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:
Bacteria and other microbes can be used to "clean up" an oil spill by breaking down oil into carbon dioxide and water. Two samples isolated from the Deepwater Horizon leak in the Gulf of Mexico were labeled A and B. The DNA of each was isolated and the percent thymine measured in each sample. Sample A contains 17.7 17.7 % thymine and sample B contains 28.9 28.9 % thymine. Assume the organisms contain normal double‑stranded DNA and predict the composition of the other bases.
Answer:
For sample A, A=17.7 %, T=17.7 %, G=32.30%, C=32.30%
For sample B, A= 28.9%, T=28.9%, G= 21.10%, C= 21.10%
Explanation:
The composition of nitrogenous bases in DNA is calculated using the suggestions proposed by the results of the experiment performed by Erwin Chargaff.
He found that in a DNA sample the composition of purine to pyrimidine is 1:1. This indicates that the amount of purine equals pyrimidine. This can be presented as purines + pyrimidines= 100. Also, adenine binds thymine and cytosine binds guanine.
In the given question,
For sample A
A= 17.7 %, Therefore, thymine will be = 17.7 %,
A+T= 35.40%
Now, G+C=100- 35.40%
G+C= 64.60%
content of G= 64.60/2= 32.30%
content of C= 32.30%
For sample B
T=28.9%, A will be= 28.9%
A+T= 57.80%
G+C=100-57.80%
G+C= 42.20%
Thus content of G will be= 42.20/2=21.10
content of C= 21.10%
Thus,
For sample A, A=17.7 %, T=17.7 %, G=32.30%, C=32.30%
For sample B, A= 28.9%, T=28.9%, G= 21.10%, C= 21.10%
In the Brookhaven beetle, straight antenna(S) are dominant over curly antenna(s). Green wings(G) are dominant over red(g).A beetle with straight antenna and green wings is crossed with a beetle that has curly antenna and red wings. 25% of the offspring will have straight antenna and green wings.What is the genotype of the parent with straight antenna and green wings?
Answer:
the genotype of beetle with straight antenna and green wings SsGg
Explanation:
Allele "S" represents straight antenna
"s" represents curly antenna
"G" represents green wings
"g" represents red wings
G is dominant over g
S is dominant over s
Genotype of beetle with curly antenna and red wings is ssgg
SG Sg sG sg
sg SsGg Ssgg ssGg ssgg
sg SsGg Ssgg ssGg ssgg
sg SsGg Ssgg ssGg ssgg
sg SsGg Ssgg ssGg ssgg
Hence the genotype of beetle with straight antenna and green wings SsGg
The parent beetle with straight antenna and green wings has a heterozygous genotype, SsGg, since 25% of the offspring express both dominant traits when crossed with a beetle having recessive traits for both antenna shape and wing color.
The question pertains to a genetic cross involving the Brookhaven beetle, where straight antenna are dominant (S) over curly antenna (s), and green wings are dominant (G) over red wings (g). Given that 25% of the offspring have straight antenna and green wings, we can deduce the genotype of the parent beetle with straight antenna and green wings. Since the other parent has a completely recessive phenotype (curly antenna and red wings, with genotype ssgg), and considering the classical Mendelian genetics, the only way to achieve a 25% occurrence of the dominant phenotype in the offspring is if the dominant parent is heterozygous for both traits. This means the genotype of the dominant parent must be SsGg.
ATP donates energy through (select all that apply)
A. Phosphate ester bond hydrolysis
B. Coupling with an endergonic reaction
C. Phospho Anhydride bond hydrolysis
D. Enzyme-mediated reactions
E. none is correct
Some anoxygenic photosynthetic bacteria are able to use _______ as an electron source.
A. light
B. CO2
C. H2S
D. H2O
E. Nitrate
__________pathogens only cause diseases under certain circumstances, e.g. in immune compromised hosts
Answer:
ATP donates energy through C phospho anhydride bond hydrolysis.
Explanation:
ATP or adenosine triphosphate is an energy rich compound which contain 3 phosphate moiety that are linked by 2 high energy phospho anhydride bonds.
The hydrolysis of phosphoanhydride bonds releases high amount of free energy which is used by cellular system of our body to exhibit various cellular,biological and physiological functions.
Some anoxygenic photosynthetic bacteria are able to use H2S or hydrogen sulfide as an electron donor.
Opportunistic pathogens only cause infection under certain circumstances eg in immune compromised hosts.
If an organism has a diploid number of 6 chromosomes, how many different combinations of maternal and paternal chromosomes are possible in its gametes?
A. 8
B. 12
C. 3
D. 16
E. 6
An organism with a diploid number of 6 chromosomes can produce 8 different combinations of these chromosomes in its gametes due to the process of independent assortment during meiosis.
Explanation:The organism in question has a diploid number of 6 chromosomes, which means it has 3 pairs of homologous chromosomes (one from each parent). Therefore, when producing gametes (the cells that combine to form offspring), this organism could create[tex]2^n[/tex] different combinations of these chromosomes, where n is the number of chromosomal pairs. This is due to the process of independent assortment during meiosis, in which the homologous pairs of chromosomes separate randomly.
In this case, n equals 3 (because there are 3 pairs of homologous chromosomes), so the number of different combinations of maternal and paternal chromosomes possible in its gametes would be [tex]2^3 = 8.[/tex] Therefore, the correct answer to your question is: A. 8
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In this case, with a haploid number of 3, there are A.8 different combinations possible.
If an organism has a diploid number of 6 chromosomes, the number of different combinations of maternal and paternal chromosomes possible in its gametes can be calculated using the formula [tex]2^n[/tex], where n is the number of haploid chromosomes. Since a diploid cell contains two homologous sets of chromosomes, one from each parent, the number of chromatids that can be combined during gamete formation is based on the different ways these chromosomes can be assorted.
Firstly, calculate the haploid number which is half the diploid number, so for this organism, it is 3 (as 6 diploid divided by 2 equals 3).Next, we apply the formula for combinations: [tex]2^n[/tex], where n equals the haploid number.[tex]2^3[/tex] gives us 8, so the correct answer is 8 different combinations.Therefore, the correct answer is A. 8 different combinations of maternal and paternal chromosomes are possible in the gametes of an organism with a diploid number of 6 chromosomes.
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:
Clusters of treats that define how people normally react in situations are called? A. Dispositions. B. Fixations. C. Schema
Answer:
Cluster of treats that define how people normally react in the situation are called disposition.
Explanation:
Disposition refers to mood swing and the attitude towards life which surrounds one person. It is also a description of ones temperament. it mean that the qualities establishing the person or a group. The persuade habit of one person and the natural mind set or an emotional outlook is disposition. It is a 'state of mind' regarding something.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.
Often the growth cycle of one population has an effect on the cycle of another. As moose populations increase, wolf populations also increase. Thus, if we are considering the logistic equation for the wolf population,ΔN/Δt=rN(K−N)K, which of the factors accounts for the effect on the moose population?
A) r
B) N
C) rN
D) K
E) dt
Answer:
The correct answer is option D) "K".
Explanation:
In this example the wolf population is described by the equation "ΔN/Δt=rN(K−N)K". Even tough the variables are not defined in the question, we can conclude that the effect of the moose population will be given by a factor that has a positive effect in the wolf's population because "as moose populations increase, wolf populations also increase". The factor "K" fits the description because it gives a positive effect on "ΔN/Δt". "K" is a factor that multiplies "rN" at two different levels, therefore the higher the value of "K", the higher value of "ΔN/Δt" will be.
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.
The mRNA codons 5'-CAA-3' or 5'-CAG-3' are translated as the amino acid glutamine by _____.a) the tRNA with an anticodon 5'-GUU-3' and glutamine at its other endb) by tRNA molecules that have been charged with glutamine by two different aminoacyl-tRNA synthetasesc) separate tRNA molecules with anticodons 3'-GUU-5' and 3'-GUC-5', respectivelyd) the same tRNA with the anticodon 3'-GUU-5'e) the small and large ribosomal units
Answer:
c) separate tRNA molecules with anticodons 3'-GUU-5' and 3'-GUC-5', respectively.
Explanation:
Glutamine would be covalently linked to two different tRNA molecules encoding the 3'-GUU-5' and one containing the 3'-GUC-5'. Only one would be allowed to bind to the P-site of the ribosome at a time.
The mRNA codons 5'-CAA-3' or 5'-CAG-3' specify the amino acid glutamine, which is translated by the same tRNA molecule with the anticodon 3'-GUU-5', due to the wobble base pairing at the third position of the codon.
Explanation:The mRNA codons 5'-CAA-3' or 5'-CAG-3' are translated as the amino acid glutamine by the same tRNA with the anticodon 3'-GUU-5'. Both codons CAA and CAG specify the amino acid glutamine according to the genetic code. The reason why a single tRNA molecule can recognize both codons is due to the phenomenon known as wobbling, where the base at the 5' end of the tRNA anticodon can form hydrogen bonds with more than one kind of base in the 3' position of the mRNA codon. This is possible here because the first two bases of the codons are the same (CA), and it's just the third base that differs (A or G), which is the 'wobble' position. Therefore, the tRNA with the anticodon 3'-GUU-5' can base-pair with both codons, carrying glutamine to be added to the growing polypeptide chain during translation.
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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.
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Fossils of ancient organisms are found in deep layers of rock, while the fossils of more recent organisms are found in shallower layers. Which two groups' common ancestor would be found in relatively shallow fossils
Answer:
animals and plants
Explanation:
Fossils are remains or traces of animals, plants or other living things preserved in rocks, sediment, ice or amber. They are preserved as molds of the body or parts thereof, as well as tracks and footprints. Fossils and their presence in rock formations and sedimentary layers are known as the fossil record.
The fossil is covered by layers of sediment, which slowly compress until they become, after centuries or millennia, into rocks. The time required to build all this is millions of years and the oldest rocks are below. Often lower fossils are older than higher fossils. For example, if trilobites are found in the deepest layers, since they are more primitive beings; whereas plants and animals are found in the shallower layers.
Final answer:
Fossils of ancient organisms are found in deep layers of rock, while the fossils of more recent organisms are found in shallower layers. This pattern can be explained by the principle of relative dating.
Explanation:
Fossils of ancient organisms are found in deep layers of rock, while the fossils of more recent organisms are found in shallower layers. This pattern of fossil distribution can be explained by the principle of relative dating, which states that lower rock layers are older than the ones above them.
Therefore, the common ancestors of the two groups would be found in relatively shallow fossils. For example, if we consider the whale and the seal, both belong to the group of mammals, and their common ancestor would be found in relatively shallow fossils.
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.
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We wish to address is whether cells we have tentatively identified as cued fear conditioning cells increase their responses to the CS because the synapses between CS-representing neurons and the conditioning cells are subject to LTP. If this is in fact the case, would you expect these cells to respond to a US priorto conditioning? (a) yes (b) no
Answer:
Yes
Explanation:
At first the response would probably be very weak or faint and not very detectable. After sometime the connection may become stronger and thus more recognizable.
A population of individuals was exposed to lead contamination in drinking water at the concentration of 27 parts per billion (ppb). The following represents a small sample set from the population indicating the level of response to the contaminant rated on a scale from 1–20. Within your response, please explain how you calculated your answer (1 indicating minimal effects and 20 indicating maximum effects to the lead exposure). Given the data set 9, 12, 7, 15, 8, 9, 10, calculate the following results:a. mean,b. mode,c. median, andd. standard deviation.Your response should be at least 300 words in length.
Answer:
a) Mean - 10
b) mode - 9
c) Median - 9
d) SD - 2.708
Explanation:
A) Mean is the average of a given set of number in which the sum of all the number is divided by the number of entries
Thus, mean is equal to
[tex]\frac{9+12+7+15+8+9+10}{7} \\= \frac{70}{7} \\= 10[/tex]
B) Mode is the most common number in a given series
Thus, mode is equal to 9 as it has appeared twice
C) Median
Number are arranged in ascending order
7, 8, 9, 9, 10, 12, 15
The middle number is the median i.e 9
D) Standard Deviation
Formula is square root of sum of [tex](X - mean )^2[/tex] divided by [tex]N-1[/tex]
[tex](9-10)^ 2+ (12-10)^2 + (7-10)^2 + (15-10)^2 + (8-10)^2+ (9-10)^2+ (10-10)^2\\= 1 + 4+ 9+ 25+ 4+ 1+ 0\\= 44[/tex]
[tex]\sqrt{\frac{44}{7-1} } \\= 2.708[/tex]
Scientist think colonies of Cyanobacteria called stromatolites were responsible for the introduction of oxygen into earths atmosphere ) true or false
Answer:
Scientist think colonies of Cyanobacteria called stromatolites were responsible for the introduction of oxygen into earths atmosphere is true.
Explanation:
In primeval earth’s atmosphere, stromatolites (colonies of Cyanobacteria) are believed to have increased the level of oxygen by their photosynthesis process. These are the first known organisms that started photosynthesis process and produced the oxygen to the earth’s atmosphere. There is a theory that, these organisms produced oxygen in the environment and the organism disappeared that couldn’t live in the atmosphere with the oxygen.Scientists gave this a name called as “great oxygenation event”.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.
David comes into the emergency room late at night with a cut that will not stop bleeding. His mother. Rose, is worried that he has a bleeding disorder, just like his grandfather. The lab is closed, so the doctor cannot order a specialized coagulation test However, the hospital does have a sample of blood from a classic hemophiliac. What is the simplest test the doctor could perform using David's blood to determine if it is missing factor VIII activity?
a. combine the samples and check for coagulation
b. determine the factor VIII activity level in Rose's blood
c. compare clotting rates between the two blood samples
d. test the samples for factor VIII mutations using PCR
Answer:
a. combine the samples and check for coagulation
Explanation:
a. combine the samples and check for coagulation
We know that hemophilia is X linked recessive disorder which arises basically due to lacking of clotting factor in the blood (Factor VIII or Factor IX). According to the condition provided if pathology is closed so doctor can not perform the specific test so only their is chance to combine the sample of patient blood with the classic hemophiliac blood sample.As here doctor is sure that the blood sample available in his hospital is of hemophiliac patients so it does not contain clotting factor. When he mix both blood and then if the blood show coagulation it means David do not have hemophilia but if after mixing both the blood sample coagulation will not occur so David may be hemophilic.
In mitochondria, the rate of electron transfer is tightly coupled to the demand for ATP. When the rate of use of ATP is relatively low, the rate of electron transfer is low; when demand for ATP increases, electron‑transfer rate increases. Under these conditions of tight coupling, the number of ATP molecules produced per atom of oxygen consumed when NADH is the electron donor, the P/O ratio, is about 2.5. Predict what would happen when an uncoupling agent, such as 2,4‑dinitrophenol, is added to an actively respiring tissue preparation. The rate of oxygen consumption
Answer:
The correct answer will be- P/O decreases
Explanation:
The uncoupling agents are the chemical drugs which uncouple the two different process of cellular respiration which is the electron transport chain from the oxidative phosphorylation or ATP synthesis.
The synthesis of one ATP requires two electrons on the basis of which ATP production/ oxygen atom or P/O ratio is calculated.
When 2,4‑dinitrophenol uncouples the ATP synthesis from electron transport chain, then demand of oxygen for the synthesis of ATP increases which result in the decrease in the value of the P/O ratio.
Thus, P/O decreases are the correct answer.
Mitochondria is the cell organelle where the electron transport chain occurs and is a series of complexes. An increase in the rate of oxygen consumption decreases the P/O ratio.
What is the ETC of the mitochondria?The electron transport chain is a sequence of complexes that occurs in the membrane of the mitochondria and is a redox reaction. This chain results in the formation of an energy molecule called ATP.
Some of the chemical drugs act as the uncoupling agent and uncouples the molecules of the cellular respiration cycles, including the ETC from the oxidative phosphorylation or the synthesis process of the ATP.
The ratio of the P/O is calculated based on the amount of ATP generated as two electrons results in the formation of one ATP molecule.
The uncoupling agent like 2,4‑dinitrophenol uncouple the synthesis of the ATP molecule from the ETC and increases the demand of the oxygen molecules for the ATP synthesis resulting in the decrease of the P/O ratio.
Therefore, the P/O ratio decreases.
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Which of the following statements about the thyroid gland is true? A. It is located anterior to the trachea and inferior to the larynx. B. The parathyroid glands are embedded within it. C. It manufactures three hormones. D. It is controlled by the anterior part of the pituitary gland. E. All of the above are true.
Answer: Option E. All the statements are correct.
Explanation:
Thyroid gland is an endocrine gland that is found in front of the neck, it consists of two lobes connected by isthimus. It is in front of the neck lying against and around the front of larynx and trachea. Thyroid gland secretes three hormones which are thyroxine, triiodothyronine and calcitonin. Thyroxine and trio doth tribune are thyroid hormone. The parathyroid glands are tiny structures found in the thyroid gland. A thick connective tissue separates the parathyroid glands from thyroid tissues. The thyroid gland is controlled by thyroid stimulating hormone called thyrotropin which is released from the anterior pituitary gland which triggers the release of thyroid hormones.
A recent study used umbilical cord blood to test for 25 hydroxyvitamin D, which is an indicator of vitamin D status of the baby. It was reported that 65% of babies tested were deficient in vitamin D in spite of the fact that the mothers consumed vitamin D supplements during pregnancy. A researcher in a northern region felt that this percentage was too high for this region because with the reduced hours of sunshine during winter months, pregnant women tended to use higher doses of supplements to compensate. A sample of 125 newborns was tested, and 72 were declared to be deficient in vitamin D. What would be the appropriate null and alternative hypotheses that the researcher should establish for a test of significance?
Answer:
Null hypothesis: p= 0.65
Alternative Hypothesis: p < 0.65
Explanation:
Below are the null and alternative hypothesis,
Null hypothesis: p= 0.65
Alternative Hypothesis: p < 0.65
Test statistic for text of significance
z = (pcap –p)/ [tex]\sqrt{ [p*(1-p)/n]}[/tex]
z = (0.576 – 0.65)/ [tex]\sqrt{[0.65×1-0.65)/125]}[/tex]
z = 1.73
P- value Approach
p-value = 0.0414
As P-value < 0.05, we reject the null hypothesis
There is sufficient evidence to conclude that the mother consumed vitamin D supplements during pregnancy. A researcher in a northern region felt that the percentage should be lower for this region because with reduced hours of sunshine during winter months, pregnant women tended to use of higher doses of supplements to compensate.
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.
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.
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
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:
Your professor (Sam) and his wife (Becca) both have normal color vision, but their son is colorblind. What do you know about Sam's and Becca's genotypes?
A. Sam = c+c+ Becca = c+c+
B. Sam = c+c Becca = c+c
C. Sam=cY Becca=cc
D. Sam = c+Y Becca = c+c
Answer:
D. Sam = c+Y Becca = c+c
Explanation:
Colorblindness is a X-linked trait. Since Sam is a male and not colorblind, there's only one possible genotype he could have which is:
[tex]Sam:\ X^{c+}Y[/tex]
Being c+ the dominant allele for not being colorblind while c is the recessive allele associated with colorblindness.
As for Becca, in order for her to not have colorblindness but her son do, she needs to have a heterozygous genotype and pass down the recessive Y allele to her son. Becca's genotype is:
[tex]Becca:\ X^{c+}X^{c}[/tex]
Therefore, the answer is D. Sam = c+Y Becca = c+c
What is Vaginosis? List some of the members of the normal microflora of the female reproductive organs and discuss factors that may contribute to an abnormal increase in these organisms.
Answer:
It is a common condition that includes overgrowth of atypical bacteria in the vagina
Explanation:
It is characterized by vaginal discharge that is when a fluid flows out from the vaginal opening with an abnormal odor or consistency and sometimes pain, the normal microflora of the female reproductive organs includes Gardenella, lactobacillus, Bacteroides, peptostreptococcus, fusobacterium , eubacterium, as well as a number of other types, some of the factors that may contribute to an abnormal increase of these bacteria or make them become unbalanced can be multiple or new sexual partners although some experts are skeptical about this; IUDs , recent antibiotic use, vaginal douching, and smoking. It is not dangerous but can cause discomfort symptoms.
The first step in the catabolism of amino acids is the removal of the nitrogen as ammonia, forming a keto acid that can enter one of the carbon catabolic pathways. The alpha-keto acid pyruvate can be formed from the amino acids alanine, cysteine, glycine, serine and threonine. Consider the route for alanine catabolism. Anine reacts with to produce pyruvate and. This reaction is catalyzed by aspartate aminotransferase. alanine aminotransferase. alanine dehydrogenase. The substrate for the first step can be regenerated by reacting with NAD^+ This reaction is catalyzed by glutamate dehydrogenase. alpha-ketoglutarate dehydrogenase. alanine dehydrogenase. aspartate dehydrogenase. The coenzyme/prosthetic group required in the first reaction is thiamine pyrophosphate. biotin. pyridoxal phosphate. lipoic acid.
Answer:
Alanine, Alanine aminotransferase,glutamate dehydrogenase,pyridoxal phosphate.
Explanation:
The alpha keto acid pyruvate can be formed from the amino acid alanine
This reaction is catalyzed by Alanine aminotransferase enzyme whch catalyzes that reversible reaction that helps in the inter conversion of alpha amino acid to alpha keto acid and vice versa.
The substrate for the first step can be reacting with NAD+.This reaction is catalyzed by glutamate dehydrogenase enzyme that catalyzes the first step of oxidative de amination process.
The coenzyme/prosthetic group required in the first reaction is pyridoxal phosphate.
Alanine is catabolized by alanine aminotransferase to pyruvate and glutamate, which then undergoes oxidative deamination by glutamate dehydrogenase. The coenzyme required for the transamination is pyridoxal phosphate. These steps are essential for the proper catabolism of alanine.
The first step in the catabolism of amino acids, such as alanine, is the removal of the nitrogen atom as ammonia. This process, known as transamination, involves the transfer of the amino group from alanine to alpha-ketoglutarate, forming pyruvate and glutamate. The enzyme that catalyzes this reaction is alanine aminotransferase.
Once glutamate is formed, it undergoes oxidative deamination, where it reacts with NAD+ to regenerate alpha-ketoglutarate and releases ammonia. This oxidative deamination step is catalyzed by glutamate dehydrogenase.
The coenzyme required in the initial transamination reaction is pyridoxal phosphate, which is vital for the enzyme's activity.
Key Points :
The conversion involves alanine reacting with alpha-ketoglutarate to produce pyruvate and glutamate.The enzyme responsible for this reaction is alanine aminotransferase.Glutamate dehydrogenase catalyzes the subsequent reaction involving NAD+, leading to the regeneration of alpha-ketoglutarate.Pyridoxal phosphate is the necessary coenzyme in the transamination reaction.
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.