Answer:
Associative learning may be defined as a type of learning in which the new response gets easily associated with the stimulus. Except habituation learning all simple learning procedure are included in this learning.
In the associative learning, reactions of an individual are based on there spank. George learns to cry on being spanked this is because this stimulus of sister cries gives response. Crying do not show any effect on ken because he does not know about his sister crying.
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.
Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey�s tail was only 6 cm while Erma�s was 30 cm. Their F1 piglet offspring all grew tails that were 18 cm. When inbred (F1X F1), an F2 generation resulted in many piglets (Erma and Harvey�s grandpigs) whose tails ranged in 4 cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). Most had 18 cm tails while 1/64 had 6 cm and 1/64 had 30 cm tails.a) Explain how tail length is inherited by describing the mode of inheritance, indicating how many gene pairs are at work, and designating the genotypes of Harvey, Erma, and their 18 cm offspring.b) If one of the 18 cm F1 pigs were mated with the 6 cm F2 pigs, what phenotypic ratio would be predicted if many offspring resulted?
Answer:
a) Tail length shows quantitative mode of inheritance. It is controlled by three pair of genes and each allele contributes a small effect. Length of tail varies according to the cumulative effect shown by all the alleles present. Let the three genes controlling the trait be A,B and C. Dominant allele contributes 5 cm of tail length whereas recessive allele contributes 1 cm of tail length.
Harvey = aabbcc = 1*6 = 6 cm
Erma = AABBCC = 5*6 = 30 cm
When Harvey and Erma mate : AABBCC X aabbbcc : AaBbCc (F1)
AaBbCc = (3*5) + (3*1) = 18 cm
b) AaBbCc ( 18 cm ) X aabbcc ( 6 cm ) = 8 type of offspring will be produced in 1:1:1:1:1:1:1:1 ratio :
abcabc, Abcabc, aBcabc, abCabc, ABcabc, aBCabc, AbCabc, ABCabc
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.
____________and ____________are chromosomal mutations that may activate cellular oncogenes by moving them to new regulatory sequences where they become over-expressed.(a) Duplications; deletions(b) Inversions; translocations(c) Duplications; inversions(d) Deletions; translocations(e) Aneuploidy; deletions
Answer:
(d) Deletions; translocations
Explanation:
Deletion mutations: It is the removal of a number of nucleotides from a DNA sequence. Deletions that result in frameshifts, that is, disrupt the 3-nucleotide reading frame can cause changes in regulatory sequences inducing over-expression of certain genes.
Translocation mutations: Are caused the the rearrangement of genes in chromosomes that are usually in close proximity. Encountering new enhancers and/or promoters can cause oncogene activation.
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.
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.
What are the correct outputs, during the citric acid cycle, from one molecule of glucose?
Answer:
no clue
Explanation:
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.
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.
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.
Which of the following is false?A. If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection.B. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce.C. Individuals who are heterozygous HbA/HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-ridden regions in Africa.D. In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
Answer:
D
Explanation:
A. If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection.
This is true. If the disease reduces fertility, the likelihood of the affected individual passing it on to children is decreased. If there is no fitness advantage, it will slowly be reduced in the population by chance, because natural selection is not acting positively upon it
B. Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce.
This is true, individuals carrying two copies of the sickle cell allele might be more resistant to malaria, but they also have a range of health problems that often cause early mortality or severely reduced quality of life, cancelling out the beneficial features of this allele.
C. Individuals who are heterozygous HbA/HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-ridden regions in Africa.
This is true. Malaria persists in certain regions of Africa because the conditions are favourable for the mosquito that carries the disease. The heterozygous HbA/HbS genotype is prevalent in African population because it protects individuals from malaria, giving them a fitness advantage. Therefore, the natural selection acts positively upon the sickle cell allele in these areas, and over time, it becomes more prevalent
D. In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
This is false. Without the selection pressure of malaria, there is no advantage of having a copy of the sickle cell allele over the normal allele.
In regions where malaria does not occur, individuals who are heterozygous
HbA/HbS have a fitness advantage over those who are hom ozygous for the
normal hemoglobin allele (HbA) is False.
The individuals who have heterozygous HbA/HbS only have an advantage in the aspect of malaria as it makes them less susceptible to the disease.
This is the only advantage it offers and it doesn't offer any special advantage
such as more fitness or increase in fertility.
Individuals with sickle cell die early thereby preventing natural selection as
they are mostly unable to reproduce before death.
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Evolutionary geneticists carefully genotype a population of saber-toothed tigers and find the following genotypes:
250 tigers are A1/A1, 500 tigers are A1/ A2, and 250 tigers are A2/A2.
Tragically, an asteroid lands in the middle of the population, killing 50% of each genotype.
A) What will the genotype frequencies be in the next generation?
Answer:
Explanation:
since 50% of each genotype was tragically destroyed, the new population will be as follows;
A1A1 = 125, A1A2 = 250, A2A2 = 125
The new genetic frequencies are as follows;
A1A1 = 125/500 = 0.25
A1A2 = 250/500 = 0.50
A2A2 = 125/500 = 0.25
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
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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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:
Place the following events in activation of a B cell by a microbe in the order in which they would occur.
Several rounds of division occur, producing many B cells that all express the same B cell receptor.
A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.
The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.
Second messenger signaling is activated in the B cell.
Gene expression is altered in the B cell, and the B cell begins to divide.
Answer:
The proper order is as follows-
1) A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.
2) Gene expression is altered in the B cell, and the B cell begins to divide.
3) Several rounds of division occur, producing many B cells that all express the same B cell receptor.
4) The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.
5) Second messenger signaling is activated in the B cell.
The activation process of a B cell by a microbe includes differentiation from a lymphocyte stem cell, antigen binding by the B cell receptor, activation of signaling pathways, alteration of gene expression, and rounds of division producing B cells with identical receptors.
Explanation:The activation of a B cell by a microbe involves a series of steps which must occur in a specific order for effective immune response. Listing these steps in the correct sequence provides a clear understanding of B cell activation:
A B cell differentiates from a lymphocyte stem cell, and the mature B cell expresses a specific type of B cell receptor on its cell membrane.The B cell's BCR (B cell receptor) binds to a foreign antigen with high specificity and high affinity.Second messenger signaling is activated in the B cell.Gene expression is altered in the B cell, and the B cell begins to divide.Several rounds of division occur, producing many B cells that all express the same B cell receptor.These steps are critical for the body's adaptive immune system to mount a specific response against the antigen presented by the microbe. The ultimate goal of these steps is the production of plasma cells and memory B cells that can produce antibodies specific to the antigen and protect the host from future infections by the same pathogen.
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.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:
Multicellular organisms have nervous systems, made up of nerve cells, which help to generate behaviors. The nerve cells throughout the body communicate with each other in order to help the nervous system carry out its function.
Which of the following behaviors requires help from the nervous system?
A.
swallowing food
B.
driving a car
C.
sensing heat
D.
all of these
Answer:
D
Explanation:
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).
"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.
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.
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.
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
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
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
Which of the following statements about regulation of eukaryotic gene expression is INCORRECT?
A) The presence of a nuclear membrane separating transcription and translation in eukaryotes led to the evolution of additional mechanisms of gene regulation.
B) In eukaryotes, most structural genes are found within operons.
C) Eukaryotic mRNAs are generally more stable than prokaryotic mRNAs.
D) The rate of degradation of mRNAs is important in regulation in eukaryotes.
E) Posttranslational regulation of histones is unique to eukaryotes.
Answer:
The answer is B
Explanation:
In eukaryotes, most structural genes are found within operons.
The incorrect statement is 'B) In eukaryotes, most structural genes are found within operons.' This is because eukaryotic genes are generally not organized in operons, unlike prokaryotic genes.
Explanation:The statement which is incorrect about the regulation of eukaryotic gene expression is 'B) In eukaryotes, most structural genes are found within operons.' This is because eukaryotic genes are generally not organized in operons. The concept of the operon, a functional unit of DNA containing clustered genes that are under control of a single regulatory signal or promoter, is more commonly associated with prokaryotic genetics.
The other options represent correct statements about eukaryotic gene expression. For instance, the nuclear membrane in eukaryotes does indeed enable additional layers of gene regulation. Additionally, the stability of eukaryotic mRNAs and the rate of their degradation are important factors in gene regulation. Lastly, posttranslational regulation of histones is a unique feature of eukaryotic cells.
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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
Glycolysis and gluconeogenesis are both pathways the body utilizes for energy production. Although both pathways employ common enzymes, they are not the reverse of one another.
They are also not usually used simultaneously.
Which pathway(s) is(are) enhanced during starvation?
gluconeogenesis
glycolysis and gluconeogenesis
glycolysis only
During starvation, the body enhances gluconeogenesis, not glycolysis, to maintain blood glucose levels by generating glucose from non-carbohydrate sources.
Explanation:During periods of starvation, the body enhances gluconeogenesis to generate glucose from non-carbohydrate sources, such as pyruvate, lactate, glycerol, and gluconeogenic amino acids. This pathway is crucial for maintaining blood glucose levels and providing energy, especially to the brain and other organs that rely heavily on glucose. Gluconeogenesis predominantly occurs in the liver and, to a lesser extent, in the kidneys and involves converting these precursors back into glucose. This process is particularly vital during starvation or fasting when dietary glucose is not available.
Glycolysis, on the other hand, is the pathway by which glucose is broken down into pyruvate, yielding small amounts of energy. Though both pathways share other enzymes and can conceptually be seen as reverses of each other, they are regulated independently to prevent both pathways from being active simultaneously, a phenomenon known as the 'futile cycle.' Therefore, during starvation, gluconeogenesis is enhanced, not glycolysis.
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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