecules for each molecule of glucose completely oxidized to carbon dioxide and water, approximately how many ATP molecules can the cell synthesize for each molecule of pyruvate oxidized to carbon dioxide and water?
possible answers:
a. 0
b.1
c. 12
d. 14
e. 15
Answer: E =15
Explanation:
Basically, 1 molecule of glucose molecule after substrate level phosphorylation in glycolysis gives 2 molecules of 3 carbon compounds called pyruvate, for complete oxdation .
1 GLUCOSE molecule ⇒ 2Pyruvates.
Each pyruvate molecule can produce 15 ATPs .( in citric acid cycle, and oxidative phosphorylation),.Since 30ATPs were produced,;each pyruvate molecule must have contributed 15 ATPs each for a total of 30ATPs
According to social identity theory
A. teams are never as productive as individuals working alone.
B. the most effective teams have a large number of members.
C. the team development process occurs more rapidly for heterogeneous teams than for homogeneous teams.
D. people define themselves by their group affiliations.
E. teams are less productive in performing complex tasks
Answer:
D. people define themselves by their group affiliations.
Final answer:
Social identity theory primarily posits that d) individuals define their selves by their group affiliations rather than focusing on team effectiveness or size.
Explanation:
According to social identity theory, the correct answer is D. people define themselves by their group affiliations. This theory suggests that people categorize themselves and others into various social groups which provide a source of pride and self-esteem. These affiliations help to define individuals and differentiate them from others. Team effectiveness, size, and development processes, as mentioned in the other options, are related to group dynamics but are not directly related to the social identity theory's central focus on self-identification within groups.
On the planet Trogus, the population of Trogites have an autosomal gene, HDZ, with two alleles, T and t. T is dominant to t. Individuals with the T allele in their genotype show a propensity for sorbet, while the remainder prefer ice cream. In multiple population studies, 60% of individuals with the TT genotype prefer sorbet, 25% of Tt individuals prefer sorbet, and 5% of tt individuals prefer sorbet.
In an F1 cross of Tt x Tt genotyped individuals, in the F2population of individuals who prefer sorbet, what is the probability that a random individual has the Tt genotype?
A. 0.991.
B. 0.141.
C. 0.435.
D. 0.044.
E. 0.522.
Answer:
The correct option is C. 0.435
Explanation:
To get the results, let's make a punnet square and check the results. The parents in the cross will be heterozygous prefering sorbet. We can say that the alleles for sorbet are dominant over the alleles which prefer ice cream.
T t
T TT Tt
t Tt tt
The results show that there is 25% (0.25) probability that the individual will be homozygous preferring sorbet (TT). 25% (0.25) will have the probability of preferring ice- cream (tt).
50% (0.5) which is nearest to 0.435 will have the probability for having heterozygous alleles (Tt) and preferring sorbet,
Identify the incorrect statement:(a) The Tropic of Cancer lies to the north of the equator.(b) The summer solstice in the Northern hemisphere occurs in the month of June.C) During the winter solstice, the sun is right overhead in the Antarctic region.(d) The Tropic of Capricorn passes through India.
Answer:
(d) The Tropic of Capricorn passes through India.
Explanation:
The tropic of cancer is the southernmost latitude that is parallel to the tropic of cancer (Tropic of Cancer- North / Tropic of Capricorn – South). It is the tropic that is the nearest to the Antarctica. The Tropic of Capricorn passes through: Namibia, Botswana, South Africa, Mozambique, Madagascar, Australia, Chile, Argentina, Paraguay, Brazil, the Atlantic Ocean, the Indian Ocean and the Pacific Ocean.
You and a friend are in line for a movie when you notice the woman in front of you sneezing and coughing. Both of you have been equally exposed to the woman's virus, but over the next few days, only your friend acquires flu-like symptoms and is ill for almost a week before recovering. Which one of the following is a logical explanation for this?
A. You have an innate immunity to that virus.
B. You have an adaptive immunity to that virus.
C. Your friend has an autoimmune disorder.
D. Your friend has allergies.
Answer:B
Explanation:
Only one of the two people gets the disease. This indicates that the non-flu person already has adaptive immunity. Option B is the correct answer.
What is immunity?Immunity is the body's defence mechanism. There are two types of immunity: adaptive immunity and innate immunity. Innate immunity is already present at birth. Cells such as neutrophils, skin, and mucus layers are examples of innate immunity.
Adaptive immunity develops after the attack of the immunogen. Immunogens are molecules or pathogens that can activate the body's immunity. T cells and B cells are examples of adaptive immunity. B cells make antibodies and memory cells after they come across immunogens like viruses, bacteria, etc. These are specific.
If a person gets the viral flu, the adaptive immune system produces antibodies against it and destroys it. When the body encounters the same virus a second time, memory cells proliferate and produce antibodies to destroy it. As here, one person gets the disease because in his body there were no prior antibodies. Another person didn't because he had the antibodies.
Hence, option b is the correct answer. The non flu person has adaptive immunity to that virus.
To learn more about immunity, refer to the following link:
https://brainly.com/question/8189807
#SPJ2
Orchid seeds are tiny, with virtually no endosperm and with miniscule cotyledons. If such seeds are deposited in a dark, moist environment, then which of the following represents the most likely means by which fungi might assist in seed germination, given what the seeds lack?
A) by transferring some chloroplasts to the embryo in each seed
B) by providing the seeds with water and minerals
C) by providing the embryos with some of the organic nutrients the fungi have absorbed
D) by strengthening the seed coat that surrounds each seed
Answer:
The correct answer will be option-C
Explanation:
Roots of Orchid trees form a symbiotic association with the Fungi which help the orchid tree seed germination.
As soon as the seed is produced and buried in the soil, the chemical produced by the seed attracts the fungal species which in result help the seed to germinate.
The fungi help the orchid seed by producing the digestive enzymes which can digest the larger compounds and breaks it into smaller elements easily absorbed by the fungi. The absorbed nutrients are supplied by the fungi to the seed embryo nourishing it and thus processing the seed to germinate.
Thus, Option-C is the correct answer.
What is true concerning a quantitative trait? Multiple Choice Individuals fall into distinct classes for comparison The phenotypic variation for the trait is continuous The phenotypic variation for the trait falls into two to three classes The frequency distribution of the trait will have an asymmetrical shape
Answer:
The phenotypic variation for the trait is continuous
Explanation:
Genetically speaking, quantitative traits are controlled by many genes, classes are not easily distinguishable and there is a continuous distribution of the phenotype. These characteristics refer to measurements of quantities (weights, volumes, measurements: kg, m, cm, g, m2, etc.).
In other words, quantitative characteristics are those that exhibit continuous variations and are partly of non-genetic origin; that is, they are greatly affected by the environment.
. You have constructed four different libraries: a genomic library made fromDNA isolated from human brain tissue, a genomic library made from DNAisolated from human muscle tissue, a human brain cDNA library, and a humanmuscle cDNA library.A) Which of these would have the greatest diversity of sequences?B) Would the sequences contained in each library be expected to overlapcompletely, partially, or not at all?
Answer:
A) Brain genomic library and muscle genomic library.
B) Brain genomic library and muscle genomic library - overlap completely.
Human brain cDNA library, and a human muscle cDNA library and other library is partially overlap.
Explanation:
A) The genomic library contains the whole genome content of the organism whereas cDNA library contains the coding genome of the organisms. Brain genomic library and muscle genomic library will constitute the all the genomic sequences of brain and muscle. The cDNA library is prepared from the mRNA and the coding regions are present in this library.
B) The overlapping in the genome library might occur due to the common sequences present in the genome. Brain genomic library and muscle genomic library might completely overlap with each other as they have more sequence common among each other. All the other library may be partially overlap with each other as they have some common DNA sequences and neither library can have unique sequences.
Which statement regarding behavioral genetics is accurate? Select one: a. The genotypes and phenotypes of behavioral problems or deviations follow Mendelian autosomal recessive inheritance patterns. b. A genetic predisposition toward a specific behavior can be modified by altering environmental influences. c. The genetic susceptibility or predisposition toward a behavioral disorder requires the trigger of an infectious disease for expression. d. Genes and gene products have been discovered that directly control behavior.
Answer:
B.
Explanation:
A genetic predisposition toward a specific behavior can be modified by altering environmental influences.
Northern blot analysis is performed on cellular mRNA isolated from E. coli. The probe used in the northern blot analysis hybridizes to a portion of the lacY sequence. Above is an example of the autoradiograph from northern blot analysis for a wild-type lac+ bacterial strain. In this gel, lane 1 is from bacteria grown in a medium containing only glucose (minimal medium). Lane 2 is from bacteria in a medium containing only lactose. Determine the appearance for northern blots of the bacteria listed below. In each case, lane 1 is for mRNA isolated after growth in a glucose-containing (minimal) medium, and lane 2 is for mRNA isolated after growth in a lactose-only medium.a. lac+ bacteria with the genotype I+P+OCZ+Y+b. lac- bacteria with the genotype I+P+O+Z-Y+c. lac- bacteria with the genotype I+P-OCZ+Y+d. lac+ bacteria with the genotype I-P+OCZ+Y+e. lac- bacteria with the genotype I+P+OCZ-Y-
Answer: the answer to this question contains pictures and can be found in the attachment below.
Explanation:
a.
In the situation, the bacteria have the genotype I+ P+ OC Z+ Y+. The bacterial organism, has all functional genes and the operator is constitutive in nature, which will not allow the action of repressor and thus, the operon will be transcribed constitutively. However, in the absence of glucose, the CAP (catabolite activator protein)-cAMP (cyclic adenosine monophosphate) complex will be formed, which will further activate beta-galactosidase production. Hence, in absence of glucose, transcription rates will be higher.
The autoradiograph for this situation can be found in the attachment below:
b.
Here, The bacteria has the genotype I+ P+ OC Z- Y+. It is clear that this bacterium has a mutated beta-galactosidase and as such, no mRNA (messenger ribonucleic acid) for beta-galactosidase will be produced in presence of glucose or lactose.
The autoradiograph for this situation can be found in the attachment below:
c.
The genotypic composition of the bacteria given is I+ P- OC Z+ Y+. In this situation, the operator is constitutive but the promotor element is mutated, which will not allow binding of RNA (ribonucleic acid) polymerase and thus, no transcription will be observed in any condition. The autoradiograph for this situation can be found in the attachment below:
d.
The genotype composition here is I- P+ OC Z+ Y+. It is clear that the operator is constitutive and thus, operon will be expressed in presence of glucose as well as lactose. However, in presence of glucose, CAP-cAMP complex will not associate with the operon and thus, the only basal level of transcription will be observed.
The autoradiograph for this situation can be found in the attachment below:
e.
The genotype of the organism is I+ P+ O+ Z- Y-. The gene for beta-galactosidase is mutated and thus no mRNA will be produced for beta-galactosidase.
The autoradiograph for this situation can be found in the attachment below: