Answer:
Part A: Pp
Part B: 1/3
Explanation:
A: The purple flower color is regulated by dominant allele "P". Since self-fertilization of the purple-flowered plant with unknown genotype gives both purple and white-flowered progeny in almost 3:1 ratio, the parent plant was heterozygous dominant for the trait. Its genotype is "Pp".
Segregation of recessive and dominant alleles during gamete formation leads to the production of two types of gametes with either "P" or "p" alleles. The random fusion of these gametes produces progeny in following phenotype ratio= 3 purple: 1 white.
B. Out of the 3 purple-flowered progeny, 1 is homozygous dominant (PP). Therefore, the probability of the purple-flowered progeny to breed true is 1/3.
Final answer:
The genotype of the original purple-flowered datura plant is determined to be heterozygous (Pp) based on the ratios of purple to white flowered progeny. The probability that a randomly selected purple-flowered progeny will breed true for the purple color is 1/4, meaning there is a 25% chance it is homozygous dominant (PP).
Explanation:
Part A: Genotype Determination
To determine the genotype of the original purple-flowered datura plant, we can look at the progeny from its self-fertilization. Given that the purple flower color is dominant and represented as allele P and the white color is recessive and represented as allele p, we can infer the genotype of the parent plant based on the ratio of purple to white flowered plants among its progeny. The progeny consists of 28 purple-flowered plants and 10 white-flowered plants. This suggests that the parent plant is not a homozygous dominant (PP) because we expect all progeny to have purple flowers in that scenario. Since white flowers (pp) are a recessive trait, they can only appear in the offspring if the parent carries the recessive allele. Therefore, the parent plant must be heterozygous Pp.
Part B: Probability of True Breeding
The probability that a randomly selected purple-flowered progeny plant will breed true (be homozygous dominant PP) can be inferred from the parent's genotype. Since the parent plant is heterozygous Pp, its possible gametes are P or p. With self-fertilization, there is a 1 in 4 chance (or 25%) of getting PP, a 1 in 4 chance of pp, and a 2 in 4 chance of Pp. The probability that the purple progeny plant selected is a homozygote (PP) is the same as finding a homozygous dominant offspring in the self-fertilization, which is 1/4 or option 'b'. The other two-thirds will be heterozygous purple (Pp).
Aniridia is a human condition in which the eye has no iris. The protein encoded by the gene responsible for Aniridia is very nearly identical to the protein product of the fly eyeless protein. What experiment could provide evidence that the two genes are functionally equivalent? A. Introduce the aniridia mutation into Drosophila embryos to look for iris formation. B. Sequence the eyeless and aniridia DNA sequence and regulators. C. Use the Eyeless mRNA as a probe in other invertebrate and non-mammalian species. D. Introduce the mouse aniridia wild-type sequence into the fly to see whether flys eyes develop. E. Introduce the mouse wild-type sequence into the fly egg to see whether mouse eyes develop.
Answer:
Option B. Sequence the aniridia and eyeless DNA sequence and regulators.
Explanation:
DNA sequencing:
It is the method of determining the sequence of nucleic acid. DNA sequence includes the method to determine the order of nucleotide sequences. Through the process of sequencing aniridia and eyeless DNA are sequenced and analysed on Gel electrophoresis. Sequencing similarity shows evidence that the two genes are functionally equivalent.
Best choice:
B. Sequence the eyeless and aniridia DNA sequence and regulators.
The experiment could provide evidence that the two genes are functionally equivalent - D. Introduce the mouse aniridia wild-type sequence into the fly to see whether fly's eyes develop.
Aniridiais a human condition in which the eye has no iris.
The protein is encoded by the PAX6 gene.this gene is responsible for Aniridia by mutation is very nearly identical to the protein product of the fly eyeless protein.if we introduce the mouse aniridia wild-type sequence or gene into a fly and check the result we would be able to find evidence that the two genes are functionally evidentif fly develops eyes then both are functionally equivalent.Thus, The experiment could provide evidence that the two genes are functionally equivalent - D. Introduce the mouse aniridia wild-type sequence into the fly to see whether fly's eyes develop.
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The vagus nerve does not innervate the ________.
a. kidneys
b. gallbladder
c. pancreas
d. parotid gland
Answer: D.the parotid gland
Explanation:
The innervation of the parotid gland is by auriculotemporal nerve and great auricular nerve as sensory innervation , and
the parasympathetic innervation originated from glossopharyngeal nerve;which synapse with otic ganglion;from this end point(the ganglion) the auriculotemporal nerve conduct parasympathetic fibres to the parotid gland.
The superior cervical ganglion , innervated the parotid gland as the sympathetic innervation. It travels along the external carotid artery to reach the parotid gland.
therefore PAROTID GLAND IS NOT INNERVATED BY VAGUS NERVE.
All other options in the questions are innervated by the vagus nerve.
Vagus nerve has the longest branch of all the cranial nerves, which extends from the brain to the abdomen. As it exist the jugular foremen , it innervates the head, the neck and abdomen to supply the above options a-c
The vagus nerve, which is part of the parasympathetic nervous system, does not innervate the parotid gland; instead, the parotid gland is innervated by the glossopharyngeal nerve.
Explanation:The vagus nerve does not innervate the parotid gland. The vagus nerve is a crucial part of the parasympathetic nervous system, which controls the function of many internal organs. The vagus nerve innervates structures such as the kidneys, gallbladder, and the pancreas. In particular, to the pancreas, the autonomic nervous system controls the secretion of insulin and glucagon, indicating that both sympathetic and parasympathetic nerve endings are found in pancreatic islets. Parasympathetic neurons, which are part of the peripheral nervous system and include the vagus nerve, leave the central nervous system via cranial nerves or sacral regions of the spinal cord. These neurons synapse in ganglia that are very close to the organs they innervate. The parotid gland, however, is innervated by the glossopharyngeal nerve, not the vagus nerve.
Calculate the average ml of oxygen molecules in 100 ml blood in the athletes at low altitude and then in athletes in high altitude while training at high altitude. Use the following information: 1.39 ml of oxygen per gram of hemoglobin.
Answer:
[tex]27.8[/tex] mL of oxygen in [tex]100[/tex] ml of blood
Explanation:
It is proven that a healthy human being has in general [tex]20[/tex] grams of hemoglobin in [tex]100[/tex] milliliters of blood.
It is given that -
Amount of oxygen found in [tex]1[/tex] grams of hemoglobin is equal to [tex]1.39[/tex] milliliters of oxygen.
Thus, the total amount of oxygen in milliliters would be equal to product of total weight of hemoglobin and the total amount of oxygen in ml in one gram of hemoglobin.
[tex]= 20 * 1.39 \\= 27.8[/tex] mL
What is a major difference between DNA polymerase I and DNA polymerase III?
(A) DNA polymerase I synthesizes DNA on leading strands and DNA polymerase III synthesizes DNA on lagging strands
(B) DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
(C) DNA polymerase I repairs DNA and DNA polymerase III synthesizes DNA in the 3׳ to 5׳ direction
(D) DNA polymerase I synthesizes DNA in the 5׳ to 3׳ direction and DNA polymerase III synthesizes on lagging strands
Answer:
(B) DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
Explanation:
Both the enzymes DNA polymerase I and DNA polymerase III involved in the process of DNA Replication with specialised functions. DNA polymerase I synthesize DNA on lagging strand where it degrades RNA primer and replace it with DNA. On the other hand, DNA polymerase III synthesize DNA from 5' to 3' end on the leading and lagging strand but stops at the RNA Primer.
Answer:
DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
Explanation:
If an important cytoplasmic determinant were missing from an animal zygote, what would you expect to happen as the zygote developed into a multicellular organism? If an important cytoplasmic determinant were missing from an animal zygote, what would you expect to happen as the zygote developed into a multicellular organism? The animal would lack certain amino acids. The animal would lack ribosomes. DNA polymerase would be unable to replicate the animal’s genome. The animal would develop normally. One of the animal’s major body axes would fail to develop normally.
Answer: Option E - One of the animal’s major body axes would fail to develop normally.
Explanation:
Cytoplasmic determinants help the organs of the embryo to develop.
Hence, absence of cytoplasmic determinants would result in one of the animal’s major body axes failing to develop normally.
The electron transport chain (ETC), or respiratory chain, is linked to proton movement and ATP synthesis. Select the statements that accurately describe the electron transport chain. Electrons generated by the citric acid cycle in the intermembrane space enter the ETC. The reactions of the ETC take place in the inner membrane of mitochondria. Consider what must be true for a proton gradient to be set up across the inner membrane. Electron carriers in the ETC include ubiquinone (coenzyme Q) and cytochrome c. Prosthetic groups, such as iron–sulfur centers, are directly involved with electron transfer. Electron carriers are organized into four complexes of proteins and prosthetic groups. Electron transfer in the ETC is coupled to proton transfer from the matrix to the intermembrane space.
Answer: Electron transfer in the ETC is coupled to proton transfer from the matrix to the intermembrane space.
Explanation:
The energy generated by the movement of electrons through protein carriers at different energy levels(ETC) is used to pump protons across the intramembrane to the matrix. Thus, if the electron transport stops, protons pumps stops, ATPs synthesis by the ATP-synthase stops
This shows a coupled reaction
Why do mountain climbers carry oxygen?
Answer:
The atmospheric pressure goes on decreasing as we go up above the sea-level. Therefore, the amount of oxygen also decreases at higher altitude. Due to this reason, mountaineers carry oxygen cylinders with them, while climbing high mountains
Explanation:
Jenner's successful use of cowpox virus as a vaccine against the smallpox virus was due to the fact that ______. (A) the cowpox virus made antibodies in response to the presence of smallpox(B) the immune system responds nonspecifically to antigens(C) cowpox and smallpox are caused by the same virus(D) there are some epitopes (antigenic determinants) common to both pox viruses
Answer: option C) cowpox and smallpox are caused by the same virus
Explanation:
Cowpox is a skin disease that affects cattle. It is caused by an Orthopoxvirus, with lesions occurring principally on the udder and teats of the animals.
Human infection may occur from touching infected cows, and thus giving immunity to smallpox (an acute infection caused by the same poxyvirus, in HUMANS)
So, Jenner's successful use was because cowpox and smallpox are caused by the same virus
Which of the following is an acceptable definition of evolution?
a. a change in the genotypic makeup of an individualb. a change in the genetic makeup of a populationc. a change in the phenotypic makeup of a populationd. a change in the environmental conditionse. a change in the species composition of a community
Answer:
a change in the species composition of a community
Explanation:
Answer:
B: A change in the genetic makeup of a population
Explanation:
Evolution can be defined as any net directional change or any cumulative change in the characteristics of organisms or populations over many generations. It explicitly includes the origin as well as the spread of alleles, variants, trait values, or character states.
Arsenate is a toxic ion that can interfere with both glycolysis and oxidative phosphorylation. Arsenate resembles inorganic phosphate (P) and can replace it in many enzymatic reactions. One such reaction is catalysed by glyceraldehyde 3-phosphate dehydrogenase in glycolysis. Arsenate competes with phosphate for entry into the active site of glyceraldehyde 3-phosphate dehydrogenase and upon completion of the reaction, instead of the normal product, 1,3-bisphosphoglycerate, the mixed anhydride 1-arsenato- 3-phosphoglycerate is formed: this undergoes rapid spontaneous hydrolysis into arsenate plus 3-phosphoglycerate, the latter being a normal intermediate in glycolysis, so glycolysis is able to proceed from the step at 3-phosphoglycerate. What would be the effect of arsenate in glycolysis? (a) Arsenate activates glycolysis so that more ATP and NADH is generated for every glucose molecule. In the presence of arsenate there is no net formation of ATP from glycolysis, but NADH generation is not directly affected Arsenate brings glycolysis to an abrupt stop with immediate lethal consequences In the presence of arsenate glycolysis produces fewer ATP and NADH molecules per glucose molecule. Arsenate does not affect the number of ATP or NADH molecules generated per glucose molecule. (b) (c) (d) (e)
Answer:
In the presence of arsenate glycolysis produces fewer ATP and NADH molecules per glucose molecule.
Explanation:
it is supposed to be 2 molecules of 3-phosphoglecerate with each producing 2 ATP making a total of 4 ATP gain but since it is just one and arsenate, only 2 ATP will be generated instead of four, thus ATP yield is reduced.
Prokaryotes and eukaryotes share some mechanisms for controlling gene expression, but all of the following are uniquely used in eukaryotes EXCEPT:a) Enhancers, silencers, cell-type specific factorsb)Operons for coordinated gene expressionc)chromatin remodelingd)RNA processing
Answer:
b) Operons for coordinated gene expression
Explanation:
Operon refers to a unit of genetic function and is mainly found in bacteria and phages. An operon consists of a promoter, an operator, and a coordinately regulated cluster of structural genes. The genes of an operon code for proteins or enzymes that mostly function in a common pathway. It allows a single promoter and operator or any other regulatory sequence to regulate the expression of functionally related genes.
For instance, lac operon consists of three structural genes that code for the enzymes involved in the catabolism of lactose sugar. The operon is expressed only when lactose is available in the medium. This pattern of gene regulation is not found in eukaryotes.
The chemical mechanisms used to avoid errors in protein synthesis are different from those used during DNA replication. DNA polymerases use a 3 ′ to 5 ′ exonuclease proofreading activity to remove mispaired nucleotides incorrectly inserted into a growing DNA strand. There is no analogous proofreading function on ribosomes. Instead, proofreading for protein synthesis is carried out by the aminoacyl‑tRNA synthetases. ldentify the mechanisms aminoacyl-tRNA synthetases use to ensure that the correct amino acid is attached to the appropriate tRNA.(A) To distinguish between two amino acids with similar structures, some aminoacyl-tRNA synthetases use a proofreading site that recognizes and hydrolyzes the incorrect amino acid.(B) The active site in each aminoacyl-tRNA synthetase has a high specificity for the correct amino acid substrate. After binding the correct amino acid, an aminoacyl-tRNA synthetase recruits the appropriate tRNA to the active site by recognizing a number of sequences and structural features in a specific tRNA.(C) Aminoacyl-tRNA synthetases have an additional active site that binds to non-cognate tRNAs. The tRNAs that bind to this second active are hydrolyzed and released from the enzyme.(D) First, the aminoacyl-tRNA synthetase binds to its cognate tRNA by recognizing a number of specific sequences and structural features in the tRNA. Next, the synthetase-tRNA complex recruits the correct amino acid to the enzyme active site.(E) Aminoacyl-tRNA synthetases can rapidly hydrolyze the ester linkage between incorrectly paired RNAs and amino acids
Aminoacyl-tRNA synthetases ensure the correct amino acid is attached to the appropriate tRNA through mechanisms such as the high specificity of the active site and the proofreading features of some synthetases.
Explanation:Aminoacyl-tRNA synthetases ensure that the correct amino acid is attached to the appropriate tRNA by using specific mechanisms. One mechanism is the high specificity of the active site in each aminoacyl-tRNA synthetase for the correct amino acid substrate. The synthetase binds to its cognate tRNA and recruits the correct amino acid to the enzyme's active site. Another mechanism is the proofreading feature in some aminoacyl-tRNA synthetases that recognizes and hydrolyzes incorrect amino acids to ensure the correct one is attached.
Enzymes interact with many different substrates a. true or b. false
Answer:
false
Explanation:
because enzyme has specific active site
Answer:
True
Explanation:
Enzymes works by binding with chemical reactants called substrates. There may be one or more substrates for each type of enzyme, depending on the particular chemical reaction.
Why is ""Climate Change"" a more accurate description of the increase in Earth’s temperature than ""Global Warming""? A. it is unclear whether or not warming of the planet is occurring B. some portions of the Earth will become colder while others become warmer C. the planet is undergoing a decrease in overall temperature
Answer: A. it is unclear whether or not warming of the planet is occurring
Explanation:
Climate change is a more accurate description of the increase in Earth’s temperature than "global warming," because it is unclear whether or not warming of the planet is occurring, which is present in Option A.
What are climate change and global warming?When the climate, such as the temperature, humidity pattern, and rain pattern, changes, this is referred to as climate change; however, global warming refers to the increase in global temperature, which does not fully explain the changed environment on Earth. Many biodiversities on Earth are affected by climate change: the seashore changes, rain patterns change, habitats change, and it is all due to climate change.
Hence, climate change is a more accurate description of the increase in Earth’s temperature than "global warming," because it is unclear whether or not warming of the planet is occurring, which is present in Option A.
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In his studies of alcoholic fermentation by yeast, Louis Pasteur noted that the sudden addition of oxygen (O2) to the previously anaerobic culture of fermentation grape juice resulted in a dramatic decrease in the rate of glucose consumption. This "Pasteur Effect" can be counteracted by the addition of 2,4-dinitrophenol (DNP), an uncoupler of oxidative phosphorylation. (6 marks) (a) Why would the yeast cells consume less glucose in the presence of oxygen? (b) Why would DNP counteract or prevent the Pasteur Effect?
Answer:
A. 38 mol of ATP is yielded during aerobic oxidation of glucose while only 2 ATP is yielded during anaerobic oxidation of glucose. This will cause the yeast cell to need 19 times more glucose anaerobically and to need less glucose aerobically.
(b) DNP is going to block the major step in aerobic ATP production causing the yeast cells to use about the same amount of glucose as anaerobic cells when compared to yeast cells without DNP present
After an exceptionally rainy spring, there is no longer a barrier between the populations, and some individuals migrate inland from the coast. Consequently, 20 % of individuals in the new inland population are originally from the coastal population. Calculate the spotted allele frequency for the inland population after migration, q combined . Round the answer to the nearest hundredth.
Answer
After inland population after migration allele frequency is 0.62 or 62%
Explanation:
Given,
Coastal striped phenotype freq. = 0.22
ss = 0.22
[tex]q_{coastal} \times q_{coastal} = 0.22[/tex]
[tex]q_{2 coastal}[/tex]= 0.22
Similarly, inland striped phenotype freq. = 0.43
[tex]q_{2inland}[/tex] = 0.43
[tex]q_{coastal} = \sqrt{q_{2coastal}}[/tex]
= [tex]\sqrt{0.22}[/tex]
= 0.4690
[tex]q_{coastal}[/tex]= 0.47 i.e. 47%
[tex]q_{inland} = \sqrt{0.43}[/tex]
= 0.655
[tex]q_{inland}[/tex] = 0.66 i.e. 66%
the migration range (m) is given as 20%
m= 0.2
allele freq. after migration = pre migration + ∆q
here,
∆q = change in the allele frequency
or
migration of allele freq. from coastal to inland
=[tex]m(q_{coastal} - q_{inland})[/tex]
= 0.2 (0.47 – 0.66)
=[tex]0.2 \times (- 0.191)[/tex]
= - 0.0382
∆q = -0.04 i.e. 4%
Final answer:
In population 1, genetic drift likely caused the change in allele frequencies, while in population 2, natural selection likely caused the change.
Explanation:
In population 1, the change in allele frequencies is likely caused by genetic drift. Genetic drift is the random change in allele frequencies due to chance events, and it is more likely to occur in smaller populations, where chance can have a larger impact. The small population size and the fact that only half of the individuals reproduce suggest that genetic drift played a role in the change in allele frequencies in population 1.
In population 2, the change in allele frequencies is likely caused by natural selection. Natural selection is a process where certain traits or alleles are favored due to their beneficial effects on survival and reproduction. The larger population size and the fact that there is differential reproduction depending on the traits suggest that natural selection played a role in the change in allele frequencies in population 2.
Clover is a plant that is found in many fields in Europe. There is a genetic polymorphism present in clover in Europe associated with the release of cyanide from the leaf tissue, when the leaf tissue is damaged. Individuals of type AA release lots of cyanide when the leaf is damaged, Aa individuals an intermediate amount, and aa individuals NO cyanide. The release of cyanide when the leaf is damaged prevents or reduces other leaves of the plant from being eaten by herbivorous animals, such as slugs, cattle, etc.
Answer:
Absolutely, Genetic Polymorphism is observed
Explanation:
In genetics, Polymorphism refers to the regular existence of two or more different genotypes within a given species or population.
Definitely, the Clover leaf as described above exhibit genetic Polymorphism, bearing that the existence of the 3 genotypes: AA, Aa, and aa in Same species of Clover has been observed.
Note that this condition occurs because that natural selection is unable to pick a single trait above another; thus diverse phenotype of Clover leaves are observed.
If Chargaff's equivalence rule is valid, then hypothetically we could extrapolate this to the combined genomes of all species on Earth (as if there were one huge Earth genome).
In other words, the total amount of A in every genome on Earth should equal the total amount of T in every genome on Earth.
Likewise, the total amount of G in every genome on Earth should equal the total amount of C in every genome on Earth.Calculate the average percentage for each base in your completed table.
Do Chargaff's equivalence rules still hold true when you consider those six species together?
Answer:
yes . A approximately equals to T and G approximately equals C in average.
Explanation:
according to Chargaff's equivalence rule
the number and concentration of adenine is similar to the number of thymine and number of guanine is similar to that of cytosine in the DNA.
so according to conditions given in the question i.e average amount of A should be equal to average amount of T and total average amount of G in every genome on Earth should equal the total average amount of C . so Chargaff's equivalence rules still hold true when you consider those six species together.
The two divisions of the ANS normally have what relationship?
a. disruptive
b. mutualistic
c. antagonistic
d. synergistic
Answer:
The correct answer will be option-C
Explanation:
The autonomic nervous system is the system of the nerves which controls the automatic response of the body.
The ANS is categorized into two types one of which is the parasympathetic and sympathetic nervous system.
The parasympathetic nervous system gets activated in the relax conditions by producing hormones that relax the human mind and body.
The sympathetic nervous system is activated in the stress conditions which produces hormones which response wither in fight response or flight response.
Since these types show antagonistic approaches in the body, therefore, Option-C is the correct answer.
Which of the following characteristics would NOT increase or decrease the fitness of the organisms that possess it? a. Certain mice in a population carry genes that cause their fur to be darker than the fur of other mice in the population that do not possess those genes. b. Certain fruit flies in a population carry genes that increase their ability to survive longer without food than other fruit flies in the population that do not possess that gene. c. Certain sterile mules in a population carry genes that allow them to carry heavier loads than other sterile mules in the population that do not possess those genes. d. Certain humans in a population carry genes that cause their eyesight to deteriorate at a more rapid rate than other humans in the population that do not possess those genes. e. Certain male birds in a population carry genes that increase the size of their tail relative to the size of the tails of the male birds in the population that do not possess those genes.
Answer:
maybe for me is answer is c
The characteristic related to sterile mules does not affect fitness since they cannot pass on their genes. Other characteristics either increase or decrease fitness based on survival and reproduction advantages or disadvantages.Option c is correct.
Let's analyze each characteristic with respect to how they affect fitness:
Certain mice in a population carry genes that cause their fur to be darker than the fur of other mice in the population that do not possess those genes. This characteristic could affect fitness, depending on the environment. Dark fur might offer camouflage in certain habitats, increasing survival rates.Certain fruit flies in a population carry genes that increase their ability to survive longer without food than other fruit flies in the population that do not possess that gene. This characteristic would likely increase fitness, as it provides a survival advantage.Certain sterile mules in a population carry genes that allow them to carry heavier loads than other sterile mules in the population who do not possess those genes. This characteristic does not affect fitness, as the mules are sterile and cannot pass on their genes.Certain humans in a population carry genes that cause their eyesight to deteriorate at a more rapid rate than other humans in the population that do not possess those genes. This characteristic would likely decrease fitness, as poor eyesight could affect survival and reproduction.Certain male birds in a population carry genes that increase the size of their tail relative to the size of the tails of the male birds in the population that do not possess those genes. This characteristic might affect fitness positively or negatively, depending on the trade-offs between attracting mates and evading predators.Therefore, the characteristic that would NOT increase or decrease the fitness of the organisms that possess it is related to sterile mules (option c).
The first organisms on Earth appeared approximately 3.8 billion years ago and were all unicellular. Multicellular organisms have only existed for about one billion years.
What evolutionary milestone made it possible for more complex, multicellular organisms to exist?
The development of specialized cells and cellular cooperation made it possible for more complex, multicellular organisms to exist.
Explanation:The evolutionary milestone that made it possible for more complex, multicellular organisms to exist was the development of specialized cells and cellular cooperation. As cells started to live together in colonies, some cells began to specialize in performing different functions, making the colony more efficient. This specialization allowed the cells to remain small while still carrying out specialized tasks, leading to the evolution of multicellular organisms. These organisms were bigger, more efficient, and capable of performing more complex functions than single-celled organisms.
Final answer:
The development of eukaryotic cells, cell specialization within colonies, and sexual reproduction were pivotal milestones in the evolution of multicellular organisms which led to the diverse array of life we see today.
Explanation:
The evolutionary milestone that made it possible for more complex, multicellular organisms to exist was the development of eukaryotic cells and later the advent of colonial organisms and cell specialization. Eukaryotic cells, which emerged about 2 billion years ago, have more complex structures like a nucleus, allowing for greater control over cellular functions. Another key development was the emergence of sexual reproduction over 1 billion years ago, which led to increased genetic variation and the potential for greater evolutionary adaptation.
As single-cell eukaryotes started forming colonies, some cells in these colonies began specializing in different functions. These specialized cells could perform tasks more efficiently, benefitting the organism as a whole and setting the stage for the evolution of true multicellularity. True multicellular plants appeared about 1 billion years ago, and true animals followed roughly 100 million years later.
Ultimately, these developments laid the foundation for the Cambrian Explosion, a period of rapid evolution around 540 million years ago during which most major animal phyla appeared. Multicellular life evolved over time through an increase in biodiversity and complexity. Organisms like Volvox illustrate the progression from solitary eukaryotic cells to more complex multicellular structures.
Your are a microbiologist. The owner of a restaurant hires you to examine a sample of preserved vegetables that appears to have spoiled in its can. The metal can only contains vegetables and water. It appears to have been sealed tightly, but the sides of the can have bulged. Could bacteria in the can have survived the canning process and caused the sides to bulge?
Answer:
Yes bacteria can survive if heating process is inadequate.
Explanation:
Most of the canned food is spoiled by thermophilic and mesophilic bacteria. These bacteria can survive the canning process if the can is inadequately heated. Most of them are able to produce gas which can cause bulging of can.
Clostridium species are mainly involved in the spoilage of canned food. C. thermosaccharolyticum is a thermophilic bacteria which spoil can food and evolve CO2 gas that cause bulging of can. Example of mesophilic bacteria that cause spoilage and swelling of can are C. botulinum, C. puterifaciens, etc.
Therefore thermophilic and mesophilic bacteria can survive the canning process if heat treatment and the temperature are not adequate.
Final answer:
Bacteria such as Clostridium botulinum could survive improper canning procedures, germinate in an anaerobic environment, and produce gas, causing the can to bulge, indicating possible botulinum toxin presence.
Explanation:
Yes, bacteria in the can could have survived the canning process and caused the sides to bulge. This is often due to the presence of anaerobic bacteria such as Clostridium botulinum, which can survive in low-oxygen environments like a sealed can. If the canning process was not properly conducted, the spores of such bacteria could remain viable, and when conditions inside the can become anaerobic, these spores can germinate and produce gas as a byproduct of their metabolism, resulting in the bulging of the can. This scenario is also a potential sign of the presence of botulinum toxin, which is highly toxic. It's important that the restaurant owner disposes of these cans safely and reviews their canning procedures to prevent potential foodborne illnesses.
Which of the following is true about Gran Dolina adult hominids? a. They were more modern than Homo erectus, and like later Homo sapiens, had a wide nasal aperture. b. They had a larger cranial capacity than later Homo sapiens. c. They were able to produce spectacular art, similar to later Homo sapiens. d. They were similar to modern humans but with a narrow, Homo erectus–like nasal aperture.
Answer:d. They were similar to modern humans but with a narrow, Homo erectus–like nasal aperture.
Explanation:
Gran Dolina adult hominids were more advanced than Homo erectus, with a wide nasal aperture similar to that of Homo sapiens supporting this fact. Their cranial capacity was not larger than that of Homo sapiens, and there is no evidence suggesting they produced art like later Homo sapiens. Moreover, their nasal aperture was wider rather than narrow like that of Homo erectus.
Explanation:The correct answer would be option 'a'. Gran Dolina adult hominids were indeed more advanced than Homo erectus, resembling later Homo sapiens in having a wide nasal aperture. This is consistent with data suggesting that these individuals were an intermediary between Homo erectus and later Homo sapiens. Such anatomical features place Gran Dolina adult hominids closer in affinity to modern humans than to Homo erectus.
However, they were not as modern as Homo sapiens in other aspects. For example, their cranial capacity was not larger than that of later Homo sapiens (option b), and there is no evidence suggesting that they were capable of producing art as spectacular as that of later Homo sapiens (option c).
Moreover, while they were similar to modern humans in many respects, they did not have a narrow, Homo erectus-like nasal aperture (option d). Instead, as mentioned earlier, they had a wide nasal aperture.
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What is the value of using peer reviews in scientific work?
A. To prove that your work is correct
B. To leverage the collective expertise of your peers and their unique perspectives
C. To make the process of writing a journal article faster
D. To brag about your accomplishments to your friends
The answer is option A "To prove that your work is correct." When you let peers review your work its making sure that your results/data are correct. When doing experiments to make sure that your results are valid you would do trials on top of that have your peers repeat the experiment with their own trails to make sure the data you get is correct.
Hope this helps.
Red flowered snapdragons are crossed with white flowered, producing all pink snapdragons in the F1. What would you expect if you crossed red with Pink?
Select correct answer:
all pink
3/4 red, 1/4 pink
1/2 red, 1/2 pink
1/2 red, 1/2 white
Answer:
It probably get pink because it has 1/2 chance it has 1/4 change of being white and 1/4 chance of being red
Explanation:
"Metamorphosis" means to change form. Metamorphic rocks directly form from A) igneous rocks. B) magma deep within the Earth. C) igneous, sedimentary, and metamorphic rock. D) sediments that are compacted and then cemented.
Answer:
C) igneous, sedimentary, and metamorphic rock.
Explanation:
Every rock can become a metamorphic rock (ignorant, sedimentary, or metamorphic). Once rocks are buried deep within the Earth at extreme temperatures and pressures, they can form new materials and textures without melting. As melting happens magma is formed, beginning the rock cycle again.
Hence, option C is correct
Answer:
C
Explanation:
Metamorphic rocks form from igneous, sedimentary, and metamorphic rock. The agents that change pre-existing rocks are heat, pressure, and chemical activity. Sediments form sedimentary rocks; magma forms igneous rocks.
Now consider a single base mutation in a codon for leucine that creates a codon for phenylalanine. A true reversion changes the phenylalanine codon back to a codon for leucine. Which of the following leucine codon(s) could be mutated once to form a phenylalanine codon, and then mutated at a second site to restore a leucine codon? Select all that apply. (Note that two different positions in the codon must be mutated.) Select all that apply. (Note that two different positions in the codon must be mutated.)
CUU
CUC
CUA
UUG
UUA
CUG
none of these codons
Answer:
CUU, CUC, UUG, UUA
Explanation:
Following codons result in leucine amino acid : UUA, UUG, CUU, CUC, CUA and CUG
Following codons result in phenylalanine amino acid : UUU and UUC
CUU : If C is mutated to U, result will be UUU which is phenylalanine codon. U at third position can be mutated to A which will form UUA which us again codon for leucine.CUC : If C at first position is mutated to U, phenlyalanine codon UUC will be formed. C can be mutated to A to form leucine codon UUA.UUG : If G is mutated to U phenylalanine codon UUU will form. U at first position can be mutated to C to form leucine codon CUU.UUA : If A is mutated to U phenylalanine codon UUU will form. U at first position can be mutated to C to form leucine codon CUU.A color blind man without hemophilia (both X-linked traits) marries a woman who is a carrier for both traits. What is the probability they will have a son with both color blindness and hemophilia? Hemophilia and color blindness are unlinked genes.
a. 1/1
b. 1/16
c. 1/8
d. 1/4
e. 3/16
Answer:
Option D.
Explanation:
Both genes are linked to X chromosome. They are recessive diseases and a heterozygous gene, which means that two different forms of a particular gene are inherited, one from each parent.
CB : Color blindness carrier, healthy
H: Hemophilia carrier, healthy
h: Disease
cb: Disease
Mom: X(CB/H) X(cb/h) (carrier)
Dad: X(cb/H) Y (color blind man, and dominant for hemophilia - healthy)
Let's make a table
MOM/DAD | X(CB/H) X(cb/h) |
X(cb/H)Y | X(cb/H) X(CB/H) | Healthy girl for H, carrier for cb
X(cb/H)Y | X(cb/H) X(cb/h) | Healthy girl for H, sick for cb
X(cb/H)Y | Y X(cb/h) | Boy sick for cb and h
X(cb/H)Y | Y X(CB/H) | Boy healthy for CB and H
1 probable case, over 4 possible cases
Imagine you are studying a population of plants and you have determined that they are all genetic clones of one another. You are interested in a particular gene found in the population that you have named LLT (low-light tolerance), that allows it to live in shady areas.
What is the frequency of this allele in the population you study?
Answer:
The frequency of the allele is high in the population
Explanation:
The selective pressure in this population is low light. Thus only plants with high tolerance for low light will be able to survive in this population. LLT gene plant has this trait.
Therefore the LLT gene plants will have higher rate of survival compare to other clones at very low light intensity .
The frequency of expression of these gene will be high due to high survival rates in the population.
the urea cycle rids in the body of excess nitrogen by converting it to a non-toxic form that can be excreted in the urine.it utilize one nitrogen atom from carbonyl phosphate and one nitrogen atom from asparate.
1.which enzyme releases urea a product? a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
2. which enzyme requires ATP a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
3- Which enzyme is located inside the mitochondrion? a) ornithine transcarbamoylase b) argininosuccinate synthetase c) argininosuccinase d) arginase
4- which intermediates of the urea cycle must cross the mitochondrial membrane? a) argininsuccinate b) ornithine c) arginine d) citrulline
Answer:
Explanation:
Number 1: The answer is A; Ariginase is responsible for the release of urea as product.
Number 2: The answer is B; Argininosuccinate synthetase requires ATP.
Number 3: The answer is D; Arginase is located in the mitochondrion
Number 4: The answer is B; Ornithine produced in the cytosol must first cross the inner mitochondrial membrane into the mitochondrial matrix where it is carbamylated