Answer:
a. add more of the enzyme.
Explanation:
Enzymes speed up the rate of reaction by lowering the activation energy for the reaction. They are not used themselves in the reaction. They are specific to substrate molecule. Substrate molecule binds to enzyme's active site and they undergo the reaction to form the product and release back the enzyme.
Rate of reaction depends on both substrate and enzyme concentration. Maximum rate of reaction is reached when all the active sites of enzyme molecules have been occupied by the substrate molecules which means that they are saturated. If more substrate is added at this point it wont have any effect on rate of reaction since there are no free active sites. Hence more enzyme is required to be added so that extra substrate can be utilized and rate of reaction can further be increased.
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.
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.
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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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.
Amino acids differ from one another by the ________________ that is present
a. amino group
b. carboxyl group
c. R group
d. R plasmid
e. core group
Amino acids, fundamental in biological processes, possess an amino group, a carboxyl group, a hydrogen atom, and a side chain or R group. Amino acids differentiate from each other based on their unique R group. This distinctive R group gives each amino acid its singular characteristics.
Explanation:Amino acids are organic compounds that play a crucial role in biological processes. They are composed of an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a side chain (R group). Each amino acid differs from another by its unique R group. So, the answer to your question is c. R group.
The R group can vary among different amino acids and this variation is what gives each amino acid its unique properties. For example, the R group can be as simple as a single hydrogen atom (as in the case of the amino acid glycine) or a more complex arrangement of atoms (as in the case of tryptophan).
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Amino acids differ from each other by their respective R groups, which represent unique side chains. These side chains could range from simple hydrogen atoms to complex multi-ring structures.
Explanation:Amino acids differ from one another by the R group that is present. Every amino acid has three main components: a central carbon atom, an amino group (-NH2), and a carboxyl group (-COOH). However, each also has a unique side chain, or R group, that sets it apart from the others.
This R group could be as simple as a single hydrogen atom or as complex as a multi-ring structure. It's due to these distinct R groups that we have 20 different common amino acids in proteins.
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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
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.
The evolution of archosaurs was heavily impacted by the climate and environmental changes that occurred in the Early Triassic. Which of these best describes the environment in the Triassic?
a. The vegetation was dominated by grasses and angiosperms which were new food sources
b. The equatorial climate became much more arid compared to that seen in the Permian
c. The polar regions were dominated by permanent polar ice caps that were linked to lower sea levels.
d. Oxygen levels were extremely high due an increase in worldwide vegetation
Answer:
Option (B)
Explanation:
The Triassic period belongs to the Mesozoic era in which the dominant species was the dinosaurs. During this time, the environment and the type of climate was relatively dry and there was no precipitation to support the vegetation. During, this Triassic period, the super-continent Pangaea was present that eventually got split up into different smaller continents by Jurassic and Cretaceous. The climate in the equatorial region during that particular time on earth was comparatively much more arid than that was in the Permian.
There were no polar ice caps during that time and tectonic activities were rapid. The earth experienced a temperature of about 10°C more than the present day temperature and oxygen level was also relatively low then. The plants that existed during this period were mostly gymnosperms.
Hence, the correct answer is option (B).
What element was pointed out for special attention in the structure of a DNA base?
a. oxygen
b. carbon
c. phosphorus
d. nitrogen
e. calcium
Answer:
The correct answer will be option-D.
Explanation:
Deoxyribose nucleic acid or DNA is the genetic material of the organisms which is composed of nucleotide monomers. Each nucleotide monomer is made up of a fiver carbon sugar, four different nitrogenous bases and a phosphate group
The DNA bases can be easily distinguished on the basis of nitrogen element as the structure of purine which is adenine and guanine contains four nitrogen molecule whereas cytosine and thymine contain two nitrogen molecules in their structure.
Thus, option-D is the correct answer.
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.
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.
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.
One oak tree cell with 14 chromosomes undergoes mitosis. How many daughter cells are formed, and what is the chromosome number in each cell?
Answer:
The correct answer will be option- 2 daughter cells with 14 chromosomes.
Explanation:
Mitosis is a type of cell division which is used by the organism to divide the somatic cells. This method is therefore considered an asexual mode of reproduction.
Mitosis proceeds in two stages: karyokinesis (a division of the nucleus) and cytokinesis (a division of the cytoplasm). Mitosis results in two daughter cells with the same amount of DNA content that is clone.
In the given question, since the tree cell contains 14 chromosomes therefore on the mitotic division will produce two daughter cells with the same amount of chromosomes that are 14 chromosomes.
Thus, 2 daughter cells with 14 chromosomes are the correct answer.
What type of lipid would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water?
a. glycerophospholipid
b. cholesterol
c. galactolipid
d. shingolipid
e. free fatty acids
Answer:
e. free fatty acids
Explanation:
Free fatty acids would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water.
Free fatty acids produced by the activity of lipase enzyme on stored triglycerides in the cytoplasm.
Micelles are the molecules of lipid which are arranged in a spherical manner which have a polar head and a hydrophobic tail. Micelles transport the fatty acids and insoluble monosaccharides.
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.
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.
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.
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.
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.
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.
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.
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.
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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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.
tRNAs:
a. have a "cloven hoof" secondary structure
b. serve as adapters between amino acids and DNA substitutions
c. have their corresponding amino acid covalently attached to their 5' end
d. are postranscriptionally modified to include abnormal bases such as dihydrosphinctosine
e. have an acceptor stem
Answer:
Have an acceptor stem.
Explanation:
RNA is present as genetic material in some viruses only. RNA is the single stranded molecules that are formed from the DNA template by the process of translation.
Different types of RNA are rRNA, mRNA and tRNA. Transfer RNA (tRNA) contains the anti codons that are complementary to the codon present on mRNA. The tRNA consists of the acceptor stem that acts as the amino acid attachment site during the process of translation.
Thus, the correct answer is option (e).
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.
Atoms of which three elements could bond together to form an organic compound?
(a) carbon, iron, and nitrogen
(b carbon, hydrogen, and oxygen
(c) oxygen, hydrogen, and nitrogen
(d) oxygen, iron, and carbon
Answer:
b
Explanation:
carbon, hydrogen and oxygen
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your answer is B) Carbon, Hydrogen, and Oxygen
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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.
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).