Answer:
a. DNA polymerase proofreading: consequence of its absence is the DNA mutation
b. Mismatch repair enzymes : consequence of its absence impedes homologous recombination resulting in the final mutation
c. Nucleotide excision repair enzymes : the absence of nucleotide cleavage repair enzymes would impede the functioning of damaged DNA repair mechanisms
Explanation:
a. DNA polymerases are the enzymes that form the DNA in cells. During DNA replication (copying), most DNA polymerases can "check their work" with each base they add. This process is called review. If the polymerase detects that you have added a wrong nucleotide (incorrectly paired), remove it and replace it immediately, before continuing with DNA synthesis
b. In homologous recombination, the information from the homologous chromosome that matches that of the damaged one (or from a sister chromatid if the DNA has been copied) is used to repair the fragmentation. In this process the two homologous chromosomes are approached and the undamaged region of the homologue or the chromatide is used as a template to replace the damaged region of the broken chromosome. Homologous recombination is "cleaner" than the union of non-homologous ends and does not usually cause 11 mutations
c. Excision repair: damage to one or a few DNA bases is usually fixed by removing (excising) and replacing the damaged region. In repair by base cleavage, only the damaged base is removed. In nucleotide excision repair, as in the mating repair we saw earlier, a nucleotide section is removed
The absence of DNA repair mechanisms such as DNA polymerase proofreading, mismatch repair enzymes, and nucleotide excision repair enzymes can lead to an elevated mutation rate, causing genomic instability and increasing the risk of cancer and genetic disorders.
Consequences of Missing DNA Repair Mechanisms
The integrity of DNA is safeguarded by various repair mechanisms. When these systems fail or are absent, the stability and function of an organism's genome are compromised. Here's what happens when certain DNA repair mechanisms are missing:
DNA polymerase proofreading: DNA polymerase adds nucleotides during DNA replication and conducts proofreading to ensure accuracy. If proofreading is compromised, incorrect bases may be incorporated, resulting in mutations which can lead to cancers or other genetic disorders.
Mismatch repair enzymes: These enzymes correct errors that escape proofreading during replication. Without mismatch repair, these errors become permanent, increasing mutation rates and contributing to diseases like cancer.
Nucleotide excision repair enzymes: These enzymes repair bulky lesions such as thymine dimers which are primarily caused by UV light. Absence of these enzymes may lead to a higher incidence of skin cancer due to the accumulation of these lesions.
In summary, the lack of these DNA repair mechanisms can result in increased mutation rates, leading to genomic instability, and may significantly raise the risk of cancer and other hereditary diseases.
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
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).
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.
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.
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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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.
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.
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.
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.
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).
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.
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.
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.
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.
A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?
The best way of solving this is to draw a Punnett square.
You know the F0 had one parent with singed bristles (s), and normal wings (L), and the other parent is normal bristles (S) with vestigial wings (l).
If you do the cross ssLL x SSll you'll find 100% of the offspring is F1: SsLl, this means, all of them show the dominant traits: normal wings and normal bristles.
If you cross two parents from F1 to have F2, you'll find:
SsLl x SsLl = SSLL + SslL + sSlL+ ssll = 25% SSLL,all dominant traits. 50% SsLl is a recessive trait carrier but shows dominant traits. 25% ssll this one has all recessive alleles, which means, it will show vestigial wings and signed bristles.
The recombinant F2 offspring in this genetic cross of Drosophila are those that do not exhibit the parental trait combinations. Therefore, singed bristles/normal wings and normal bristles/vestigial wings are considered recombinant offspring. Option C is correct.
A researcher crossed true-breeding Drosophila flies with singed bristles (s) and normal wings (L) with true-breeding flies having normal bristles (S) and vestigial wings (l). The F1 offspring all had normal wings and normal bristles, indicating that these traits are dominant. When these F1 flies were crossed with flies having singed bristles and vestigial wings, the F2 generation showed various combinations of these traits.The F2 progeny would exhibit the following combinations: normal bristles/normal wings, singed bristles/vestigial wings, normal bristles/vestigial wings, and singed bristles/normal wings. The recombinant offspring are those that do not exhibit the parental combinations of traits. Thus, the recombinant F2 offspring would be singed bristles/normal wings and normal bristles/vestigial wings.Complete question as follows:
A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?
A.Singed bristles/vestigial wings only
B.Singed bristles/normal wings only
C.Singed bristles/normal wings and normal bristles/vestigial wings
D.Singed bristles/vestigial wings and normal bristles/normal wings
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.
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.
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.
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.
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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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.
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.
Besides a high voltage shock, what is another method to make E. coli competent to take up "naked" DNA?
a. high concentrations of calcium ions followed by high temperature
b. high concentrations of calcium ions and several hours on ice
c. large amounts of DNA added directly to a bacterial culture growing at 37C
d. high concentrations of minerals followed by high temperature
e. a high voltage shock is the only way to make E. coli competent
Answer:
The correct answer is a. high concentrations of calcium ions followed by high temperature.
Explanation:
Apart from high voltage shock, calcium chloride heat shock method can be used to make E. coli competent to take up naked DNA. As both DNA and lipids contain phosphorus and phosphorus contains negative charge, therefore, ice-chilled CaCl₂ makes cation bridge between phosphorylated lipid of cell membrane and phosphorus present in the DNA.
This allows the naked DNA to attach to the cell surface of bacteria. Then cells are given heat shock which creates gaps(holes) in the membrane and makes the cells competent to take up the naked DNA by these holes.
Then the transformed cells can be selected by screening methods like growing transformed cells on media containing a suitable antibiotics.
E. coli cells can be made competent to take up DNA by incubating with calcium ions and then applying heat shock, which allows DNA to enter the cells through pores formed in the membrane. So the correct option is a.
Explanation:Besides a high voltage shock, another method to make E. coli competent to take up DNA is by using high concentrations of calcium ions followed by a heat shock. This process involves incubating E. coli cells with an ice-cold solution of 50 mM calcium chloride at 4°C for 30 minutes, which allows the DNA molecules to attach to the cell exterior. Following this, the cells are briefly incubated at 42°C, which causes the formation of pores in the membrane, allowing DNA to enter the cell.
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.
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.
Microphylls are found in which plant group?
a. lycophytes
b. liverworts
c. ferns
d. hornworts
Answer:
a. lycophytes is the correct answer.
Explanation:
Microphylls are found in lycophytes.
lycophyte is a vascular plant, they have a unique type of leaves called microphylls.
The lycophytes belong to the division of Lycophyta, they are seedless plants and have vascular tissue.
The leaves(microphylls) present in the lycophyte have a single vein that is unbranched and narrow, which provides the water to the leaf and supplies nutrients to other parts of the plant.
Microphylls, which are characterized by a single vascular vein, are found primarily in lycophytes. Lycophytes are an ancient group of vascular plants and include species like club mosses and quillworts.
Explanation:Microphylls are a specific type of leaf that are typically characterized by only having a single vascular strand or vein. This contrasts with megaphylls, which usually have multiple veins. The answer to which plant group microphylls are found in is a. lycophytes.
Lycophytes are one of the oldest groups of vascular plants and are recognized by their microphyllous leaves. Notable examples of lycophytes include club mosses, quillworts, and spike mosses. The presence of microphylls is a defining characteristic of this plant group.
Learn more about Microphylls and Lycophytes here:https://brainly.com/question/32107182
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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.
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
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.