Answer:
The main substrate of chymotrypsin includes tryptophan, tyrosine, phenylalanine and methionine
Explanation:
1.
Histidine yields a proton to aspartate and recovers it from serine.
Seen in another way: aspartate captures a proton from the serine through histidine.
2.
a) The serine (deprotonated) is thus capable of attacking the peptide bond (nucleophilic carbonyl attack) and forms a tetrahedral intermediate; The substrate is thus covalently bound to the enzyme (now it is a transition state).
b) The peptide bond is broken and the released amino terminus (R) recovers a proton from histidine.
c) Histidine, in turn, recovers it from aspartic.
3.
a) Aspartate captures a proton of histidine again, so that it can capture it in turn from water.
b) This generates a hydroxide anion that attacks the ester intermediate between the serine and the carboxyl part (R ′) of the substrate peptide.
c) A new tetrahedral intermediate bound to the enzyme is formed (via serine residue).
4.
a) The carboxyl group of the peptide is regenerated, the serine being separated and the other peptide fragment being free (the R ′ part with a free carboxyl end)
b) The serine recovers the proton at the expense of histidine, which in turn captures it from aspartic acid.
c) The catalytic triad (Asp, His, Ser) has been regenerated in its original state.
The net reaction is:
R–NH – CO –R ′ + H2O ⟶ R– NH2 + HOOC –R ′ ⟶ R – NH3 + + −OOC –R ′
The active site or catalytic center of chymotrypsin is formed by several amino acid residues, among which the essential role corresponds to the "catalytic triad".
Which of the following does not
describehemoglobin?
a.) It binds less tightly to oxygen
thanmyoglobin.
b.) It is a multi-subunit protein with four heme
groupsfor oxygen binding.
c.) 2,3-bisphosphoglycerate increases its affinity
foroxygen.
d.) Maternal hemoglobin has a lower affinity for
oxygenthan fetal hemoglobin.
e.)none of the above.
Answer:
C
Explanation:
All statements describe hemoglobin for the exception of option C.
2,3 biphosphoglycerate (2,3 BPG) its used by erythrocytes to DECREASE affinity for oxygen to unload it to the cells. When the erythrocyte goes to high-metabolic demand areas (the ones that in most need for oxygen) synthesis of 2,3 BPG increases, affinity for oxygen then decreases, and oxygen detach from hemglobin and goes into the tissues.
In four-o'clocks, the allele for red flowers is incompletely dominant over the allele for white flowers, so heterozygotes have pink flowers. What ratios of flower colors would you expect emong the offspring of the following crosses: (a) pink x pink, (b) white x pink, (c) red x red, (d) red x pink, (e) white x white, and (f) red x white? If you specifically wanted to produce pink which of these crosses would be most efficient?
Answer: Crossing red with white is the most efficent way to produce pink flowers.
Explanation:
Let's call the "red" allele R and the "white" allele r. A red flower would have both copies of R, and its genotype would be RR, a white flower would have both copies or r, being rr, and a pink flower would be Rr (heterozygote).
We can represent crosses between individuals by Punnet squares. From pink flowers you can get gametes R and r, from red flowers you can get only R gametes and from white flowers only r. If we put male gametes in the first row and female gametes on the first column, we get the following cases:
a: Crossing pink with pink, we get a ratio of a red flower to a white flower to two pink flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & R & r \\ \ R & RR & Rr \\ \ r & Rr & rr \\ \end{tabular}\end{center}[/tex]
b: Crossing white with pink, we get two pink flowers, and two white flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & r & r \\ \ R & Rr & Rr \\ \ r & rr & rr \\ \end{tabular}\end{center}[/tex]
c: Crossing red with red, we get only red flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & R & R \\ \ R & RR & RR \\ \ R & RR & RR \\ \end{tabular}\end{center}[/tex]
d: Crossing red with pink, we get a ratio of two pink flowers to two red flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & R & R \\ \ R & RR & RR \\ \ r & Rr & Rr \\ \end{tabular}\end{center}[/tex]
e: Crossing white with white, we get only white flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & r & r \\ \ r & rr & rr \\ \ r & rr & rr \\ \end{tabular}\end{center}[/tex]
f: Crossing red with white, we get only pink flowers.
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & R & R \\ \ r & Rr & Rr \\ \ r & Rr & Rr \\ \end{tabular}\end{center}[/tex]
Thus crossing red with white is the most efficent way to produce pink flowers.
In four-o'clock flowers, incomplete dominance results in pink flowers when red is crossed with white. The offspring ratios vary with different crosses: pink x pink yields 1 red : 2 pink : 1 white, red x white produces all pink offspring, which is the most efficient cross for producing pink flowers.
Explanation:In four-o'clocks, there is a pattern of inheritance known as incomplete dominance, where the allele for red flowers is incompletely dominant over the allele for white flowers, resulting in pink flowers in heterozygous plants. According to Mendelian genetics, we can predict the outcomes of various crosses.
(a) pink x pink (RW x RW): The expected ratio would be 1 red (RR) : 2 pink (RW) : 1 white (WW).(b) white x pink (WW x RW): The expected ratio would be 1 pink (RW) : 1 white (WW).(c) red x red (RR x RR): All offspring would be red (RR).(d) red x pink (RR x RW): The expected ratio would be 1 red (RR) : 1 pink (RW).(e) white x white (WW x WW): All offspring would be white (WW).(f) red x white (RR x WW): All offspring would be pink (RW).If you specifically wanted to produce pink four-o'clock flowers, the most efficient crosses would be red x white (RR x WW) because all offspring would be pink (RW), and there would be no variation in flower color among the progeny.
All the organisms on your campus make up
a. an ecosystem.
b. a community.
c. a population.
d. a taxonomic domain.
All the organisms on your campus make up a community. The correct option is (b).
Understanding Community of OrganismsA community refers to all the populations of different species that coexist and interact in a particular area. It includes all the living organisms, such as plants, animals, fungi, and microorganisms, that inhabit a specific location or ecosystem.
An ecosystem, on the other hand, encompasses both the community of organisms and their physical environment, including abiotic factors like soil, water, and climate.
A population refers to a group of individuals of the same species that live in the same area and have the potential to interbreed.
A taxonomic domain, also known as a superkingdom, is a high-level classification category in the hierarchical system of taxonomy. It is a broader classification than the others mentioned and does not specifically refer to the organisms present on a campus.
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When considering all the organisms on your campus, you're referring to a biological 'community'. This includes diverse species interacting within the same location. It does not refer to an ecosystem, a population, or a taxonomic domain.
Explanation:The organisms found on your campus would comprise what we refer to as a community. This term implies that various species live together and interact in many ways. For example, this community includes everything from students and teachers (humans), to birds, insects, plants, and microbial life forms. It fits perfectly the concept of a community as per biology, because it encompasses an assortment of life forms sharing the same geography. On contrast, an ecosystem would also include inanimate elements like temperature, humidity, soil, etc. A population refers to a group of individuals from the same species living in a specific area. Whereas, a taxonomic domain is a high-level classification of life forms.
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Because of gall stones, Fred had his gallbladder surgically removed. Which of the following is the most likely side effect?
a. difficulty with protein digestion
b. difficulty with feces production
c. difficulty with vitamin B12 absorption
d. difficulty with fat digestion
e. difficulty with bile production
Answer:
d. difficulty with fat digestion
Explanation:
The gallbladder is an accessory gland that stores bile produced in the liver, so option "e" is not possible. Bile is a type of enzyme that helps to break down fats which makes option "d" as the best option. Protein digestion is mainly by different enzymes known as proteases. Feces are formed in the short intestine.
One theory of human evolution suggests that Homo erectus evolved directly into Homo sapiens (humans), after which point Homo erectus became extinct. If this is true, humans arose by ________.
a. allopatric speciation
b. cladogenesis
c. anagenesis
d. parapatric speciation
Answer:
The correct answer will be option-C.
Explanation:
The evolution of human species in which one species transformed into another species evolves by a mechanism known as anagenesis.
Anagenesis is the mechanism of evolution which transform one species into a different species within a lineage. This process is slow and takes time to form species, therefore, is also known as gradualism or phyletic transformation.
The Homo sapiens evolved from Homo erectus where Homo sapiens overwrites the ancestral species and caused the species to become extinct.
Thus, Option-C is the correct answer.
Beadle and Tatum used which of the following organisms to support their one gene - one enzyme concept?
a. escherichia coli
b. drosophila
c. neurospora
d. salmonella
e. homo sapien
Answer:
Neurospora.
Explanation:
Beadle and Tatum experiment shows one gene one enzyme hypothesis. According to this, a single enzyme is encoded by each gene. This idea is not accepted in today's world.
Beadle and Tatum performed experiment on the neurospora. They chosed neurospora in their experiment because neurospora shows the fast life cycle with alternation of generation. The genetic experiments can be easily performed on neurospora.
Thus, the correct answer is option (c).
What is the total number of ATPs produced by cellualar metabolism (per glucose molecule)?
Answer:
38 ATP
Explanation:
One molecule of glucose on complete breakdown yields 38 ATP. The break up of APT production at different steps of cellular respiration is given below:
Glycolysis yields 4 ATP and 2 NADH is produced, but 2ATP is used. Net gain of 2ATP. During formation of Acetyl CoA, 2 NADH is produced. During Krebs cycle 2 ATP, 6 NADH, 2 FADH₂ are produced. During Electron transport chain, reduced coenzymes NADH and FADH₂ oxidised to release ATP.Each NADH releases 3 ATP and each FADH₂ releases 2 ATP. Altogether 10 NADH is produced which yield 30 ATP and 2 FADH₂ yields 4 ATP.
Therefore, on complete oxidation of one molecule of glucose yields 38 ATP.
If you are about to fight or flight a dog that wants to bite you, the cellular signal inside your body that will trigger a signaling network for this response is called:
a. NADH
b. connexon
c. estrogen
d. adrenaline
Answer:
D. Adrenaline
Explanation:
Adrenaline hormone is secreted by adrenal medulla during stressful conditions. The hormone is involved in preparing the body for stressful conditions by intensifying the sympathetic response.
If a person is chased by a dog, the adrenaline hormone is released from the adrenal medulla. The hormone triggers the preganglionic neurons of the sympathetic division of the autonomous nerve system to generate the flight or flight response.
Adrenaline increases the heart rate, blood pressure, and blood flow to skeletal muscles and adipose tissues. The hormone also triggers the dilation of airways and increases the blood levels of glucose and fatty acids.
After the first meiotic cell division ___________
a. two haploid gametes are produced
b. cells are produced that contain twice the same number of chromosomes as somatic cells from which they came.
c. the number of chromosomes will vary depending on how the paternal and maternal chromosomes align at the metaphase plate.
d. DNA replication occurs.
e. None of the above
After the first meiotic cell division, a) two haploid cells are produced, which have half the number of chromosomes compared to the original diploid cell. These cells are not gametes yet; DNA replication does not occur after the first division.
After the first meiotic cell division, a) two haploid cells are produced. This division, known as Meiosis I, separates homologous chromosomes into two new cells. It's important to note that these two cells are not yet gametes, as a second division (Meiosis II) is required for that. Meiosis I is preceded by a phase of DNA replication during the S-phase of the cell cycle, and this replication is crucial as it ensures that each chromosome consists of two identical sister chromatids which are later separated during Meiosis II. The number of chromosomes in the daughter cells after Meiosis I is halved compared to the original diploid cell, and the resulting cells are haploid. The distribution of maternal and paternal chromosomes during Meiosis I is random, leading to genetic variation. As for DNA replication, it does not occur after the first meiotic division; it occurs before Meiosis I begins.
NADPH is produced in the light independent reaction.
a. True
b. False
Answer:
Given statement is true.
Explanation:
The energy molecules NADH and ATP are produced in light reaction which comprises of series of chemical and photo chemical reactions. These light dependent reaction takes place in the stroma of the chloroplasts. With in the stroma, carbon dioxide is produced from carbohydrates along with formation of glyceraldehyde 3-phosphate there by producing two energy molecule such as ATP and NADH.
Hence, the given statement is true.
Explain the importance of Mendel's inclusion of reciprocal crosses within his controlled breeding program of pea plants.
Answer:
Reciprocal cross may be defined as the cross done by reversing the parent genotype. The reciprocal cross was performed by Mendel's during the pea plant experiment.
Mendel's reciprocal cross is important as it determines the contribution of the male or female in the particular trait. He reversed the male and female trait, cross them, whether the male or female is responsible for the transmission of the trait. He found that the progeny of the reciprocal cross are similar as the normal cross. Thus, he concluded that both the parents contribute equally in the transmission of trait.
Lysosomes are membranous orgahelles that contain digestive enzymes. Lysosones can function inside the cell, where their enzymes digest particles taken in by endocytosis or worn-out cell components Lysosomes can also release their enzymes outside the cell, where the enzymes break down extracellular material.
Answer the following questions about conditions related to lysosomal function.
l-cell disease is a lysosomal storage disease that results in the buildup of carbohydrates, lipids, and proteins as inclusion bodies within the cell. Which is the probable cause?
(A) The cell does not take up food molecules by endocytosis.
(B) Lysosomes accumulate extra lipid-digesting enzymes
(C) The lysosome cannot fuse to the cell membrane to release enzymes extracellularly
(D) The Golgi apparatus cannot target the appropriate enzymes to the lysosome.
Answer:
(D) The Golgi apparatus cannot target the appropriate enzymes to the lysosome.
Explanation:
Enzymes are a type of protein build in ribosomes. The rough endoplasmic reticulum works with the golgi apparatus where proteins can be modified, sort and packed according to cell needs. Lysosomes are formed by budding of the Golgi apparatus.
The Golgi apparatus cannot target the appropriate enzymes to the lysosome is correct answer for this question.
Explanation:l-cell disease, also known as lysosomal storage disease, is typically caused by defects in the lysosomal enzymes that are necessary for the breakdown of certain molecules. Thus, the probable cause for the accumulation of carbohydrates, lipids, and proteins in the cell is option (D) The Golgi apparatus cannot target the appropriate enzymes to the lysosome. The Golgi apparatus has a crucial role in the sorting and modification of proteins, including lysosomal enzymes. If it is unable to correctly direct these enzymes to the lysosome, digestion of these molecules within the lysosome will not occur, resulting in their accumulation within the cell.
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Which of the following is an example of post-transcriptional control of gene expression?
a. the addition of methyl groups to cytosine bases of DNA
b. the binding of transcription factors to a promoter
c. the removal of introns and alternative splicing of exons
d. gene amplification contributing to cancer
Answer:
Option (c).
Explanation:
Transcription may be defined as the process of formation of RNA molecule from the template DNA with the help of enzymes and transcription factors. The transcription occurs in 5' to 3' direction.
Post trancriptional modification occurs in the RNA molecule that plays an important role in transcription as well as translation. The introns are removed from the RNA transcript and exons joined together is known as splicing. The alternative splicing occurs in which different protein isoforms are formed by the single exons.
Thus, the correct answer is option (c).
While carbohydrates and proteins are necessary for a healthy diet, fats should be avoided whenever possible.
a. True
b. False
Answer:
false
Explanation:
as long as you don't have too much fat you are fine
Which of the following does not occur during mitosis?
a. condensation of the chromosomes
b. replication of the DNA
c. separation of sister chromatids
d. spindle formation
Answer:
Option (b).
Explanation:
Mitosis may be defined as the process of cell division in which a single parent cell divides into two cells. This division is known as reduction division because the chromosome number remains the same.
The chromosome condensation, spindle formation and sister chromatid separation occurs in mitosis. The replication of DNA occurs in the synthesis phase of cell cycle not during the process of mitosis.
Thus, the correct answer is option (b).
Replication of the DNA does not occur during mitosis.
Explanation:The correct answer is b. replication of the DNA. DNA replication occurs during interphase, which is the phase before mitosis. During mitosis, the DNA has already been replicated and is condensed into visible chromosomes. The other events listed in the options all occur during mitosis: condensation of the chromosomes, separation of sister chromatids, and spindle formation.
Describe the four ways that drugs can affect a person's nervous system.
Answer:
Four ways by which drugs can affect the nervous system are as follows:
Drugs like heroin can mimics the chemical structure of the neurotransmitters. They causes the excess activation of neurons in the brains and affect the nervous system.
Some drugs also mimics the neurotransmitter but causes the abnormal messages sent through the brain. Their mimicry of neurotransmitters cannot activate the neuron but causes the interruption and wrong messaging in the nervous system.
Drugs like cocaine causes the stimulation of release of large number of neurotransmitters. This causes the prevention of recycling of brain chemicals and disturbs the neuron communication.
Some drugs causes the inhibition of release of neurotransmitters in the brain. This stops the signalling of the nervous system and interrupts the chemical messages of the body.
This family of bacteria is often associated with urinary tract infections?
A) Listeria
B) Salmonella
C) Proteus
Answer:
The correct answer is C) Proteus
Explanation:
Proteus comes under the family Enterobacteriaceae and is often associated with urinary tract infections in animals and humans. These bacteria are often the part of intestine but become pathogenic when enter into urinary tract.
Proteus mirabilis is a gram negative anaerobic bacteria present as normal flora of intestine known to cause urinary tract infections in humans but most of the urinary tract infection is caused by E. coli.
When these bacteria get into the blood stream they can cause systemic inflammatory response syndrome and sepsis which has very high mortality rate.
Final answer:
The family of bacteria often associated with urinary tract infections is Proteus. While not included in the options, Escherichia coli is the most common cause of UTIs.
Explanation:
The family of bacteria often associated with urinary tract infections (UTIs) is Proteus. Among the options given, Proteus specifically refers to a genus of bacteria that can cause infection in the urinary tract. While not included in the given options, it is noteworthy that Escherichia coli is actually the most common cause of UTIs. Bacteria from the genus Proteus are known for their ability to produce a strong odor and have the ability to form stones in the kidney due to the production of urease, causing further complications.
Uterine muscle responsiveness to oxytocin
a. is enhanced by an increased in estrogen/progesterone ratio
b. is prevented by high prostaglandin levels during pregnancy
c. is decreased by down-regulation of oxytocin receptors during the third trimester
d. is enhanced by decreased estrogen/progesterone ratio
e. is independent of estrogen/progesterone ratio
Answer:
a. is enhanced by an increased in estrogen/progesterone ratio
Explanation:
The average duration of human pregnancy is about 9 months or we can say 38 weeks/266 days. This period is called as the gestation period. At the end of pregnancy vigorous contraction of the uterus result in expulsion or delivery of the baby. It is called parturition. Parturition is a neuroendocrine mechanism.
During pregnancy uterine contractions are inhibited due to the high progesterone levels. Progesterone maintains the endometrium and prevents contraction of myometrium. At the end of last trimester, the progesterone levels plateau and then drops whereas estrogen levels continue to rise.
As a result the E/P ratio increases which makes the myometrium more sensitive to contraction stimuli.
The decreases levels of progesterone may lead to Braxton Hicks contraction which is nothing but false labor. Meanwhile, oxytocin hormone is secreted by the posterior pituitary gland which induces the contraction of myometrium.
What are potential hypotheses (explanations) regarding why there are ecological equivalents between many metatherians and eutherians? Discuss your hypothesis in terms of the processes that could lead to the current ecologies and distributions of metatherians and eutherians.
Answer:
by the needs of each specie
Explanation:
Biologist think that the metatherians are close related with marsupials, not just for the adaptive conditions, but also because their teeth.
The eutherians are close related with the placentarian animals, the main reason of the differences is the disposition of teeth.
The hypothesis will be related to the teeth, depending the shape and the position in the mouth, make that the animals prefer to eat different food, and with the passing of time they evolve in one or other group.
How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... CTTACAGTTTATTGATACGGAGAAGG... 3' 3'... GAATGTCAAATAACTATGCCTCTTCC... 5'
According to the DNA sequence shown in the question above, we can see that there are 3 reading frames without stop codons.
You can find this answer as follows:
Isolate the 5'-3' Sequence: You will not need the 3'-5' Sequence, so you will need to isolate the 5'-3' Sequence. In this, you should observe the sequence of nitrogenous bases, identify the codons (set of three nitrogenous bases) and create three sequences. The first sequence will exclude the first nucleotide, the second sequence will exclude the second nucleotide, and the third sequence will exclude nothing. Based on this, you will have the three sequences below:5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3' (no stop codon)
5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3' (no stop codon)
5'... CTT ACA GTT TAT TGA TAC GGA GAA GG... 3' (with stop codons)
Create complementary sequences: You should do this based on the complementarity of the nitrogenous bases. In this case, it is necessary to remember that Adenine (A) is complemented by Thymine (T), Guanine (G) is complemented by cytosine (C), and vice versa. In this case, you will have the following sequences:3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'
3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'
3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'
Identify the inverse complement: Still taking into account the complementarity of the nitrogenous bases, you should find the inverse complementary sequences. In this case, you will find the sequences:CCT TCT CCG TAT CAA TAA ACT GTA AG (with stop codon)
CC TTC TCC GTA TCA ATA AAC TGT AAG (no stop codon)
C CTT CTC CGT ATC AAT AAA CTG TAA G (with stop codons)
From this, we can see which sequences have one of the stop codons, which are TAA, TAG, TGA. As you can see, only the first two frames of the 5'-3' sequences and the first two frames of the inverse sequences do not have these codons, so it is possible to observe 3 open reading frames without stop codons.
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The query's genetic sequence potentially has three different open reading frames, starting from the first, second or third position. However, without specific information about the position of the start codon AUG and the stop codons, we can't definitively verify how many open reading frames extend through this sequence.
Explanation:In genetics, the concept of an open reading frame (ORF) is of immense importance. An ORF is a sequence of DNA or RNA that could be potentially translated to give a protein. Understanding the open reading frames in a genetic sequence helps us predict how that sequence might be translated into amino acids.
For the given sequence, we can have three ORFs since you can start reading the sequence from three different positions (first, second, or third nucleotide) until you run into a stop codon. However, without knowing the exact position of the start and stop codons in this specific sequence, we cannot accurately say exactly how many open reading frames might extend through it.
The concept of AUG start codon used to initiate translation and stop codons that terminate protein synthesis are essential for determining open reading frames. The sequence AUG signifies the beginning of an ORF.
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Variant Creutzfeldt-Jakob disease, listeriosis, anthrax, and E.coli 0157-H7 infections all have which of the following in common?
a. All four diseases can be transmitted from cattle to humans
b. All four diseases are caused by bacteria
c. All four diseases can be treated with antibiotics
d. All of the above are true for all four diseases.
Answer:
A. All four diseases can be transmitted from cattle to humans
Explanation:
Variant Creutzefeldt Jakob disease (VCJD) is a brain disease caused by a mutated protein (prion).
This particular type of CJD can be caused by eating beef from animals that were infected with bovine spongiform encephalopathy.
Listeriosis is caused by a bacteria called Listeria monocytogenes
Listeriosis often occurs through digesting contaminated foods such as raw meat, beef or dairy products.
E.coli-0157:H7
E. coli is a bacteria that occurs naturally in the digestive system of humans and animals.
However, it can be disease causing once it spreads to other parts of the body.
There are different strains of E.coli.
E.coli 0157:H7 is a dangerous strain to humans and is found in the manure of cattle, dogs and geese.
People can become sick with this by eating raw contaminated meat.
Anthrax
Anthrax is caused by a bacteria called Bacillus Anthracis.
Anthrax mainly affects livestock (cattle) and wild animals.
Anthrax is transmitted to humans by direct contact with an infected animal.
Light energy is converted into chemical energy in the:
a. Antenna complex
b. Reaction center
c. Stroma
d. Inner membrane
Answer:
The correct answer will be option-B.
Explanation:
Photosynthesis is the process which converts the light energy to chemical energy which is utilized in the formation of the glucose molecule.
The light-dependent reaction of photosynthesis begins when the photon of light is captured by the chlorophyll pigment molecules of the antenna complex. The chlorophyll electron gets excited which gets transferred via resonance pathway to a trans-membrane protein-pigment complex called photochemical reaction centre.
The photochemical reaction centre traps the excited electrons and passes the electron immediately to electron acceptors and an electron transport chain starts. Therefore it is the petrochemical reaction centre of the photosystem which converts light energy to chemical energy and thus, Option-B is the correct answer.
The amount of net secondary productivity cannot exceed the amount of net primary productivity.
a. True
b. False
Answer:
True
Explanation:
The net secondary productivity refers to the net amount of energy available to produce biomass at the consumer level. The energy available in the plant tissues after cellular respiration represents the net primary productivity. The net primary productivity represents the rate of incorporation of organic matter into the plant tissues. Hence, net primary productivity is the source of energy and organic matter for the net secondary productivity.
A part of the net primary productivity consumed by a herbivore is used in cellular respiration and other processes and the rest becomes available for biomass accumulation.
Due to the consumption of the part of the net primary productivity for vital processes at the consumer level, the amount of energy available for biomass accumulation, the net secondary productivity, is always less than the NPP.
Which of the following statements is not true about mRNA?
a. prokaryotic mRNA may contain multiple structural genes on the same transcript, known as polycistronic mRNA.
b. eukaryotes only transcribe one gene at a time on mRNA, called monocistronic mRNA.
c. some eukaryotes are capable of having polycistronic mRNA.
d. eukaryotes almost always produce polycistronic mRNA.
e. the genes for metabolic pathways in bacteria are typically located close together and transcribed on one mRNA.
Answer:
The correct answer is d. eukaryotes almost always produce polycistronic mRNA
Explanation:
In eukaryotes one transcription unit contains the information of only one gene which codes for only one protein or polypeptide therefore eukaryotic mRNA is called monocistronic mRNA.
In prokaryotes transription unit contains set of genes adjacent to each other which are transcribed together and codes for multiple proteins. So prokaryotic mRNA is called polycistronic mRNA.
Almost all messenger RNA present in eukaryotes are monocistronic mRNA because eukaryotes are more complex than prokaryotes and require modification at many stages which is easily possible with monocistronic mRNA.
Explain how a genetic map (in map units) is related to actual physical distance (in base pairs of DNA).
Explanation:
A map unit [or centimorgan (cM)] is a unit for measuring genetic linkage. It is defined as the distance between chromosome positions for which the expected average number of intervening chromosomal crossovers in a single generation is 0.01. Its relation with actual physical distances is inconsistent because the number of base pairs to which it corresponds varies widely across the genome, it also depends on whether the meiosis in which the crossing-over takes place is a part of male or female genes (female genome is 4782 cM long, while the male genome is only 2809 cM long).
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Genetic maps show the relative positions of genes using centimorgans as a measure of the likelihood of recombination, while physical maps show actual distances in base pairs. The conversion between cM and bp varies by organism and chromosome region. Three main methods—cytogenetic, radiation hybrid, and sequence mapping—are used to create physical maps.
A genetic map indicates the relative positions and distances between genetic markers or genes on a chromosome. These distances are expressed in map units or centimorgans (cM), which approximate the probability of recombination occurring between these markers during meiosis. A physical map, on the other hand, provides the actual physical distance between genetic markers, measured in the number of base pairs (bp).
It is essential to understand that one map unit does not correlate to a fixed number of base pairs across all species or even all regions of a chromosome. For example, in Arabidopsis, a model organism in plant biology, 1 cM is roughly equivalent to 150,000 base pairs. This variation is due to differing recombination frequencies across the genome; areas known as 'crossover hot spots' experience more frequent recombination, while regions of heterochromatin may show less. To construct physical maps, methods like cytogenetic mapping, radiation hybrid mapping, and sequence mapping are employed.
Do the Rickettsia bacteria harm or injure the tick or mite host, as we see in lice and Typhus? Explain.
Answer:
No. Rickettsia use an arthropod as a vector host to cause the disease in their final host, usually vertebrades. However, they do not harm vector host as ticks.
Explanation:
Rickettsia are strict parasites, they are bacteria that must live inside the cells of their hosts. Specifically, they are found in mammals and at some point in their life cycle, they are associated with arthropods (fleas, lice or ticks) that transport the parasite from one animal to another, without getting any harm. There are different diseases in humans associated with different species of Rickettsias and the arthropods that carry them:
Typhus is caused by Rickettsia prowazekii and transmitted by body or head lice.
Murine typhus is caused by Rickettsia typhi, transmitted by fleas.
Rickettsia rickettsii is normally transmitted by ticks causing Rocky Mountain spotted fever
Josey wants to look at the cells of an onion under a light microscope. She peels off a thin, transparent layer of onion and places it on a
microscope slide. Josey places the sample on the microscope stage, looks through the eyepiece, and adjusts the stage to focus the image
However, she is unable to see any details of the cells.
Which of the following possible solutions could possibly help Josey to see the onion cells in detail?
A. Josey should use a thicker piece of onion for her sample
B. Josey shouldhplace her onion sample in a vacuum.
C. Josey should place a drop of water on her sample
D. Josey should apply a stain, such as iodine, to her sample
Answer:
D
Explanation:
A cell and its organelles are usually transparent hence hard to discern without staining. There are different staining techniques that are useful for viewing various type of cells and organelles depending on which stain they pick up well. In this case of an onion, the starch molecules in the onion will take up the iodine and improve visualization. This due to the fact that iodine molecules get emended in the structure of the amylose and amylopectin molecules of starch causing starch molecules to turn blue lack (due to a change in light absorption spectra of the complex).
Final answer:
Josey should try applying a stain, such as iodine, to her onion sample to see the cells in detail. Using a thicker piece of onion, placing the sample in a vacuum, or adding a drop of water can also improve visibility, but applying a stain is the most effective solution.
Explanation:
To see the onion cells in detail, Josey should try applying a stain to her sample. Stains like iodine can help to make the cells more visible under the microscope. By adding a stain, the cells will appear darker or have a different color, making it easier to see their structures and details.
Using a thicker piece of onion for the sample or placing the sample in a vacuum will not help Josey see the onion cells in detail. A thicker piece of onion will not necessarily enhance the visibility of the cells, and placing the sample in a vacuum is not necessary for observing cells under a light microscope.
Adding a drop of water to the sample can actually help improve the visibility of the cells. Water can help to enhance the contrast between the cells and the microscope slide, making the cells easier to see.
If someone shakes your hand, the proteins that change shape and are responsible for initiating the feeling of this hand touch pressure are called:
a. stretch receptor proteins
b. carboxy receptor proteins
c. cardiac receptor proteins
d. dynamin receptor proteins
Answer:
d. dynamic receptor proteins
Explanation:
In the skin there are several receptor cells that can pick up various stimuli such as temperature, pressure and pain. They can be classified into mechanoreceptor, thermoreceptor and pain receptor cells. When we are dealing with mechanical stimuli like pressure, mechanoreceptors respond to stimuli by sending the signal to the brain.
Light touches on the skin are received by nerve endings called Meissner corpuscles and Merkel discs. Already more intense pressures (such as a handshake) are received by receptors called Pacini corpuscles.
In a nucleosome, the DNA is wrapped around
a. histones.
b. ribosomes.
c. polymerase molecules.
d. a thymine dimer.
Answer:
a. histones. is the correct answer.
Explanation:
In a nucleosome, the DNA is wrapped around eight histones proteins to create strong loops termed nucleosome.
A nucleosome is the structural subunit of the chromatin.
Chromatin is a fiber polymer of double-stranded DNA which is composed up of a nucleosome core. The nucleosome core is made of histone proteins
Each nucleosome is composed of DNA enclosed around eight histone proteins which are called histone octamer.
Tom's brother suffers from phenylketonuria (PKU), a recessive disorder. Tom and the brothers' parents do not have PKU. What are the chances that Tom is a carrier of the recessive PKU allele?
a. 1/4
b. 1/3
c. 1/2
d. 2/3
e. 4/3
Final answer:
Tom has a 2/3 chance of being a carrier of the recessive PKU allele.
Explanation:
To determine the chance that Tom is a carrier of the recessive PKU allele, we need to understand the inheritance pattern of PKU. PKU is a recessive disorder, which means that an individual needs to inherit two copies of the recessive allele to have the disorder. Since Tom's parents do not have PKU, they must be carriers of the recessive allele. This means that each parent has one copy of the recessive allele and one copy of the dominant allele.
If both parents are carriers, there are four possible combinations of alleles that they can pass on to their offspring: two dominant alleles (NN), one dominant allele and one recessive allele (Nn), one recessive allele and one dominant allele (nN), and two recessive alleles (nn).
Since Tom does not have PKU, we know that his genotype is either NN or Nn. The chance that Tom is a carrier of the recessive PKU allele is 2/3, which corresponds to the possibility of him having the Nn genotype. Therefore, the correct answer is d. 2/3.