Answer:
True
Explanation:
The thylakoid membrane of chloroplast has electron carriers embedded in it. During light-dependent reactions of photosynthesis, the electrons move from water molecules to the PSII and then through electron carriers to the PSI and finally to NADP+.
Movement of electrons through the intermediate carrier is accompanied by the pumping of protons from stroma to the thylakoid lumen. This creates an electrochemical gradient along the thylakoid membrane.
The protons are moved back from the thylakoid lumen into the stroma down the concentration gradient through proton channels known as "CFo".
As the protons move down their concentration gradient, the energy is used to phosphorylate the ADP into ATP.
Which of the following is present in a prokaryotic cell?
a. mitochondrion
b. ribosome
c. nuclear envelope
d. chloroplast
The correct answer is B. Ribosome
Explanation:
In biology, cells are mainly classified as eukaryotic or prokaryotic. Each of these types of cells has different features, to begin with, eukaryotic cells are those that contain a defined nucleus and are part of both unicellular and multicellular organisms. On the opposite, prokaryotic cells do not have a defined nucleus or a nuclear envelope and are mainly present in unicellular organisms, besides this, they lack mitochondrion, Golgi apparatus or chloroplasts that are present in eukaryotic cells. However, prokaryotic cells still contain DNA, ribosomes, vesicles, and vacuoles. According to this, the one that is present in a prokaryotic cell is ribosome.
The only option from the list present in a prokaryotic cell is a ribosome. Prokaryotic cells do not contain mitochondria, nuclear envelopes or chloroplasts which are associated only with eukaryotic cells.
Explanation:The correct answer to the question 'Which of the following is present in a prokaryotic cell?' is (b) ribosome. Prokaryotic cells, such as bacteria, do not contain mitochondria, nuclear envelopes, or chloroplasts as these are structures associated only with eukaryotic cells. However, they do contain ribosomes which play a crucial role in protein synthesis.
It's important to note that eukaryotic cells also contain ribosomes, the primary difference being that the ribosomes in eukaryotic cells are larger and more complex than those found in prokaryotic cells.
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Which of these is true regarding prokaryotes?
a. It takes about 6-8 hours to duplicate DNA
b. Microtubules bind to the centromeres and pull the chromosomes apart
c. Separated DNA is attached to the cell membrane before the cell divides
d. The nucleus dissolves
Answer:c. Separated DNA is attached to the cell membrane before the cell divides.
Explanation:
The prokaryotes are single celled organisms. These are simple organisms which reproduce through asexual mode of reproduction that is cell division. They do not posses well define nucleus. Thus the genetic material remain in the cytoplasm of the cell. On cell division the genetic material (DNA) is distributed into halves for development of two daughter cells. Due to lack of nucleus and it's associated membrane the separated DNA get attach to the membrane before the cell actually divides.
A mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with this plant by manually opening and pollinating the flowers with pollen from another plant. When closed x open crosses were done, all the F1 progeny were open. The F2 plants were 145 open and 59 closed. A cross of closed x F1 gave 81 open and 77 closed. How is the closed trait inherited? What evidence led you to your conclusion?
Answer:
This trait is inherited according to Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.
Explanation:
First, there are closed-flowers plants and open-flowers plants. A cross between them produces only open flowers at F1. With this information, we can say that open condition is dominant over closed, and probably, parental plants were homozygous, like this:
- Open flowers: FF
- Closed flowers: ff
- F1 open flowers: Ff
When F1 was crossed, the progeny was:
- 145 open
- 59 closed
When we cross two heterozygous, we obtain a phenotypical radio of 3:1, like this:
- (1) FF Open
- (2) Ff Open
- (1) ff Closed
This ratio is very similar to the F2 above reported (145 open and 59 closed).
When F1 (Ff) plants were crossed with closed plants the progeny was:
- 81 open
- 77 closed
This is alike to a ratio 1:1.
In the Punnett Square, it is possible to notice that in this cross it is expected a ratio of 1:1 closed/open.
In conclusion, this trait is inherited according to Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.
Briefly describe the functional recovery of brain after brain injury or stroke.
Answer:
The brain is made up of billions of neural connections that transit and gather signals that are processed into information. Brain injury or stroke may cause damage to some of these connections. Neurons have a limited ability to repair themselves, unlike normal cells hence making brain injury difficult to recover fully from. The brain recovery after such as injury is are mainly driven by the brain finding new neural pathways around the damaged ones.
However difficult, there is a chance that a neuron can be repaired especially if the cell body is not damaged in the injury. If the axon is regenerated, it needs to make the right connections to re-establish signaling. The problem is that glial cells, that are integral to neurons structure and function, reduce the chances of recovery because they clean up damaged neurons and produce molecules that inhibit recovery
Which of the following is not considered evidence supporting the evolutionary relationship among all plant taxa?
a. Plants share a significant number of genes
b. Plants and plant ancestors have chlorophyll
c. Fossils of plants show us transition series and evolutionary patterns of plants
d. Structures/characteristics among most plants are very similar
Answer:
The correct answer is option d. "Structures/characteristics among most plants are very similar".
Explanation:
There are extensive evidence to support the evolutionary relationship among all plant taxa, however, similarity of structures and characteristics among most plants is not one of them. Plants are one of the most diverse kingdoms having very different species such as flowering plants, conifers, lichens, cycads, algae, among others. The structures and characteristics of these plants are very different among them, and this diversity makes difficult to classify and clarify the evolutionary relationship that they share.
The mitochondrial inner membrane form a series of infoldings known as cristae to:
a. Increase surface area
b. Decrease surface area
c. Increase production of H+
d. Decrease production of H+
Answer: Increase the surface area.
Explanation:
The mitochondria is an important cell organelle that is found in both the plant cell and animal cell.
It basically aids in the production of ATP, hence also known as the power house of the cell.
The mitochondria is a double membrane system in which the inner membrane of mitochondria helps in increasing the surface area for integral proteins.
It helps in embedding the proteins in the folding known as cristae.
Energy pyramids, which display graphically the relative amounts of energy in each trophic level of an ecosystem, are very similar across a wide range of different ecosystems, but biomass and organism abundance at different trophic levels are more variable between different ecosystems.
a. True
b. False
Answer:
a. True
Explanation:
Energy pyramids are a simple representation of the flow of energy at each trophic level in an ecosystem, based on the well known fact that a lot of energy gets lost between consumers (approximately only around 10% is made into tissues, while 90% is used up and is lost as heat).
While the biomass and abundance of the organisms at a trophic level, even the identity of that organisms (species), is highly dependent on the characteristic of the environment. So very different environments (for example, a dessert and a rain forest) will have also very different species with different biomass and abundance fulfilling the same trophic role.
Each DNA strand contains bases but also a structure that can be described as "sugar- lipid-sugar-lipid-sugar-lipid...", repeating this simple sequence from one end to the other
a. True
b. False
Answer:
False.
Explanation:
DNA is present as genetic material in all the organism except some viruses. RNA is present as genetic material in some viruses only. The DNA strands are complimentary with each other.
DNA is made of the polymers of nucleotides. The nucleotides consists of the nitrogenous bases (adenine, guanine, thymine and cytosine), deoxyribose pentose sugar and phosphate group. The DNA molecule doesnot contain sugar..lipid in its structure.
Thus, the answer is false.
If you were using cladistics to build a phylogenetic tree of cats, which of the following would be the best outgroup?
a.wolf c.lion
b.domestic cat d.leopard
Answer: a. Wolf
Explanation:
A phylogenetic tree is a diagrammatic representation in which the premitive living species are compared with the modern organisms on the basis of morphological as well as genetic similarities and differences. Hence, they are separated into distinct groups.
All the given organisms in the options (wolf, domestic cat, lion, and leopard) belong to the vertebrata phyllum. Among the options given all the organisms, except wolf, the other organisms are morphologically close to one other. Thus for constructing a phylogenetic tree for these organisms, wolfs will be an out-group. The other organisms will be the descendent belonging to the same group.
When using cladistics to build a phylogenetic tree of cats, then a WOLF would be the BEST outgroup (Option a).
Cladistic is an evolutionary approach to classify taxonomic groups (i.e., clades) based on the most recent common ancestor of these clades.In cladistics, phylogenetic trees are graphical representations that show lines of evolutionary descent of different clades from a common ancestor.An outgroup is a more distantly related clade that helps to determine the level of evolutionary relationships among organisms or taxonomic groups (e.g., species) within a phylogenetic cladistic tree.In conclusion, when using cladistics to build a phylogenetic tree of cats, then a WOLF would be the BEST outgroup (Option a).
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The technique used when a biologist separates tissues by gently placing the closed tips of a pair of blunt-ended scissors between the two tissue layers and separating them by opening the scissors is called __________.
Select one:
a. sharp dissection
b. blunt dissection
c. normal dissection
d. tissue dissection
The correct answer is B. Blunt dissection
Explanation:
The word dissection refers to the process of separating tissues or different anatomical structures to analyze them or as part of medical processes such as surgeries. Although there are multiple techniques in dissection the two main ones are sharp dissection that uses sharp objects to cut through tissues and blunt dissection in which blunt-ended object such as special scissors or fingers are used to separate tissue without cutting through them. For doing this, fingers or scissors are placed between tissues and then open gently. According to this, the technique described is blunt dissection not only because a blunt-ended object is used but because this is used to separate tissue layer rather than to cut through them.
The correct answer is b. blunt dissection.
Blunt dissection is a technique used in surgical and laboratory settings to separate tissues or layers within tissues without cutting them. This method involves the use of blunt instruments, such as the closed tips of blunt-ended scissors, to gently pry apart the layers. The process minimizes damage to the tissues and preserves their integrity, which is particularly important when the tissues need to remain viable for further study or when preserving the function of the tissues is crucial, as in surgery.
In contrast, sharp dissection involves cutting through tissues with a scalpel or sharp scissors, which can cause more damage and is typically used when a precise incision is needed or when the tissue being cut will not be used for further study or function. Normal dissection and tissue dissection are not specific terms used to describe a particular technique like blunt dissection. Therefore, the correct term for the technique described in the question is blunt dissection.
Cite the most common molecular explanations for dominant and recessive alleles.
Answer: Dominant allele codes for a functional protein. Recessive allele codes for a less functional or nonfunctional protein, or it does not code for any protein.
Explanation:
Alleles are different forms of a gene. There are two types, dominant and recessive allele. Heterozygous organisms have one of each, and the dominant allele is expressed. Homozygous organisms have either two dominant alleles or two recessive alleles.
A dominant allele produces a functional protein, even in the presence of a recessive allele, because only one copy of the allele is enough and it masks the effects of the recessive allele. This dominant trait is shown in individuals who are homozygous dominant or heterozygous,
A recessive allele produces a less functional or nonfunctional protein, or it does not code for any protein at all. A recessive allele does not become a trait unless both copies of the gene are present.
Which of the following innervates the superior, medial and inferior rectus muscles of the eye and is a constrictor of the pupil?
a. Trochlear
b. Abducens
c. Oculomotor
d. Trigeminal
e. Facial
Answer:
The correct option is: c. Oculomotor
Explanation:
Oculomotor nerve is one of the twelve pairs of cranial nerves, emerging from the brain. Oculomotor is the third cranial nerve that originates from the third nerve nucleus in the midbrain.
The oculomotor cranial nerve (CN III), innervates the extrinsic eye muscles including the superior rectus, inferior rectus, and the medial rectus; that are responsible for raising the eyelid and movements of the eye.
The fibers of this nerve innervates the intrinsic eye muscles that are responsible for the constriction of the pupil.
Describe the structure and function of haemoglobin (Hb). You should include a description of both tertiary and quaternary structures of the protein and the location and function of non-protein components.
Answer:
Heamoglobin is the structure present in red blood cells which is involved in carrying of oxygen from the blood to all parts of the body.
Explanation:
Tertiary Structure : A molecule of heamoglobin consists of four groups of heme and a group of globulin.A molecule of heamoglobin contains four iron atoms to which oxygen molecules get attached. As there are four iron atoms in a molecule of heamoglobin so a single heamoglobin molecule can carry four oxygen atoms.
Quaternary structure: Multi sub unit globular proteins combine to make the Quaternary structure of heamoglobin. Alpha helices are made by the amino acids present in heamoglobin. Non helical segments, short in length, connect the alpha helices. The helical segments are stabilized by hydrogen bonds which results in folding of the polypeptide into the specific Quaternary shape.
Function:
1.The basic function of heamoglobin is that it transfers oxygen from the lungs to different parts of the body.
2. CO2 ions and hydrogen ions are also transported by the heamoglobin.
Humans have 46 chromosomes in each somatic cell. a. How many chromosomes does a child receive from its father? b. How many autosomes and how many sex chromosomes are present in each somatic cell? c. How many chromosomes are present in a human ovum? d. How many sex chromosomes are present in a human ovum?
Answer:
Explanation:
a. How many chromosomes does a child receive from its father?
Every person receives half of their chromosomes from their father and half from their mother. For that reason, if humans have 46 chromosomes in each somatic cell, 23 of them come from the father.
b. How many autosomes and how many sex chromosomes are present in each somatic cell?
Each somatic cell has 46 chromosomes in total; of which 22 pairs are autosomes and 1 pair are sex chromosomes. Therefore, 44 chromosomes are autosomes and 2 chromosomes are sex chromosomes (the X and/or Y chromosomes).
c. How many chromosomes are present in a human ovum?
The human gametes (ovum in women and sperm cell in men) have half of a somatic cell's genetic material, so that when they combine in a zygote to create their child, they form an initial cell with 46 chromosomes in total. The ovum therefore has 23 chromosomes: 22 autosomes and 1 sex chromosome.
d. How many sex chromosomes are present in a human ovum?
As I mentioned in the previous question, a human ovum has 1 sex chromosome. That way, when it combines with a sperm cell, which also has 1 sex chromosome, they form a zygote with a pair of sex chromosomes (one that comes from the mother and one from the father).
a. A child receives 23 chromosomes from its father. b. Each somatic cell has 44 autosomes and 2 sex chromosomes. c. A human ovum contains 23 chromosomes. d. A human ovum has 1 sex chromosome.
Explanation:a. A child receives 23 chromosomes from its father.
b. Each somatic cell has 44 autosomes and 2 sex chromosomes, for a total of 46 chromosomes.
c. A human ovum contains 23 chromosomes.
d. A human ovum has 1 sex chromosome.
Compare the means of sex determination in different organisms.
Answer:
Sex determination may be defined as the phenomena of determining the sex whether the organism is male, female or hermaphrodite. Different methods of sex determination are present in different organism.
In case of humans, the sex chromosome of an individual with XY is male whereas XX is female. Humans male are heterogametic. In case of birds the chromosome with ZZ are males (homogametic) whereas ZW is female. In case of Drosophila XY and XO flies are males , while XX, XXY, and XXYY flies are females.
A paleontologist has recovered a bit of tissue from the 400-year-old preserved skin of an extinct dodo (a bird). To compare a specific region of the DNA from a sample with DNA from living birds, which of the following would be most useful for increasing the amount of dodo DNA available for testing?
a. SNP analysis
b. polymerase chain reaction (PCR)
c. electroporation
d. gel electrophoresis
Answer:
b. polymerase chain reaction (PCR)
Explanation:
PCR is the technique that generates millions of copies of DNA starting from a small amount of template.In this reaction, a molecule of double stranded DNA is denatured and the chains separated. Then, the single strands of DNA are hybridized with a small complementary fragment of DNA called a primer, which is then extended by an enzyme called DNA polymerase by adding complementary nucleotides to the template strand. The result will be two double stranded DNA molecules. If you repeat the process many times, you can amplify the original DNA material exponentially, and the final PCR product will be millions of DNA molecules that can be used for testing.
A SNP analysis is a method that allows scientists to detect variations of a single nucleotide in the DNA sequence of different individuals.Electroporation is a technique that creates temporary pores in a cell membrane, allowing the delivery of drugs or DNA to a cell.Gel electrophoresis is a method to analyze the PCR product in which the DNA molecules are separated by size.The polymerase chain reaction (PCR) would be the most useful technique for increasing the amount of dodo DNA available for testing, as it allows rapid amplification of specific DNA sequences.
Explanation:To increase the amount of dodo DNA available for testing when comparing a specific region of the DNA from a sample with DNA from living birds, the most useful technique would be polymerase chain reaction (PCR). This method allows for the rapid amplification of specific DNA sequences, thereby increasing the number of copies of the dodo DNA for further analysis. PCR is a highly effective tool in molecular biology for the amplification of DNA from limited samples, such as those obtained from extinct species like the dodo or other ancient remains.
Which test provides an operational definition of a gene?
a. complementation test
b. biuret test
c. gram stain
d. neurospora test
e. fluctuation test
Answer:
The correct option is: A) Complementation test
Explanation:
Complementation test is used for determining that the mutations in the two strains, associated with a particular phenotype, are present in different genes.
The mutations are said to be complementary, if present in different genes.
Therefore, the complementation test is the process for determining the function of the gene in recessive allelism.
Some scientists have proposed that the earliest forms of life may have existed in an "RNA World" where RNA was both the genetic material and responsible for enzymatic activity. Imagine that you have recreated such a life form. What would demonstrate that RNA, but not protein or DNA, is necessary and sufficient for these functions?
(A) The organism lives and replicates despite RNase and DNase treatment, but the organism dies when treated with protease.
(B) The organism lives and replicates despite protease and DNase treatment, but the organism dies when treated with RNase.
(C) The organism lives and replicates despite protease treatment, but the organism dies when treated with DNase and RNase.
(D) The organism lives and replicates despite DNase treatment, but the organism dies when treated with RNase and protease.
Answer:
(B) The organism lives and replicates despite protease and DNase treatment, but the organism dies when treated with RNase.
Explanation:
DNA is the genetic material in present day organisms while some of the proteins serve as enzymes and catalyze the metabolic reactions.
An organism having RNA with the ability to serve as genetic material would not require DNA to survive. Likewise, if RNA molecules in these organisms also serve as the enzyme, proteins would not be required for survival.
Therefore, when treated with DNase and protease that digest the DNA and proteins respectively, these organisms would be able to survive. However, treatment with RNase enzymes that digest the RNA would kill them as RNA was their biocatalyst and genetic material.
A researcher wanted to study the effect of low light environments on the growth of alfalfa plants, so she grew six plants and measured their heights after two weeks. She calculated a sample mean of 5.35 cm and a sample standard deviation of 0.90 cm. If the heights of the first five plants in her sample were 6.0 cm, 5.9 cm, 4.0 cm, 5.5 cm, and 6.2 cm, what is the height of the other plant?
Answer:
Height of the other plant is [tex]4.5[/tex] cm
Explanation:
- Given -
Mean of the sample set of plants [tex]= 5.35[/tex]cm
Mean of a given set of data is equal to the sum of integral values of each data divided by the number of such data points
[tex]M = \frac{x}{n}[/tex]
Where M signifies the Mean values
X signifies the sum of all unit data with in a sample
and n signifies total number of data units with in a sample
On substituting the given values in above equation, we get -
[tex]\frac{6.0+5.9+4.0+5.5+6.2+A}{6} = 5.35\\5.35 * 6= 27.6 + A\\A = 32.1-27.6\\A=4.5[/tex]
Hence, height of the other plant is [tex]4.5[/tex] cm
The height of the sixth alfalfa plant is 4.5 cm, calculated by subtracting the total of the first five plant heights from the overall sum required to achieve the sample mean.
The student requires the calculation of the height of the sixth alfalfa plant given the sample mean, standard deviation, and the heights of the first five plants. To find the height of the sixth plant, we will use the formula for the sample mean: mean = (sum of all values) / number of values. Given that the sample mean is 5.35 cm and there are six plants, the sum of all the plant heights is 5.35 cm * 6 = 32.1 cm. The sum of the heights of the first five plants is 6.0 cm + 5.9 cm + 4.0 cm + 5.5 cm + 6.2 cm = 27.6 cm. Subtracting this from the total sum we get the height of the sixth plant: 32.1 cm - 27.6 cm = 4.5 cm.
According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly
a. spread in a continuous layer over the inner and outer surfaces of the membrane.
b. confined to the hydrophobic interior of the membrane.
c. embedded in a lipid bilayer.
d. randomly oriented in the membrane, with no fixed insideoutside polarity
Answer:
Embedded in a lipid bilayer. (Ans. C)
Explanation:
Fluid mosaic model explains different observations such as the structure of functional cell membranes. According to this model, there is a thin polar membrane composed of a two-layer of lipid molecules called lipid bi-layer or phospholipid bi-layer in which proteins molecules are embedded. This layer provides fluidity and elasticity to the membrane structure.
These membranes form a continuous barrier around all cells and they are a flat sheet in structure. The lipid bilayer is working as a barrier that keeps proteins, ions and other molecules where they are needed to be and preventing diffusing them to other areas. They are impermeable to most hydrophilic molecules. They are particularly impermeable to ions, which allow cells to regulate pH and salt concentration in membrane through the transport ions across their membranes using proteins knows as ion pumps.
According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer and have a specific inside-outside polarity.
Explanation:According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer. They can be found throughout the membrane, both on the inner and outer surfaces, as well as in the hydrophobic interior. However, they are not randomly oriented in the membrane, but have a specific inside-outside polarity.
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Plants are more readily manipulated by genetic engineering than are animals because
a. plant genes do not contain introns.
b. more vectors are available for transferring recombinant DNA into plant cells.
c. a somatic plant cell can often give rise to a complete plant.
d. plant cells have larger nuclei
Answer:
The correct answer is option c. "a somatic plant cell can often give rise to a complete plant".
Explanation:
One characteristic that makes plants readily manipulated organisms for genetic engineering studies is that a somatic plant cell can often give rise to a complete plant. This process is known as somatic embryogenesis, and is widely used with relative ease in biotechnology laboratories. Somatic embryogenesis helps to quickly test genetic manipulation tests in complete plants allowing to check for the final phenotype easily.
Plants are more readily manipulated by genetic engineering than animals due to their ability to regenerate, the availability of vectors for gene transfer, and the direct insertion of synthetic genes into plant genomes.
Explanation:Plants are more readily manipulated by genetic engineering than animals because:
Plant cells have the ability to regenerate and grow into complete plants, making it easier to select and cultivate cells with desired qualities.There are more available vectors for transferring new genes into plant cells, such as plasmids and microscopic pellets.Plant transformation allows for the direct insertion of synthetic genes into plant genomes without the need for generations of breeding.In contrast, animal cells do not have the same regeneration capacity, and the process of introducing new genes and traits is more complex.
In humans, a. How many sperm develop from 100 primary sper matocytes? b. How many sperm develop from 100 secondary spermatocytes? c. How many sperm develop from 100 spermatids? d. How many ova develop from 100 primary oocytes? e. How many ova develop from 100 secondary oocytes? f. How many ova develop from 100 polar bodies?
Answer:
In males primary spermatocytes undergo two meiotic divisions. First meiotic division give rise to two secondary spermatocytes and secondary meiotic division give rise to four spermatids from two spermatocytes.
Therefore a. from 100 primary spermatocytes 400 sperms will develop (100*4= 400)
.
b. From 100 secondary spermatocyteS 200 sperms will develop(100*2= 200).
c. Spermatids do not undergo any further division rather they undergo differentiation to form mature sperms. Therefore from 100 spermatids, 100 mature sperms will develop.
d. One primary oocyte gives rise to one secondary oocyte and one secondary oocyte give rise to one mature ova so only one mature ova will develop from 100 primary oocytes.
e. 100 ova will develop from 100 secondary oocytes.
f. Polar body do not develop into ova and degenerates after some time so no ova will develop from 100 polar bodies.
Approximately 200 sperm develop from 100 primary spermatocytes.
b. Around 400 sperm develop from 100 secondary spermatocytes.
c. Roughly 100 sperm develop from 100 spermatids.
d. Only one ovum develops from 100 primary oocytes.
e. Only one ovum develops from 100 secondary oocytes.
f. No ova develop from 100 polar bodies.
What is the numbers of spermIn people, spermatogenesis makes sperm and oogenesis makes ova, but they make different amounts of each. Here are the estimated numbers of sperm and eggs that develop from certain cell stages:
When 100 cells called primary spermatocytes go through a process called meiosis, they turn into 200 cells called secondary spermatocytes. Each cell called primary spermatocyte breaks into two smaller cells named secondary spermatocytes.
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Expression of the A and B antigens on red blood cells is influenced by the FUT1 gene. This gene encodes fucosyl transferase, an enzyme that helps produce the H substance. Which of the following statements about the H substance is incorrect?
(A) Individuals that cannot produce the H substance appear to be type O even if they have functional A and/or iB alleles.
(B) Individuals that fail to produce the H substance are said to have the Bombay phenotype.
(C) Individuals that are heterozygous for the FUT1 gene cannot produce the H substance
(D) The H substance is a substrate for the enzymes produced by the n and β genes. These enzymes add the appropriate terminal sugar to the H substance producing the A and B antigens, respectively.
Answer: The correct answer is option C
Explanation:
Bombay phenotype refers to a rare genetic trait in which a person appears as blood group O even in presence of functional A, B, or AB alleles.
It is because of recessive mutation that occurs in other locus, that is, FUT1 gene. Normal gene produces H antigen or protein in the cell. This antigen is essential for the final expression of the ABO alleles.
If a person is homozygous recessive then he or she is not able to produce H antigen due to which the blood group appears as O.
However, if a person carries even a single functional copy of FUT1 gene, that is, heterozygous or homozygous then he or she can efficiently produce H antigen and then ABO alleles are expressed accordingly.
Thus, option C contains incorrect statement.
8) Each time a molecule of glucose is completely oxidized via aerobic respiration, how many oxygen (O2) molecules are required?
A) 1
B) 2
C) 6
D) 12
E) 38
Answer:
C) 6
Explanation:
The balanced equation for aerobic cellular respiration is:
C₆H₁₂O₆ + 6 O₂ ----> 6 CO₂ + 6 H₂O + Energy (ATP)
For each glucose molecule that's oxidized, 6 oxygen molecules are used in order to produce ATP. Carbon dioxide and water are byproducts of aerobic respiration..
Each complete oxidation of a glucose molecule through aerobic respiration requires 6 molecules of oxygen (O2). The process, represented by the equation C6H12O6 + 6O2 -> 6CO2 + 6H2O + ATP, shows this clearly.
Explanation:In the process of aerobic respiration, glucose (C6H12O6) is oxidized, while oxygen is reduced. The overall equation for this process can be represented as C6H12O6 + 6O2 -> 6CO2 + 6H2O + ATP. From this equation, we can see that each molecule of glucose requires 6 molecules of oxygen (O2) to be completely oxidized. Therefore, the correct answer to the question 'Each time a molecule of glucose is completely oxidized via aerobic respiration, how many oxygen (O2) molecules are required?' is C) 6.
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Show how two phenotypically normal parents can produce a male child with color-blindness.
Answer:
The female (XcX) is carrier for color blindness while the male (XY) is normal. The carrier female transmits its one X chromosome with gene for color blindness to one of her sons making him color blind.
Explanation:
Color blindness is X linked recessive disorder and followed the criss-cross inheritance wherein the color blind sons get the gene from their carrier/affected mother while the color blind father would transmit the gene to his daughters along with X chromosome.
Since both the parents are normal, the mother (XcX) should be carrier for the disease to get the color blind son. A cross between carrier mother (XcX) and normal father (XY) would make 50% of their sons color blind.
The drug cytochalasin B blocks the function of actin. Which of the following aspects of the animal cell cycle would be most disrupted by cytochalasin B?
a. spindle formation
b. spindle attachment to kinetochores
c. cell elongation during anaphase
d. cleavage furrow formation and cytokinesis
Answer:
d. cleavage furrow formation and cytokinesis
Explanation:
Cytochalasin B is a cell permeable toxin produced by fungi. It inhibits actin polymerization and interaction of actin filaments. Cytokinesis is the process in which the cell undergoes cytoplasmic division to form two new cells. It begins in late anaphase and is completed by the end of telophase. It begins by the production of cleavage furrow on the cell surface. This furrow is formed by the contractile ring which in turn is composed of actin and myosin filaments along with other regulatory proteins.
When actin polymerization is inhibited by cytochalasin B, the contractile ring is not able to form so the cleavage furrow does not appear which ultimately inhibits cytokinesis.
When is high density liproteins (HDL) most prevalent in the body?
Answer:
HDL are most prevalent in the body when the person has high physical activity and a healthy diet.
Explanation:
Physical activity and a diet rich on fruits, fiber, omega 6 and esential fat acids increases the level of HDL, which responsible on drecreasing the levels of cholesterol in blood.
Genetic factors might also impact on the levels of HDL.
Base-pair substitutions involving the replacement of a purine with a pyrimidine and vice versa are called:
a. isomers
b. transitions
c. transversions
d. inversions
e. translations
Answer:
The answer is: c. tranversions
Explanation:
A mutation is the result of changes in the structure of a gen, these variations happen in the nucleotide sequence of a genome, and can be as a consequence of DNA damage.
One type of DNA mutation is the substitution of one base pair for another. Two type of substitutions can happen. One of them are the transversions, which are interchanges of a purine (A or G) for a pyrimidine (C or T) and vice versa.
The other type of substitution are the transitions that are interchanges of one purine (A) for another purine (G) or a pyrimidine (C) to another pyrimidine (T).
The tissue(s) that is/are considered excitable because of the ability to generate electrical signals is/are called ________ tissue. The tissue(s) that is/are considered excitable because of the ability to generate electrical signals is/are called ________ tissue. muscle neural epithelial muscle tissue and neural muscle tissue, neural tissue, and epithelial
Answer:
Muscle tissue and neural tissue
Explanation:
Excitability refers to the ability of muscle and nerve cells of the respective tissues to respond to a stimulus and generate an action potential. Both muscle cells and neurons respond to a stimulus and convert it into the action potential.
Action potential refers to the electrical signal. Propagation of action potential along the membranes of these cells results in muscle contraction and functioning of neurons.
The membrane potential of these cells changes in response to the stimulus and the changed potential is propagated to the other cells.
Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.
Answer:
Linkage chromosomes may be defined as the chromosome present at the same chromosomes and has the ability to undergo the process of recombination. The linkage results in the production of recombinant gametes that differ from their parents.
Map distance may be defined as the distance between the two gene on a chromosome. The recombination frequency is equal to the map distance. For example: Gene A and B has the recombination frequency of 10%, this means there map distance is 10 mu.