Answer: It is difficult to kill clostridium botulinum in food because the organism grow under low oxygen condition and produce spores and toxin. The bacterium can survive under harsh condition and produce spores that are resistant to heat, drying and chemicals. These spores germinate into bacteria in food when conditions is favourable.
Explanation:
Botulisim is a serious and fatal illness caused by the neurotoxin secreted by clostridium botulinum. This toxin block nerve function and can lead to respiratory and muscular paralysis. The toxin is found in improper processed can food.
If you are performing site-directed mutagenesis to test predictions about which residues are essential for a protein’s function. Which of each pair of amino acid substitutions listed below would you expect to disrupt protein structure the most? And why? how do you go abouts knowing this?
O Pro replaced by Gly or His,
O Gln replaced by Glu or Asn,
O Lys replaced by Asp or Arg
Answer:
O Gln replaced by Glu or Asn
Explanation:
Gln replaced by Glu or Asn, is a non-conservative substitution. Gln has a much shorter side chain or R-group than Glu or Asn. Additionally both Glu and Asn R-groups are polar and negatively charged while Gln is non-polar.In a protein, replacement with Glu or Asn would greatly destabilize the protein by introducing a charge and disrupting hydrophobic interactions formed between Gln and other non-polar residues. By comparing the R-groups of the residues, their length, hydrophobicity, charge and shape. One can predict if the substitution is disruptive or not, if residues are very similar then the interactions that they form are preserved. For example replacing Gln with Ala would be less disruptive.You treat cells with the ionic detergent sodium dodecyl-sulfate (SDS) to disrupt the membrane. Will integral membrane proteins be affected by this treatment and how?
Answer:
Yes. the Integral membrane will be affected.
Explanation:
Generally, Ionic detergents have hydrophilic head group which can be either negatively charged or positively charged. If treated with plasma membrane, it masks the native charge of the protein, colonizing the entire charges of the protein, converting the overall charge to its own.
To be specific SSD is anionic. It therefore adds overall negative charge(anion) to the entire protein charges, even if they have isoelectric points. ''just like the HIV-RNA genome which undergo retro-transcription of the entire human DNA genome to its own HIV- genome of the infected host!''.
This disrupts the hydrogen bonds, hydrophobic interactions and other non-covalent bonds of protein , thus the 3-Dimensional protein structure of tertiary protein and therefore of the integral membrane proteins . Denaturation of protein structure, and therefore the conformation protein function results.With resulting separation of the protein molecules based on their sizes.
what are the drawbacks to only regulating at the transcriptional level and not the translational level or protein level?
Answer:
Energy and Cellular Space
When gene Gene expression at( transnational level ) is regulated it ensures energy and cellular space conservation, If a gene were to be regulated at each time after transcription, then significant amount of ATPs beyond the synthetic capacity of the cells will be needed, the cells will need large surface to accommodate the mechanism s. Thus regulation saves more energy and space to regulate gene expression. by turning on genes when needed to expressed and off when not needed.
Sequence of Gene
In addition, for the timely synthesis of exact copies of protein needed by the cells for various cellular activities; it is important for cells to regulate and control how the DNA is translated to the required proteins. If this mechanisms were not regulated deletion , addition of nitrogenous bases in gene sequence during translation may lead to mutations and therefore wrong coding of the needed protein in the cells.
Protein Quantity
Furthermore, the need to know the quantity of protein to synthesize, when to stop the synthesis, necessitated regulation of the process.If required amount is not expressed wrong amino acids units will be synthesized leading to abnormalities in hormones and enzymes.
Explanation:
Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Decreasing the salt concentration of the solution lowers DNA's melting point (Tm). Increasing the G-C content of DNA raises its melting point (Tm). Single-stranded DNAs can only anneal to one another if they are 100% identical in nucleotide sequence. Melting point (Tm) is the temperature at which the DNA is one-half double-stranded and one-half single-stranded. DNA is more likely to denature when exposed to a high pH.
Answer: "Decreasing the salt concentration of the solution lowers DNA's melting point (Tm)" is not a true statement
Explanation:
Increasing salt concentration would lower the DNA's melting point (Tm), not otherwise.
For instance:
- In 8M urea (8M means 8 Moles per dm3), Tm is decreased by nearly 20°C.
- 95% formamide at room temperature would completely denature the double stranded DNA.
Thus, higher concentration of salts like urea or formamide lowers Tm, not otherwise
Which of the following specialized structures/inclusions would aquatic photoautotrophic bacteria likely possess? 1. Thylakoids 2. PHB Granules 3. Carboxysomes 4. Gas vacuoles 5. Chloroplasts
Answer:
1. Thylakoids 2. PHB Granules 3. Carboxysomes
Explanation:
1. Thylakoids are membrane bounded compartments present in bacteria for light dependednt reactions.
2. PHB Granules help in carbon fixation
3. Carboxysomes help to retain carbon when enough supply isn't available.
Answer: Option 1,2 and 3.
Thykaloids, PHB granules and carboxysomes.
Explanation:
Carboxysomes consist of polyhedral protein and enzymes Rubisco which is important to supply carbon and fix it.
Thykaloids are membrane bound part of chloroplast which is the site of light dependent reaction during photosynthesis.
PHB granules are important component which help to fix carbon .
A dominant mutation in Drosophila called Delta causes changes in wing morphology in Delta/+ heterozygots. Homozygosity for this mutation (Delta / Delta) is lethal. In a population of 150 flies, it was determined that 60 bad normal wings and 90 had abnormal wings. a. What are the allele frequencies in this population? b. Using the allele frequencies calculated in part (a), how many total zygotes must be produced by this population in order for you to count 160 viable adults in the next generation? c. Given that there is random mating, no migration, and no mutation, and ignoring the effects of genetic drift, what are the expected numbers of the different genotypes in the next generation if 160 viable offspring of the population in part (a) are counted? d. Is this next generation at Hardy-Weinberg equilibrium? Why or why not?
How much ATP (in solid form) do you need to weigh, in order to make 1 mL of stock solution with a concentration of 100 mM ATP? The molecular weight of ATP is 551.14
Answer:
mass(g) = 0.55114g
Explanation:
Given that;
Molarity (M) 100mM = 0.1M
Volume (V) = 1mL
= 1.0 × 10⁻³ Litre
numbers of moles = Molarity × Volume
= 0.1M × 1.0 × 10⁻³ L
= 0.001
= 1.0 × 10⁻³ mol
Now, to determine the mass of ATP (in solid form) that we need to weigh;
we have our numbers of moles to be =[tex]\frac{mass(g)}{molarmass(g)}[/tex]
where;
numbers of moles = 1.0 × 10⁻³ mol
molar mass = 551.14
mass(g) = numbers of moles × molar mass
mass(g) = 1.0 × 10⁻³ mol × 551.14
mass(g) = 0.55114g
∴ 0.55114g ATP(in solid form) is required to be weighed in order to make 1 mL of stock solution with a concentration of 100 mM ATP.
Do all animals share a common ancestor?
Answer:
The most recent common ancestor of all currently living organisms is the last universal ancestor, which lived about 3.9 billion years ago. ... 6,331 groups of genes common to all living animals have been identified; these may have arisen from a single common ancestor that lived 650 million years ago in the Precambrian.
A researcher identified a bacterial enzyme that is essential in the breakdown of glucose. The researcher wants to test a potential antibiotic that acts on the newly identified enzyme. She finds that glucose is indeed broken down at a slower rate when the potential antibiotic is present. Addition of a higher concentration of glucose does not have any impact on Vmax. Which of the following properly characterizes this finding?
a. The potential antibiotic may be a competitive or noncompetitive inhibitor of the enzyme. These possibilities could be tested by adding more enzyme.
b. The potential antibiotic is a competitive inhibitor of the glucose-converting enzyme and could be an effective treatment for bacterial infections.
c. The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
d. The potential antibiotic reduces the number of available enzymes and is, therefore, a competitive inhibitor.
Answer:
The potential antibiotic is a competitive inhibitor of the glucose converting enzyme and could be an effective solution for bacterial infections.
Explanation:
Enzyme kinetics mainly deals with the study of the rate of reaction of the enzyme. Different factors that can affects the enzyme rate are also studied in enzyme kinetics.
The competitive inhibitors of the enzyme do not show any effect on the Vmax value of the enzyme and binds to the enzyme only. These inhibitors do not have ability to binds with enzyme substrate complex. The increase in the enzyme and substrate concentration can remove the inhibitor affect.
Thus, the correct answer is option (b).
The potential antibiotic is acting as a noncompetitive inhibitor since higher glucose concentrations do not affect the Vmax, suggesting the antibiotic changes the enzyme's conformation rather than competing with glucose for the active site. So the correct option is c.
Explanation:The finding that the potential antibiotic slows down the breakdown of glucose by a bacterial enzyme but that a higher concentration of glucose does not affect the Vmax (maximum rate of enzyme activity) indicates the antibiotic is likely acting as a noncompetitive inhibitor. A competitive inhibitor would have a different effect: the Vmax could potentially still be reached if sufficient additional substrate (in this case, glucose) were added, as competitive inhibitors can be outcompeted by high concentrations of the substrate. Since adding more glucose did not change the Vmax, this suggests that the potential antibiotic is not a competitive inhibitor but rather implies that the antibiotic binds at a site other than the active site, thereby altering the enzyme's conformation and affecting its activity without directly blocking the substrate from binding. Therefore, the correct characterization of this finding would be option c: The potential antibiotic is a noncompetitive inhibitor of the enzyme and likely changes the shape of the active site.
An RFLP can be created:_________. a) only by a mutation creating a new restriction enzyme recognition site b) only by a basepair change within the restriction enzyme recognition site c) only by a deletion removing a restriction enzyme reognition site d) by any mutational event that cretes or deletes a restriction enzyme recognition site e) only in RNA
Answer:
Option B, only by a basepair change within the restriction enzyme recognition site
Explanation:
A restriction fragment length polymorphism (RFLP) consists of alternative alleles with varying sizes of restriction fragments. In the traditional RFLPs, base pairs were changes at the restriction sites. These restriction sites comprises of nucleotide sequences that are identified by restriction enzymes. In RFLP analysis, the restriction fragments are created when restriction enzyme divide DNA into fragments. These restriction fragments are then separated while gel electrophoresis.
Before the advent of PCR, RFLPs (which were predominant form of DNA variation) were used to analyze linkages.
Hence, option B is correct
Cells of a normally rod-shaped bacterium (e.g., Bacillus subtilis) that have completely lost the ability to produce the MreB protein would mostly likely be Choose one: A. nonflagellated. B. stalked (like Caulobacter). C. coccoid-shaped. D. filamentous in form. E. unable to divide symmetrically.
Answer:
C. coccoid-shaped.
Explanation:
MreB protein serves as function as actin proteins do in the eukaryotic cells. MreB protein is involved in maintaining the cell shape. It is observed in many rod-shaped bacteria and archaea such as Bacillus subtilis. MreB polymerizes to form a spiral around the inside periphery of the cell and maintain the elongated shape of the cell. A bacterium that is not able to produce functional MreB protein cannot maintain the rod shape of its cell and is coccoid-shaped.
Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where T is tall and t is short, and for color, where R is red and r is tan. If a short, tan male goat mates with a tall, red female goat of an unknown genotype, what is the probability that they would produce short, tan offspring?
and....
One of a pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of green and round offspring.
Answer:
(A) 0.0625
(B) 25%
Explanation:
A
Given that ;
T is tall
t is short
R is red
r is tan.
If a short, tan male goat mates with a tall, red female goat of an unknown genotype;
Let's take it one after the other;
A short tan male goat = ttrr
The female goat is an unknown genotype, we will have to determine that but we were told that it is tall and red
∴ For tall it is T and red is R;
Definitely, the female genotype should be T_ R_;
the probability genotype of the female for each allele is either:
- TT or Tt
- RR or Rr
However, from above;
the probability that the female will be Tt = [tex]\frac{1}{2}[/tex]
the probability that the female will be Rr = [tex]\frac{1}{2}[/tex]
the probability that the recessive allele 't' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]
the probability that the recessive allele 'r' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]
∴ the probability that they would produce short, tan offspring (ttrr) will be;
= [tex](\frac{1}{2})^{4}[/tex]
= [tex](\frac{1}{16})[/tex]
= 0.0625
(B)
One of a pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of green and round offspring.
For Color, Let;
Y = Yellow
y = Green
For shape, Let;
R = Round
r = wrinkled
If a cross occur between Yy Rr × yy rr. (i.e yellowround and green wrinkled)
the percentage of green and round offspring will be;
If Yy Rr self crossed; we have the following traits (YR, Yr, yR, yr)
yy rr will result to (yr, yr)
YR Yr yR yr
yr YyRr Yyrr yyRr Yyrr
yr YyRr Yyrr yyRr Yyrr
Yellow Yellow Green Green
Round Wrinkled Round Wrinkled
Therefore, the percentage of Green and Round offsprings from above is:
= [tex]\frac{2}{8}*100%[/tex]%
=[tex]\frac{1}{4}*100[/tex]%
= 25%
An organism is classified as a heterotroph or autotroph based on the type of ________ it utilizes.. A. respiration source (e.g., oxygen or other) B. nitrogen source C. carbon source D. none of the above
Answer: Option C) carbon source
Explanation:
The type of Carbon source is the distinction between heterotroph or autotroph. As autotrophs like green plants generate their food (glucose) from simple inorganic molecules like carbon dioxide (CO2) in the atmosphere, while heterotroph like man utilize carbon in carbohydrates such as starch that have been stored in the tissues of autotrophs.
Thus, the type of carbon source utilized distinguished heterotroph and autotrophs
Which of the following is true of both starch and cellulose? a. They are both polymers of glucose. b. They are geometric isomers of each other. c. They can both be digested by humans. d. They are both used for energy storage in plants. e. They are both structural components of the plant cell wall.
Answer: Option A. They have both polymers of glucose
Explanation:
Starch and cellulose are both polymers of glucose. They have similar structure, Starch and cellulose are two similar polymers. They are both made from the same monomer, glucose, and have the same glucose-based repeat units. They consists of long chains of glucose molecules connected to (1-4)-glycosidic bonds. Glycosidic bonds are the standard way of attaching things to glucose. The (1-4) bit simply means that the glucose molecules in the chain are connected to each other to the 1st and 4th carbon in the glucose ring opposite each other.
The key similarity between glycogen and cellulose is that both are polymers of glucose.
Glycogen:
It is a polymer of glucose that is stored in animals as a storage of energy.
Cellulose:
It is also a polymer of glucose. It forms the cell wall of glucose.In plants starch work as the storage of energy.Therefore, the key similarity between glycogen and cellulose is that both are polymers of glucose.
To know more about glucose polymer:
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Which of the following statements are accurate? (MORE THAN ONE ANSWER)
A. Invertebrates are animals that lack a backbone
B. Ecdysozoans typically have an exoskeleton
C. Bilaterians include three clades:
1) Deuterstomia
2) Lophotrochozoa
3) Ecdysozoa
D. Givens about animal "A'
I. the animal has three prominent germ layers
II. an animal has bilateral symmetry
III. the animal lacks a backbone
You would be able to classify this animal, animal A, as a bilaterian
E. All animals share a common ancestor
Answer:
The statements in options A,B,C,D and E are all correct.
A. Invertebrates lack a back bone
B. Ecdysozoans typically posses a flexible exoskeleton that protects these animals from harmful external factors like water loss.
C. Deuterstomia, Lophotrochozoa and Ecdysozoa are classes of Bilaterians.
D. Animal A is a bilateral because bilateral embryos are triploblastic possessing three germ layers, are bilaterally symmetrical and lack a back bone.
E. All animal from studies are found to have a common ancestry.
Two autosomal genes control horn color in dragons. Pure-breedinggold-horned dragons were mated to pure-breeding silver-horneddragons. All of the F1 were gold. The F1 were intermated and the F2generation consisted of 147 gold, 17 silver and 92 bronze.
Fill in the blanks with whole numbers to indicate the geneticallybased phenotypic ratio that should be hypothesized to explain theF2 data.
gold: silver: bronze
Conduct a chi-square test to test the appropriate type ofepistasis.
In the chi-square test,
The expected number of gold-horned dragons is
The expected number of silver-horned dragons is
The expected number of bronze-horned dragons is
Answer:
The genetically based phenotypic ratio for of Gold: Silver: Bronze = 9:1:6
The expected number of gold-horned dragons is = 144
The expected number of silver-horned dragons is = 16
The expected number of bronze-horned dragons is = 96
Explanation:
Given that; F2 generation consisted of:
147 gold
17 silver
92 bronze.
Total autosomal genes control horn color in dragons = 256
When we conducted the punnet square for the Pure-breedinggold-horned dragons were mated with the pure-breeding silver-horneddragons, then intercrossed them for the F2 generation, It is seen that;
The genetically based phenotypic ratio for of Gold: Silver: Bronze = 9:1:6
∴
The expected number of gold-horned dragon = [tex]256*\frac{9}{16}[/tex]
= 144
The expected number of silver- horned dragon = [tex]256*\frac{1}{16}[/tex]
= 16
The expected number of Bronze-horned dragon = [tex]256*\frac{6}{16}[/tex]
= 96
Null Hypothesis = The observed ratio of Gold:Siver:Bronze
=144:16:96
=9:1:6
Observed Expected (OF - EF)² [tex]\frac{(OF-EF)^2}{EF}[/tex]
Frequency Frequency
(OF) (EF)
147 144 9 0.062500
17 16 1 0.062500
92 96 16 0.166667
Total: 0.291667
Since Degree of Freedom = 2
Our P-Value = 0.864
∴ Our P. Value is very high that we accept the null hypothesis
An elephant had a long, powerful trunk. According to the ideas of Lamarck, how did the trait of long, powerful trunks develop develop in elephants
Answer:
D
Explanation:
Lamark believed in the inheritance of acquired characteristics. That is, that if an individual's body or organs changed in some way during the course of their life time (due to the environment or some other pressure), that these characteristics could be passed on to the next generation.
This was his theory of evolution. Therefore, if, slowly elephants started using their trunks to an advantage more, they would start to develop and grow. Lamark believed these acquired traits would be passed on to the next generation. We now know this not to be true, and random, selectively advantageous mutations in DNA are what cause evolution over time.
A cross is done between two mutant T4 phage (a- b+ c+) and (a+ b-c)(no order implied by these genotypes). The genotypes of the progeny from this cross and their relative frequencies are given below: a-b+ c+ (30%) a+D c. (30%) a+ b+ c-(2%) a+ b+ c+ (8%) a- b-e+(2%) a-b-c-(8%) e+b-c+ (10%) a-b+ c-(10%) What is the distance between "a" and "b? Please give your answer in % recombination, but L number please! EAVE OUT the units (NO % sign), such as 20, or 3,
Since, the relative frequency is incorrectly mentioned in the question. The correct relative frequency table is attached below.
Answer:
20.
Explanation:
The genotype with the least recombinant frequency will represent the double cross overs.
The double cross overs progeny are a+ b+ c-(2%) and a- b-c+(2%).
The percentage of the recombinant frequency determined the distance between the genes. In the double crossovers, the b gene gas been exchanged and present in the middle.
The single recombinant crossovers as compared with the parents are a+ b+ c+ (8%) and a-b-c-(8%).
Distance between the gene a and b = Single cross overs + double cross overs
Distance between the gene a and b = 8 + 8 +2 +2 = 20.
Therefore, the distance distance between the gene a and b is 20 centi morgan.
Which of the following statements is true regarding DNA replication? A. The 5' to 3' orientation of the new strand will depend upon where it is in the replication bubble relative to the position where replication first starts. B. The new strand is the opposite of the template strand. C. The new strand is an exact complement to the template strand. D. The new strand is exactly the same as the template strand.
Answer:
These options are true about DNA.
A. The 5' to 3' orientation of the new strand will depend upon where it is in the replication bubble relative to the position where replication first starts.
B. The new strand is the opposite of the template strand.
C. The new strand is an exact complement to the template strand.
Explanation:
Since the two strands of a DNA double helix are antiparallel, this 5′-to-3′ DNA synthesis can take place continuously on only one of the strands at a replication fork (the leading strand). On the lagging strand, short DNA fragments must be made by a “backstitching” process. The opposite strand with a base sequence directly corresponding to the mRNA sequence is refereed as coding strand. See images for clarification.
If you want to study more here is the reference:
Alberts B, Johnson A, Lewis J, et al. Molecular Biology of the Cell. 4th edition. New York: Garland Science; 2002. DNA Replication Mechanisms.
The new strand formed during DNA replication is an exact complement to the template strand.
Explanation:The correct statement regarding DNA replication is C. The new strand is an exact complement to the template strand. DNA replication is a semi-conservative process where each of the original parent DNA strands acts as the template for the synthesis of a new complementary strand. The parent DNA strands separate, and each strand serves as a template for the synthesis of a new strand. Each base on the parent strand pairs with its complementary base on the new strand, following the base-pairing rules (A with T and G with C).
Therefore, the new strand formed during DNA replication is an exact complement to the template strand. In other words, it has the opposite nucleotide sequence of the template strand, with the same specific adenine (A) to thymine (T) and guanine (G) to cytosine (C) ratios.
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. A normal vision male marries a female who is a carrier of the color-blind gene and she becomes pregnant. What will a genetic counselor tell the parents regarding the a) percentage of having a daughter who is color-blind; b) what is the percentage of having a son who is color-blind? Consider a different couple with a female with normal vision and a male with normal vision. Would it be possible for their daughter to be a carrier for color-blindness? Explain your answer.
Answer:
Explanation:
a, The genetic counselor should explain that color-blindness is a sex-linked disorder, that is its expression is related to a particular sex-cells.
a) The percentage is 0% . This is because your husband is normal, so no sex-linked alleleis attached to his only X-chromosomes.However either of your X-chromosomes can bear the allele for color blind. Since a girl child sex-chromosomes combination is XX, and your husband X chromosomes is normal,hence your daughters can only be carries (XCX, XCX)like you if they inherited any of your X -chromosomes color blind allele.But none will be colorblind.
b)Because you are carrier for this alllele, and your husband is normal; chances are all your sons be colourblind is 50% proportion(YXC/YXC). This is because your husband Y chromosomes determines a male child, and you can bear the defective(carrier allele) for colourblind on either of your X-chromosmes to give a male child as XY. Thus either of the male child assuming with Mendelian fashion of inheritance, will be colour blind in 50% proportion.
No. A girl child can only be a carrier for color blindness if the inherited the color blind gene from her carrier mother or the father is color-blind. None of this is present in this scenario. The two parents are normal Therefore there is no chance of their daughter emerging a carrier.
A genetic counselor would tell the parents that (a) the percentage of having a daughter who is color-blind is 0% and (b) the percentage of having a son who is color-blind is 50%.
When considering the genetic possibilities for a couple where the female is a carrier of the color-blindness gene and the male has normal vision, genetic counselors would provide the following insights:
The percentage of having a daughter who is color-blind is 0%. This is because for a female to be color-blind, she must inherit two color-blindness genes – one from each parent. In this scenario, the father can only pass on a normal vision gene.
The percentage of having a son who is color-blind is 50%. Since males have only one X chromosome, inherited from their mother, they will be color-blind if that X chromosome carries the color-blindness gene.
For a different couple, both with normal vision, their daughter can still be a carrier for color-blindness if the mother is a carrier or if the father's mother is a carrier. This is because the mother can pass a carrier X chromosome to the daughter, or the father can pass an X chromosome that he inherited from his color-blindness carrier mother.
What kind of evidence has recently made it necessary to assign the prokaryotes to either of two different domains, rather than assigning all prokaryotes to the same kingdomA) molecularB) behavioralC) ecologicalD) nutritionalE) anatomical
Answer:
A) molecular
Explanation:
Prokaryotes have 70S type of ribosomes. The small ribosomal subunit of ribosomes has some conserved sequences. This subunit is similar in the closely related species. On the other hand, the structure of the small ribosomal subunit is different in the organisms that are distantly related.
Microbiologist Carl Woese analyzed the small ribosomal subunits of some bacteria and Archaeans and found distinct differences. On the basis of the composition of this biomolecule, prokaryotes were divided into eubacteria and archaea.
Final answer:
Molecular evidence, especially differences in cell membrane structure and SSU rRNA sequences, has led to prokaryotes being classified into two separate domains: Bacteria and Archaea, which differ significantly from each other and from Eukarya.
Explanation:
The evidence that has recently made it necessary to assign the prokaryotes to either of two different domains, rather than assigning all prokaryotes to the same kingdom, is molecular evidence. Advances in genetic analysis have revealed significant genetic differences between two groups of prokaryotes, leading to the classification into two separate domains: Bacteria and Archaea. These domains are based on differences in the structure of cell membranes and in ribosomal RNA (rRNA), specifically the sequences of small-subunit ribosomal RNA (SSU rRNA). While both domains consist of organisms with prokaryotic cells, lacking a nucleus and true membrane-bound organelles, they are as different from each other as they are from the third domain, Eukarya, which includes all eukaryotic organisms with membrane-bound organelles and a nucleus.
Early twentieth century Russia was in great turmoil as the Bolshevik Revolution ended 300 years of monarchical rule. Tsar Nicholas Romanov, his wife, Tsarina Alexandra, and their five children were placed under house arrest, but disappeared in the summer of 1918. Rumors abounded about their possible execution or escape from Russia.
Two years later, a young woman named Anna Anderson claimed that she was Anastasia, the youngest daughter of Tsar Nicholas II. There was much controversy surrounding Anna's claim. Some believed she was a fraud, others that she was indeed the lost Princess Anastasia. Anna remained steadfast in her claim until her death in 1984.
In 1991, the remains of nine skeletons, five male and four female, were exhumed from a shallow grave east of Moscow. Evidence from nuclear DNA showed that three of the young women were related and that one of the men and one of the women were their parents.
Further evidence from mitochondrial DNA (mtDNA) showed that one of the women could be positively identified as Tsarina Alexandra and that one of the men was indeed Tsar Nicholas II. The other three women had mtDNA that matched that of the Tsarina's and were identified as three of the Tsar's children. Anastasia and her younger brother, Alexei, were not among those found in the grave. This led to further speculation about Anastasia's possible escape from Russia after her parents were killed.
In 2007, an additional grave was found near that exhumed in 1991. It contained the remains of a teenage girl and boy. These remains were mtDNA tested, as well as samples from Anna Anderson, uncovered from hospital storage long after her death.
Analysis of the mtDNA supported the hypothesis that Princess Anastasia had been killed with her family in 1918. Which of the following statements support this hypothesis? Select all that apply.
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra.
b. The mtDNA from the remains of the teenage girl matched that of Tsar Nicholas II.
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
d. The mtDNA from Anna Anderson matched that of Tsarina Alexandra.
e. The mtDNA of the teenage girl and boy matched each other.
Answer:
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
Explanation:
Mitochondria contain their own DNA known as mtDNA, a strand which is inherited maternally. mtDNA can be used to find maternal ancestry and cannot be passed down by a paternal parent.
This information provides evidence given that the remains of the teenage girl had mtDNA that was identical to the mother, Tsarina Alexandra. It also gave evidence as to why Anna Anderson's claim of being Anastasia were false as the mtDNA of her and Tsarina Alexandra were not similar.
As to why the other answers cannot support the hypothesis, mtDNA is a maternally passed-down strand making the father, Tsar Nicholas Romanov, from having identical mtDNA to his children. On the other hand, if the teenage girl were to not have matching mtDNA to that of Tsuarina's mtDNA, this would disprove the theory of the girl being related to her.
While the teenage boy also had matching mtDNA with the girl, this only proves that the two were related to one another but not necessarily related to Tsarina unless the mtDNA of all three were proven identical.
The statements supporting the hypothesis concerning the killing of Princess Anastasia with her family in 1918 are stated below:
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra.
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
e. The mtDNA of the teenage girl and boy matched each other.
Clearly, these statements negated Anna Anderson's claim of being the youngest daughter of Tsar Nicholas II.
The matching of Anastasia's mtDNA with Alexei's mtDNA in 2007 confirmed that Tsarina Alexandria was indeed their mother. Recall that teenagers were not among the first nine skeletons exhumed in 1991.
Thus, the 2007 discovery unequivocally confirmed the hypothesis that Princess Anastasia was killed alongside her family in 1918 during the Russian revolution.
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A scientist discovers a new organism living deep beneath the earth surface this organism thrive despite intense pressure heat lack of water and sunlight this organism is probably a
Answer:
The organism that can live deep beneath the earth surface despite intense pressure heat lack of water and sunlight might be Nematodes.
Explanation:
Nematodes are able to cope extreme heat or extreme cold and dehydration. They have adopted by learning technique that allows them to survive. They can transform into a hardy form called the dauer stage. They can survive harsh conditions for longer durations at this stage. And again awaken themselves when conditions are favourable again. They can be found in hot springs, deserts, high up mountains and in the deepest oceans.If a cell contains a set of duplicated chromosomes, does it contain any more genetic information than the cell before the chromosomes were duplicated?
Answer:
No.
Explanation:
When the set of chromosomes are duplicated, it means the chromosomes will be double the number.
However the genetic information will not change. It will remain the same.
Duplication of chromosomes is usually occurs among the interphase stage of mitosis where chromosomes will duplicate to ensure the formation of two identical daughter cells.
In these cells there will be no addition or variation in genetic information.
Final answer:
A cell with duplicated chromosomes doesn't contain more genetic information than before; it simply has two identical copies of each chromosome, called sister chromatids. The total DNA in the cell is increased, but genetic content remains consistent. This replication ensures that each daughter cell gets a complete genetic set after cell division.
Explanation:
If a cell contains a set of duplicated chromosomes, it does not contain more genetic information than it did before the chromosomes were duplicated. During DNA replication, each chromosome in the cell creates an exact copy of itself, known as sister chromatids, which are attached at a region called the centromere. While the chromosome number remains the same, for instance in humans (n = 46), the physical amount of DNA within the cell is actually doubled. However, in terms of genetic content, the information stays the same, it's just that each chromosome now consists of two identical chromatids.
During the cell cycle, specifically in the S phase, the DNA replication occurs to prepare a cell for division into two cells. This process is crucial because it ensures that each daughter cell receives a complete set of the organism's genome. In human cells, after replication, there still are 46 chromosomes, but each chromosome is now made up of two sister chromatids. Therefore, after cell division during mitosis, each cell will have the same amount of genetic material, identical to the original parent cell, assuring the continuity of genetic information.
In response to stress, cells in the adrenal cortex produce cortisol, a steroid hormone. Therefore, these cells would specifically exhibit large amounts of this organelle: Select one a. lysosomesb. rough ER C. smooth ER d. ribosomes e. vaults
Answer:C. smooth ER
Explanation:the endoplasmic reticulum is a membrane bound organelle found in the cytoplasm of the cell.it has a highly folded appearance.
Its function in the cell is the transportation of cellular products.
There are two types of endoplasmic reticulum;the rough endoplasmic reticulum and the smooth endoplasmic reticulum.
The rough endoplasmic reticulum have ribosomes attached to them ,giving them a granular appearance.the ribosomes synthesis proteins,which the rough endoplasmic reticulum transports outside or within the cell.
The smooth endoplasmic reticulum do not have ribosomes attached to it.its functions is to synthesis lipids.
It is also involved in the production of steroid hormones in the adrenal cortex and endocrine glands
Large amounts of this organelle C. Smooth ER Therefore , C. smooth ER is correct .
Smooth Endoplasmic Reticulum (Smooth ER):
The smooth ER is an organelle involved in various functions, one of which is the synthesis of lipids and steroids, including cortisol.
Steroid hormones like cortisol are lipids, and their synthesis primarily occurs in the smooth ER of cells.
In the adrenal cortex, cells are specialized to produce steroid hormones such as cortisol.
Cortisol is a crucial steroid hormone involved in the body's response to stress and helps regulate various physiological processes.
Cortisol Production:
The production of cortisol starts with cholesterol, which is transformed into cortisol through a series of enzymatic reactions, predominantly occurring in the smooth ER of the adrenal cortex cells.
The enzymes responsible for the synthesis of cortisol are predominantly located in the smooth ER membranes.
During stressful situations, the body signals the adrenal cortex to increase cortisol production.
The cells in the adrenal cortex respond by synthesizing more cortisol, leading to an increase in the smooth ER activity and volume to accommodate the higher demand for steroid hormone synthesis.
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A population is made up of individuals where 104 have the A1A1 genotype, 55 have the A1A2 genotype, and 37 have the A2A2 genotype. What is the allele frequency of A1? Answer to 2 decimal places.
Answer:
0.67
Explanation:
In this population there are:
104 A1A1 individuals55 A1A2 individuals37 A2A2 individualsN: 196
The frequency of the A1 is the amount of times that allele appears divided by the total number of alleles in the population.
The A1 allele is present twice in the A1A1 individuals and once in the A1A2. Each individual has 2 alleles, so the total number of alleles is N × 2.The frequency of the A1 allele can be thus calculated using the following formula:
[tex]freq(A1)=\frac{2*A1A1\ + A1A2}{2*N} \\\\freq(A1)=\frac{2*104\ + 55}{2*196} \\\\\\freq(A1)= 0.67[/tex]
A woman with Turner syndrome is found to be color-blind (an X-linked recessive phenotype). Both her mother and her father have normal vision. a. Explain the simultaneous origin of Turner syndrome and color blindness by the abnormal behavior of chromosomes at meiosis. b. Can your explanation distinguish whether the abnormal chromosome behavior occurred in the father or the mother? c. Can your explanation distinguish whether the abnormal chromosome behavior occurred at the first or second division of meiosis? d. Now assume that a color-blind Klinefelter man has parents with normal vision, and answer parts a, b, and c
Answer:
A) The turner syndrome is characterized by lacking half chromosome numbers which are also termed as the monosomy as there is only half the number of X linked chromosomes that are responsible for the colorblindness will be greater. The chances are 50% of having this disorder in such a case.
B) The correct answer would be no, as the error will be shown in only one of the due to n number of chromosomes present in both parents.
C) Anaphase II of meiosis II it takes places as here only the strands are pulled to opposite poles of the cell
.
D) Assuming it as Klinefelter man with colorblindness has normal parents then,
a)double XX +Y , the colorblind gene is recessive in XX and Y. or recessive in XX but dominating Y only rules out the other recessive.
b) Explanation is similar as B)
c) Due to a nondisjunction event during meiosis, I or meiosis II in the female X chromosome is still present in it.
What causes the myosin head to move into the 'cocked' position after it is released from actin? Group of answer choices ATP binding to myosin. ATP hydrolysis into ADP and Pi on the myosin head. ADP and Pi release from the myosin head Ca2+ binding to troponin.
Answer:
Explanation:
The molecular and cellular mechanisms and processes that explain muscle contraction in striated muscle occur in the myofibril sarcomere. Their understanding depends on the organization's understanding of the structure of the sarcomere. In an imaginary experiment we first assemble an ideal sarcomere.
Remember that the myofibril is a set of cylindrical compartments that are located next to each other, constituting an elongated cylinder. Each of these cylinders is a sarcomere and borders its neighbor on a line or band called, line or band z.
On each side of the z line, thin cylindrical filaments that are actin filaments are inserted. Each actin filament is formed by a double strand of actin molecules that are rolled over each other. In this organization, actin is called actin F.
The myosin head is 'cocked' into a high-energy state due to ATP hydrolysis into ADP and Pi, which occurs after ATP binds to myosin and it detaches from actin.
Explanation:The cause of the myosin head moving into the 'cocked' position after it is released from actin is ATP hydrolysis into ADP and Pi on the myosin head. When ATP binds to myosin, it results in the release of the myosin head from actin. Subsequently, the hydrolysis of ATP to ADP and inorganic phosphate (Pi) occurs due to the ATPase activity intrinsic to myosin. This process releases energy that changes the angle of the myosin head into the 'cocked' configuration. In this high-energy state, the myosin head has potential energy and is prepared for further movement upon binding to actin again, which is essential for muscle contraction.
Which statement is TRUE? a. All fatty acids in a triglyceride are unsaturated. b. All lipids contain fatty acids. c. All steroids contain a 4-ring region. d. All phospholipids have one saturated fatty acid tail and one unsaturated fatty acid tail.
Answer: Option C.
All steroids contain 4 ring region.
Explanation:
Steroids are drug like or biological compound that look like cortisol hormone. Steroids have four linked carbon ring and their rings are fused together. Steriod are hydrophobic and are not soluble in water. They have short tail. Steroids are important cell components and they are signaling molecules.
An mRNA molecule has a sequence 5'- CAGAUCUAAUGCUUAUCGGAU-3'. When translated in a laboratory setting in which translation can be initiated anywhere along the molecule, how many reading frames are possible?
Answer:
Three reading frames
Explanation:
Translation of mRNA always happen in one direction from the 5' end to the 3' end of the RNA strand.
Reading frame refers to the grouping of three consecutive bases to form a codon that can constitute an amino acid.
There are six possible reading frames in any nucleotide sequence.
Three from the 3' to the 5' end and three possible reading frames from the 5' to the 3' end.
As mentioned earlier, translation in mRNA happens in one direction therefore the three possible reading frames are;
5'- C AGA UCU AAU GCU UAU CGG AU-3'. 5'- CA GAU CUA AUG CUU AUC GGA U-3'. 5'- CAG AUC UAA UGC UUA UCG GAU-3'.An mRNA sequence can be translated into a protein in three possible reading frames, therefore, for the given sequence that can be initiated anywhere, there are three possible reading frames.
Explanation:In the realm of molecular biology, an mRNA sequence can be translated into a protein in three possible reading frames. Each reading frame will read the sequence from a different starting point, thereby creating a unique series of codons, and potentially, a different protein sequence. Hence, for an mRNA sequence such as 5'- CAGAUCUAAUGCUUAUCGGAU-3' that can be translated anywhere in a laboratory setting, it means there are 3 possible reading frames. Each frame begins from a different nucleotide within the first three (frame 1 from C, frame 2 from A, frame 3 from G) and continues in triplets from there on.
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