Answer: In this type of cross between MmDdPp and mmddpp, the phenotypic ratios in the eight possible phenotypes are as follow; a. solid, normal, smooth = 1:8
b. solid, normal, peach = 1:8
c. solid, dwarf, smooth = 1:8
d. solid, dwarf, peach = 1:8
e. mottled, normal, smooth = 1:8
f. mottled, normal, peach = 1:8
g. mottled, dwarf, smooth = 1:8
h. mottled, dwarf, peach = 1:8
Explanation: The number of gametes that can be formed from a particular genotype is given as 2^n where n is the number of heterozygous loci. In the genotype MmDdPp, n = 3. Therefore, 8 gametes can be formed.
In genotype mmddpp, n = 0. Therefore, only one type of gamete can be formed.
Crossing the two genotype (MmDdPp and mmddpp) as shown in the attached image, 8 possible offsprings can be achieved. Phenotypes of these offspring had been stated earlier. The phenotypic frequency of each of these offspring is one out of eight (⅛).
The eight possible phenotypes resulting from a cross between an F1 plant (MmDdPp) and a homozygous recessive plant (mmddpp) are equally probable (1:1:1:1:1:1:1:1) due to the principle of independent assortment. This principle indicates that each trait segregates independently.
Explanation:The tomato plant question deals with the principle of independent assortment in genetics, which states that the alleles for different genes get sorted into gametes independently of one another. In the testcross with an F1 plant (MmDdPp) and a plant homozygous recessive for all traits (mmddpp), the F1 plant can provide either the dominant or recessive allele for each trait, resulting in eight possible phenotypes in the F2 generation.
To predict the phenotypic ratio of the offspring in the F2 generation, we could use the Punnett Square method or probability concepts. As each trait has a 1:1 ratio (for example, solid:mottled leaves), this results in equal chance (50%) for each trait to be either dominant or recessive. Applying this to all three traits and considering them independently, the expected phenotypic ratio would be 1:1:1:1:1:1:1:1 for the eight phenotypes:
solid, normal, smoothsolid, normal, peachsolid, dwarf, smoothsolid, dwarf, peachmottled, normal, smoothmottled, normal, peachmottled, dwarf, smoothmottled, dwarf, peachLearn more about Independent Assortment here:https://brainly.com/question/32921835
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Order the steps from arrival of transcription-promoting factors at the eukaryotic DNA gene promoter to synthesis of an mRNA copy of the gene. Include the terms RNA polymerase II, elongation, promoter, TATA box, nucleotides, and transcription factor.
Answer:
Explanation:
There are 3 different types of DNA dependent RNA polymerase. RNAP II is the main enzyme synthesizing mRNAs.
1. Signals for initiation of transcription
Promoters: there are certain specific areas on the DNA that act as starting signals for initiation process.
The DNA polymerase attaches to the promoter site on the template DNA strand.
TATA Box: upstream of the transcription start site is 5'-TATAAA-3' it is referred to as the TATA Box. It acts as a signal for the start region.
2. Initiation of Transcription. The DNA helix unwind, and the RNAP binds to the promoter site on the 5' end DNA with the help of sigma factor. RNAP reaches the appropriate site of the gene and add the first nucleotide of the mRNA.
3. Elongation of Transcription: the RNAP moves along the DNA template, new nucleotides are incorporated in the nascent mRNA, one by one, according to the base pairing rule.
4. Termination of transcription: specific signals are recognized by the termination protein (Rho factor). When it attaches to the DNA, rnap can't move further. So the enzyme dissociate from DNA and consequently newly formed RNA is released.
Final answer:
The eukaryotic transcription process begins with the assembly of transcription factors at the gene promoter, including the TATA box, forming a transcription initiation complex, followed by RNA polymerase II binding, initiation, elongation, and termination to create an mRNA copy of the gene.
Explanation:
Steps of Eukaryotic Transcription:
Transcription factors arrive at the eukaryotic DNA gene promoter, which includes the TATA box, and help in the initial binding of the transcription machinery.The transcription initiation complex is formed when the TATA-binding protein (TBP) and other components of TFIID bind to the TATA box.RNA polymerase II binds to the promoter with the help of transcription factors, signaling the start of transcription.Initiation of transcription occurs as RNA polymerase II unwinds the DNA and begins reading the template strand.Elongation occurs as RNA polymerase II adds complementary nucleotides to the growing mRNA strand.The process continues until RNA polymerase II reaches a termination signal, at which point the completed mRNA strand is released.This sequence ensures that a precise mRNA copy of the gene is synthesized, which can then be processed and translated into a protein.
discuss what
you know about Earth's seasons.
Answer:
Explanation:
The Earth's seasons are not caused by the differences in the distance from the Sun throughout the year (these differences are extremely small). The seasons are the result of the tilt of the Earth's axis.
The Earth's axis is tilted from perpendicular to the plane of the ecliptic by 23.45°. This tilting is what gives us the four seasons of the year - spring, summer, autumn (fall) and winter. Since the axis is tilted, different parts of the globe are oriented towards the Sun at different times of the year.
Summer is warmer than winter (in each hemisphere) because the Sun's rays hit the Earth at a more direct angle during summer than during winter and also because the days are much longer than the nights during the summer. During the winter, the Sun's rays hit the Earth at an extreme angle, and the days are very short. These effects are due to the tilt of the Earth's axis.
Solstices
The solstices are days when the Sun reaches its farthest northern and southern declinations. The winter solstice occurs on December 21 or 22 and marks the beginning of winter (this is the shortest day of the year). The summer solstice occurs on June 21 and marks the beginning of summer (this is the longest day of the year).
Equinoxes
Equinoxes are days in which day and night are of equal duration. The two yearly equinoxes occur when the Sun crosses the celestial equator.
The vernal equinox occurs in late March (this is the beginning of spring in the Northern Hemisphere and the beginning of fall in the Southern Hemisphere); the autumnal equinox occurs in late September (this is the beginning of fall in the Northern Hemisphere and the beginning of spring in the Southern Hemisphere).
Answer:
Nearly everyone enjoys the change of seasons on Earth – from winter to spring, from summer to fall. ... Instead, Earth has seasons because our planet's axis of rotation is tilted at an angle of 23.5 degrees relative to our orbital plane – the plane of Earth's orbit around the sun
A type of neuron that stimulates action by conveying signalsto muscles or glands and initiating a body's response tostimuli is a(n):
a. afferent
b. efferent
c. sensory neuron.
d. interneuron
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You purchase two identical houseplants and place them side by side on your windowsill. You water both plants equally. You leave one plant, plant A, alone, but you inject florigen into the cells of the shoot apical meristem of plant B. What would you expect to occur?
Answer:
Both plants will have equal conditions in these experiments including the water and other components except florigen. Florigen is a hormone that is responsible for triggering the flowering in a plant. It acts in the apical meristem of growing tips and buds.
As it is injected in the apical meristem of the plant so it will act on the plant b and flower would be induced at its apical meristem. Once the flower will be gone plant will grow taller but only from the lateral meristem of the plant b.
Answer:
the shoot apical meristem of plant B will develop into a flower
Explanation:
Which of the following is something you should do after severe weather occurs?1. clean up flood waters2. keep away from windows3. create a family safety plani4. nvite friends over to see the damages
Answer:
do in this way: 1 , 3 , 2 , 4
Explanation:
please give me brainliest as i just need 3 more until expert. plz plz plz
The correct answer is 1. Clean up flood waters. After severe weather, cleaning up flood waters is essential to prevent health hazards, mold growth, and structural damage.
After experiencing severe weather, it's critical to know the appropriate steps to ensure your safety and begin the recovery process. Here's a breakdown of what you should do:
Clean up flood waters: Ensuring that flood waters are cleaned up is crucial to prevent waterborne diseases, mold growth, and further structural damage to your home.Keep away from windows: Although keeping away from windows is important during severe weather, it's not a necessary step immediately afterward.Create a family safety plan: While creating a family safety plan is essential, it is something you should do before disaster strikes, not immediately after.Invite friends over to see the damages: This is not advisable as it can lead to additional hazards and delays in the recovery process.In conclusion, cleaning up flood waters is the most immediate and critical action to take after severe weather to ensure safety and prevent further damage.
The process of translation in both prokaryotes and eukaryotes requires tRNAs, amino acids, and ribosomal subunits. Which of the following components must also be present for translation to proceed?
Translation in both prokaryotes and eukaryotes requires tRNAs, amino acids, ribosomal subunits, mRNA template, ribosomes, and enzymatic factors. These components are necessary for the formation of bonds between amino acids and the synthesis of polypeptides.
Explanation:In order for translation to proceed, three components must be present: tRNAs, amino acids, and ribosomal subunits. Additionally, several other components are required, including an mRNA template, ribosomes, and various enzymatic factors. These components are necessary for the formation of bonds between sequential amino acids specified by the mRNA template, as well as for the catalysis of the bonding between new amino acids and the growing polypeptide chain. Translation occurs in three-nucleotide steps and is facilitated by the ribosome.
Molecule 1 has the nitrogenous base sequence TCA AGT. Which set of bases in Molecule 2 can bond to that sequence in a complementary way?
The complementary sequence to TCA AGT in Molecule 2 is AGT TCA, based on the rules of complementary base pairing where A pairs with T, and C pairs with G.
Explanation:The question deals with understanding how DNA strands pair together based on complementary base pairing. In DNA, adenine (A) always pairs with thymine (T), and cytosine (C) pairs with guanine (G). Given the nitrogenous base sequence TCA AGT in Molecule 1, the complementary sequence in Molecule 2 would therefore be AGT TCA.
This is because T pairs with A, C pairs with G, and vice versa. This complementary base pairing is crucial for the structure of DNA and the transmission of genetic information.
The set of bases in Molecule 2 that can bond to the sequence TCA AGT in Molecule 1 in a complementary way is AGT TCA. The correct answer is A. AGT TCA.
The complementary base pairing rule in DNA dictates that adenine (A) pairs with thymine (T) and cytosine (C) pairs with guanine (G). Therefore, to find the complementary sequence in Molecule 2 for the given sequence TCA AGT in Molecule 1, we need to find the sequence that pairs with each base:
T (thymine) pairs with A (adenine)
C (cytosine) pairs with G (guanine)
A (adenine) pairs with T (thymine)
A (adenine) pairs with T (thymine)
G (guanine) pairs with C (cytosine)
T (thymine) pairs with A (adenine)
Based on these pairings, the complementary sequence to TCA AGT in Molecule 2 would be AGT TCA.
Therefore, the correct answer is a) AGT TCA
Complete Question
Molecule 1 has the nitrogenous base sequence TCA AGT. Which set of bases in Molecule 2 can bond to that sequence in a complementary way?
a) AGT TCA
b) GAT TCA
c) AGT TGA
d) TCA AGT
Which of these is required for natural selection to occur?
a. un-equal reproductive success.
b. inheritance.
c. individual variation.
d. all of the answers are correct.
Describe the phenotypes and genotypes of the rabbits.
Answer:
1. BB x bb
2. 100% black
Explanation:
Heterozygous offspring would have the genotype Bb. In order to produce offpsring that are 100% heterozygous, that means each parent must be homozygous. Therefore, the genotypes of the parents are BB (black rabbit) and bb (white rabbit). See punnet square attached.
The phenotype refers to the expression and display of the genotype. We will assume that because black is denoted by a capital B, and white by a small b, that black is the dominant allele. Therefore, if 100% of the offspring have the genotype Bb, then 100% of the offspring have the phenotype of black fur.
What process did Watson and Crick use to develop their model of DNA, and how did it differ from the controlled experiments used by Griffith, Avery, and Hershey and Chase?
Answer:
Watson and Crick already knew DNA's composition, based on the work of other researchers (Franklin). G A H and C manipulated possible components of DNA that would be able to identify DNA's material.
Explanation:
Watson and Crick is known for discovering the chemical compositions of the DNA.In 1953, they found that DNA has the double helical structure along with the stored information of the human gene.Griffith, Avery, and Hershey and Chase did an experiment to find that DNA is transferred through the generations and concluded that instead of protein, DNA carried the human genes.Final answer:
Watson and Crick used X-ray diffraction data, notably from Rosalind Franklin, and combined it with Chargaff's rules to develop their double helix model of DNA, without conducting their own experiments.
Explanation:
Discovery of the DNA Structure
In the 1950s, Francis Crick and James Watson utilized X-ray diffraction data from Rosalind Franklin to piece together the double helix structure of DNA. Unlike earlier experimenters who conducted controlled experiments like Griffith, Avery, and Hershey and Chase, Watson and Crick did not perform their own experiments but instead built upon existing research and information.
Franklin had shown through her X-ray diffraction images the helical nature of DNA, which Watson and Crick interpreted to model the double-stranded structure. Their use of Chargaff's rules, which indicated that certain DNA nucleotides always appeared in equal proportions, further guided them to propose the pairing now known as base pairing.
Rosalind Franklin's work was critical but often under recognized, whereas Watson and Crick's fame largely comes from the publication of the DNA model in their original Nature article. This model provided the foundation for understanding how DNA encodes information and how it is replicated, which underpins all modern biotechnologies.
One such technology is the polymerase chain reaction (PCR), a technique for amplifying DNA sequences. PCR and other biotechnological innovations like gene sequencing, DNA fingerprinting, and genetically modified foods rely on the fundamental understanding of DNA structure and function that Watson and Crick's model provided.
Which kind of media contains a dye or other indicator that responds to metabolic reactions during growth to identify a specific organism?
Answer:
Eosin methylene blue agar (EMB) is a selective and differential medium used to isolate fecal coliforms. Eosin Y and methylene blue are pH indicator dyes which combine to form a dark purple precipitate at low pH; they also serve to inhibit the growth of most Gram positive organisms.
Explanation:
A typical PCR program is listed below. Define what is happening to promote DNA in steps 2‐4 below amplification (i.e., what happens to DNA and enzyme in those steps). PCR program – C487 Amplification 1. 95° C for 1 minute 30 seconds 2. 95° C for 30 seconds (melting) 3. 52° C for 30 seconds (annealing) 4. 72° C for 60 seconds (DNA elongation) 5. Repeat steps 2‐4, 29 times (30 cycles total) 6. 72° C for 5 minutes 7. Hold at 4° C
In step 2: Denaturation of the double-strand occurs.
In step 3: Annealing of the primer to the single strands.
In step 4: Extension or elongation takes place in this step.
Explanation:
In the PCR program an enzyme Taq Polymerase is used because it can withstand high temperature without altering its functions.
PCR is required for the amplification of DNA into multiple copies for experimental purpose. The artificial environment is created to form new DNA molecules from the sample in question.
The first step in replication is the opening of the double helix which is done by temperature treatment in PCR. The temperature would be 90 degrees for some 30 sec to two minutes.
The next step of primer annealing would be done at 52 degrees, this is the primer melting temperature.
The elongation of the DNA strand to be synthesized will take place at 72 degrees as Taq Polymerase can withstand that temperature.
Nearly one million copies of DNA will be made after 30 cycles of PCR.
PCR products can be stored at 4 degrees for some two months.
Significant damage to which of the following parts of the brain will most likely cause a person to fall into a deep coma from which the person will be unable to awaken?
A. Hypothalamus
B. Frontal lobe
C. Reticular formation
D. Amygdala
E. Thalamus
Damage to the Reticular Formation, a part of the brain involved in maintaining consciousness, is most likely to cause a person to fall into a deep coma from which they can't awaken.
Explanation:Significant damage to the Reticular Formation of the brain is most likely to cause a person to fall into a deep coma from which he or she will not be able to awaken. The reticular formation plays a crucial role in maintaining consciousness and damage to this area can lead to a state of comatose. The other parts of the brain listed, such as the Hypothalamus, Frontal lobe, Amygdala, and Thalamus also have significant roles, but their damage doesn't predominantly lead to a coma.
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You analyze a DNA sample and find that its base composition is 30% A, 20% T, 30% G, and 20% C. What can you conclude about the structure of this DNA?
Because A equals G, and T equals C, this must be a double-stranded DNA molecule.
Because A does not equal T, and G does not equal C, this cannot be a double-stranded DNA molecule. It must therefore be a single-stranded DNA.
Because A is half as much again as T, and G is half as much again as C, this must be a mutated double-stranded DNA molecule.
There is not enough information to determine whether it is a single-stranded or a double-stranded DNA molecule.
Answer:
Because A does not equal T, and G does not equal C, this cannot be a double-stranded DNA molecule. It must therefore be a single-stranded DNA.
Explanation:
The base composition in the observed DNA sample does not follow the Chargaff rule. If it was a double-stranded DNA, the percent proportion of adenine base and thymine base should be equal (either 30% or 20% for both). Similarly, the percent proportion of guanine and cytosine bases must also be equal (either 30% or 20% for both). Since the given sample does not follow it, it is not a double-stranded DNA molecule but is present as a single strand.
Sickle cell disease involves the red blood cells, or hemoglobin, and their ... or capillaries and stops the movement of healthy, normal oxygen-carrying blood. .
Answer: The statement is true
Explanation:
The sickle-cell disease is characterized by red blood cells with sickle shape (like a semi-circle). The iron-containing pigment, HEMOGLOBIN in these sickle-shaped red blood cells is
- inefficient in transporting oxygen
- likely to break down in environment with low oxygen concentration.
Such abnormally shaped red blood cells block the cavities of small blood vessels like CAPILLARIES, hindering the free movement of healthy, normal oxgen-carrying blood in them.
A sample of radioactive material has a half life of 20 years and an activity level of 4 curies. After how many years will the activity level of the sample be at 0.25 curie?
The activity level of a radioactive material decreases by half with each half-life. In this case, with the half-life being 20 years, the activity level would reach 0.25 curies after 80 years, equivalent to four half-lives.
Explanation:The subject in question involves the concept of half-life, a term commonly used in nuclear physics to denote the time it takes for half of a certain radioactive substance to decay. It's given that the sample has an activity level of 4 curies and a half-life of 20 years. The activity level after some years is to be 0.25 curie.
As a rule, each time a sample goes through one half-life, its activity level gets halved. So, from 4 curies to 2 curies would be one half-life (20 years), from 2 curies to 1 curie would be another half-life (another 20 years), from 1 curie to 0.5 curie would be yet another half-life (another 20 years), and finally from 0.5 curie to 0.25 curie would be a fourth half-life (another 20 years).
Therefore, it would take 4 half-lives—totaling 80 years—for the activity level of the sample to drop from 4 curies to 0.25 curie.
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A 5 year old child announces to her parents that she is going to hold her breath until she is allowed to watch more TV. The parents are not worried. Explain why the parents are not worried. Explain how the child's blood carbon dioxide levels and blood hydrogen ion and bicarbonate ion levels change while she is holding her breath.
Explanation:
Parents are not worried because they know that when the child holds her breath, the level of carbon dioxide will increase and activate breathing centers forcing her to breathe.
A child’s blood carbon dioxide levels, blood hydrogen ion, and bicarbonate ion levels all increase during breath-holding increasing pCO2 (partial pressure of carbon dioxide).
PCO2 reflects how much carbon dioxide is dissolved in the blood.
If the concentration of salt in the fluid surrounding cells decreases and the concentration of other solutes remains constant?
Answer:
The cell will swell.
Explanation:
The cells react distinctly when placed in different solutions like hypertonic, isotonic and hypotonic solutions. In the mentioned question, that is, in the fluid surrounding the cells, the concentration of salt reduces, which makes the solution hypotonic. Hypotonic solution exhibit high water potential and low solute concentration.
This makes the water move from the hypotonic solution to the inside of the cell as the osmotic movement occurs from high solvent concentration to low solvent concentration, thus, swelling of the cell takes place.
Suppose a cell has the following molecules and structures: enzymes, DNA, ribosomes, plasma membrane, and mitocondria. It could be a cell from:_______.a) bacterium. b) animal cell. c) plant cell. d) virus.
Answer:
A plant cell and an animal cell
Explanation:
Nearly all eukaryotic cells contain or possess enzymes, DNA, plasma membrane and mitochondria.
Enzymes, DNA, ribosomes, plasma membranes, and mitochondria are all components of eukaryotic cells that are necessary for cellular function. Thus option B and C is correct.
What is the eukaryotic cells?Organisms called eukaryotes have nuclei and membrane-bound organelles in their cells.
The majority of algae, all animals, plants, fungi, and protists are eukaryotic organisms. Eukaryotes are multicellular or unicellular organisms.
Organelles that are bound to membranes are present in eukaryotic cells, including the nucleus, but not in prokaryotic cells.
The presence of mitochondria and chloroplasts, the cell wall, and the makeup of chromosomal DNA are all examples of differences between the cellular structures of prokaryotes and eukaryotes.
Therefore, all given the component of cells is present in Eukaryotic cells.
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A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of the normal phenylalanine. The consequence of this for the cell will be that:_____________a) none of the proteins in the cell will contain phenylalanine.b) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.c) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons.d) the ribosome will skip a codon every time a UUU is encountered.e) none of the options will occur; the cell will recognize the error and destroy the tRNA.
Answer:
Proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.
Explanation:
Mutation may be defined as the sudden or heritable change in the genetic sequence of the organisms. The mutation can change the phenotype and genotype of the organisms.
The AAA codes for the lysine. The mutation attaches the anticodon AAA instead of normal phenylalanine. The aminonacids AAA will code for the lysine and no phenylalanine is observed in the cell as phenylalanine is coded by UUU and not by AAA.
Thus, the correct answer is option (b).
Constructing phylogenies is not as simple as grouping organisms with similar characters together. Identify situations in which similar characters must be IGNORED.
-similar characters are derived
-similar characters evolved owing to convergent evolution
-similar characters evolved owing to gene flow
-similar characters are ancestral
Answer
similar characters evolved owing to convergent evolution; similar characters are ancestral
Explanation:
Convergent evolution happens when two organisms belonging to different species, living at different times or different continents but the same biome develop analogous structures hence making them appear similar. The analogous structures were not present in the known ancestors of both groups.
Therefore convergent evolution is ignored during constructing phylogenies
Also similar characteristics that are ancestral are ignored.
Hormones such as calcitonin, parathyroid hormone, and the pituitary hormones are protein-based and water-soluble. How does a hormone that cannot penetrate the lipid membrane effect change in the cell?
Answer: Substances that are protein based cannot cross the lipid cell membrane, but can exert their effect by binding to the proteins based receptors present on the lipid bilayer.
Explanation:
Hormones such as calcitonin , Parathyroid hormone and pituatry horomones are proteins and water soluble.They cannot cross lipid bilayer but can bind to ligand binding protein receptors.
These receptors are protein macromolecules that are specific in nature for specific hormones.
After binding to the receptors, these hormone-recpetor complexes activate second messengers and regulate the downstream effects.
Answer:
Hormones that cannot penetrate the lipid membrane are known as Water-soluble or protein-based.
They effect change in the cell by binding to a receptor protein on the plasma membrane of the cell.
Following an interaction with the hormones, a cascade of secondary effects within the cytoplasm of the cell is triggered, often involving the addition or removal of phosphate groups to cytoplasmic proteins, changes in ion channel permeability, or an increase in the concentrations of intracellular molecules that may act as secondary messengers, such as cyclic AMP.
If you were putting together a SWOT analysis for your department's latest production cycle, in which category would you list a competitor that offers lower prices? (1 point) • strengths • weaknesses • opportunities • threats
Option D
In threats category would you list a competitor that offers lower prices.
Explanation:
SWOT analyses can render extents of knowledge about the factors that a business or enterprise discovers itself in. A threat to one company is a visible factor, something that you not able to guide, which could negatively influence one's firm. This could involve anything from different organizations, to fulfill curtailments.
Threats are abrogating and obvious. This indicates that threats do not profit your business, but there is annihilation you can do to prevent them from proceeding about. Recognizing threats to your market is a paramount chief start to lessening their venture.
What are donut-shaped structures in the limbic system that jointly control a variety of functions relating to emotions and self- preservation?
Answer:
Hippocampus and amygdala.
Explanation:
The hippocampus and the amygdala are part of the limbic system. This is responsible through these structures of emotions, memory and is decisive for the survival (self- preservation) of the human being. The hippocampus participates in the process of consolidation of some types of memory, it receives multiple afferents, through the neurons responsible for receiving sensations to transmit them to the brain, especially of the amygdala, which is considered as the main element or nucleus central involved in the management and control of emotions and feelings. It receives cortical afferences, is located in the depth of the temporal lobe and is connected to the entire brain, which denotes its integrative role in emotional processing.
in pea plants t is the allele for tall plants while t is the allele for dwarf plants. If you have a tall plant, demonstrate with a testcross how it could be determained if the plant is homozygous tall or heterozygous tall.
Answer:
The tall plant is homozygous (TT) if all the progeny of the test cross is "tall". Presence of both tall and short progeny makes the "tall" plant heterozygous (Tt).
Explanation:
The "tall" plant with unknown genotype would be crossed with a homozygous short plant (tt) of the same species. This is called the testcross. If the progeny of test cross includes both tall and short plants, the "tall" plant with unknown genotype was heterozygous for the gene. On the other hand, the presence of only tall progeny in the testcross reflects the fact that the "tall" plant with unknown genotype was homozygous for the allele for "tallness".
To determine if a tall pea plant is homozygous tall or heterozygous tall, a testcross can be performed by crossing the tall plant with a dwarf plant.
Explanation:In pea plants, the allele for tall plants is represented by uppercase letter T, while the allele for dwarf plants is represented by lowercase letter t. To determine if a tall plant is homozygous tall or heterozygous tall, you can perform a testcross by crossing the tall plant with a dwarf plant. If the tall plant is homozygous tall (TT), all the offspring will be tall. If the tall plant is heterozygous tall (Tt), half of the offspring will be tall and half will be dwarf.
In neuronal communication, ________ involves moving a neurotransmitter from the synapse back into the axon terminal from which it was released.
Answer:
In neuronal communication, _reuptake_ involves moving a neurotransmitter from the synapse back into the axon terminal from which it was released
Reuptake is the reabsorption of a neurotransmitter by a neurotransmitter located along an axon terminal's plasma membrane (i.e., the pre-synaptic neuron at a synapse) or glial cell after performing its role of transmitting a neural impulse.
the scientific method is used by scientists to explain a certain natural phenomenon, and it involves the formation of a ______ as a tentative explanation of the observed or measured phenomenon
Answer:
Hypothesis
Explanation:
In the scientific method used by scientist, The Hypothesis has a tentative or proposed explanation that gives report of an observed or measured phenomena. The hypothesis is usually based on previous observations which fails to be explained by available scientific theories. It is the starting point for further investigation that can be tested through experimentation.
Label the genotypes of the parents (P1) and two offspring generations (F1 and F2). Labels can be used once, more than once, or not at all.
Answer: the question is not detailed enough
Explanation:
For instance if the genotype of the parent is AS and AS there would be a cross breeding of the genotype to give the First filial generation (F1) then any other crossbreeding is known as the F2
Note: the parent genotype is known as P1
A loss of 90% of the population in the Pingelap atoll from a typhoon in 1775 resulted in a high percentage of the current population being affected by achromatopsia. As a result, all of these affected individuals _____.
a. have normal vision
b. are descendants of one of the 20 original survivors
c. receive genetic testing and genetic counseling
d. have exceptional hearing
e. are descendants of the 20 original survivors
Answer:
Option-B
Explanation:
Pingelap is an atoll consisting of three islands: Daekae, Sukoru and Pingelap Island.
The present population of the atoll is 250 residents, the proportion of which has the color blindness called achromatopsia caused by the loss of the cone cells which respond to a different wavelength of light.
The achromatopsia on the atoll can be traced back in 1775 when a catastrophic typhoon swept the island and left only 20 individuals. One of these individuals named the Doahkaesa Mwanenihsed was the carrier of the disease and the inbreeding in the descendants of this individual resulted in the increase in the frequency of the recessive allele and is the cause of the affected individuals.
Thus, Option-B is the correct answer.
Existing landfilles are generally considered sufficient to serve as the sole means of disposing of the nations waste for the foreseeable future.True/false
Answer:
False.
Explanation:
The waste may be defined as any undesirable substance that left out after using the particular substance. The waste substance can be defined into two main classes - biodegradable and non biodegradable waste.
The one way to dispose out the waste product is the landfilles. The substance that cannot be degraded by any other means can be land filled. But this waste disposal method is not enough to contain the whole nation waste and all the waste cannot be land filled.
Thus, the correct answer is false.