Answer:
The most recent common ancestor of all currently living organisms is the last universal ancestor, which lived about 3.9 billion years ago. ... 6,331 groups of genes common to all living animals have been identified; these may have arisen from a single common ancestor that lived 650 million years ago in the Precambrian.
How do you plant a lavender? What are the best conditions to grow it
Answer:
Lavender is best planted in the spring as the soil is warming up. If planted in the fall, use bigger plants to ensure survival over the winter
2. Plant lavender 2 to 3 feet apart
3.It thrives in any poor or moderately fertile soil.
4.Keep away from wet, moist areas.
Answer:
Lavender is grown in garden beds, pots and requires maximum sunlight and a soil that is well drained for good germination. The soil require for lavender growth should be moderately fertile. It can grow in both arid and humid lands but climate affects the its growth rate.
Addition of ________causes the induction of the ______________ polymerase. This polymerase then binds to the T7 ________________ upstream of the rGFP gene, resulting in the _____________________ of rGFP gene to produce rGFP mRNA. This mRNA is then ______________________ to produce the rGFP protein .
Answer:
Addition of HEAT causes the induction of the T7 RNA polymerase. This polymerase then binds to the T7 RNA POLYMERASE PROMOTER upstream of the rGFP gene, resulting in the TRANSCRIPTION of rGFP gene to produce rGFP mRNA. This mRNA is then TRANSLATED to produce the rGFP protein
Explanation:
T7 RNA polymerase is a warmth inducible compound that is initiated inside the nearness of a warmth source. This polymerase ties to a chose grouping alluded to as the T7 RNA Polymerase advertiser succession. It moves along the DNA grouping prompting the amalgamation of mRNA during a procedure alluded to as translation.
This mRNA is changed over into a protein item through a procedure alluded to as interpretation.
You are running a nursery for garden peas at Baton Rouge, and you have two pure-bred garden pea strains: Yellow Wrinkled [YYrr] and Green Round [yyRR]. You know that Yellow is dominant to Green, and also that Round is dominant to Wrinkled. One customer dropped by and asked about one particular pure-bred strain, Yellow Round [YYRR]. You promised to that customer that you could generate that strain next year through the following breeding experiments.
I. Crossing pure-bred Yellow Wrinkled with Green Round
II. Self-crossing F1 progeny
III. Screening pure-bred Yellow Round among the F2 progeny
What is the genotype for F1 progeny?
A. YYrr
B. rrYY
C. YyRr
D. yyrr
E. none of the above
Answer:
C. YyRr
Explanation:
The cross was done between pure-bred Yellow Wrinkled and Green Round plants. The genotype of the pure breeding yellow wrinkled parent plant: YYrr. The genotype of the pure breeding green round parent plant: yyRR.
A cross between YYrr and yyRR would obtain the progeny in the following ratios:
YYrr x yyRR= All YyRr (yellow and round)
If you allowed your dilution tubes to incubate for 24 hours before plating them on the TSA agar plates, do you think the results of the experiment would be impacted? Assume that unlimited resources are present in the tubes. Explain your answer.
Answer:
The TSA or the tryptic soy agar is formed of casein and soybean meal, this formation helps in the appropriate growth of a huge array of non-fastidious and fastidious microbes. This combination of soy and casein provides organic nitrogen in the form of polypeptides and amino acids, which makes the medium more suitable for growth.
In the given case, if one permits the incubation of the dilution tubes for 24 hours prior to plating them on the TSA agar plates than there is a more probability of the result to get affected. As if unlimited resources are already present in the tubes, it will provide more favorable conditions for the formation of more colonies and thus will influence or change the colony-forming units per milliliter.
Even if the dilution is performed in a hood and in an autoclaved medium then also there will be an increase in the colonies of the microbes as in the time of 24 hours interval more microbes will get differentiated and will increase in number.
Final answer:
Allowing dilution tubes to incubate for 24 hours with unlimited resources before plating on TSA agar plates would impact the results by potentially causing confluent growth, making it difficult to distinguish individual colonies. TSA plates would typically have more colonies compared to EMB plates as TSA is non-selective and supports a broader range of bacterial growth.
Explanation:
If you allow dilution tubes to incubate for 24 hours before plating them on TSA agar plates, the results of the experiment would likely be impacted. Given unlimited resources in the tubes, this prolonged incubation could lead to an increase in bacterial growth, which may result in confluent growth (a mass of bacteria) in the test tube itself. Therefore, when you eventually do the plating, distinguishing individual colonies could be difficult or impossible, which would complicate or invalidate the results meant to measure bacterial concentration.
Regarding the comparison between TSA and EMB plates, due to the selective and differential properties of EMB agar, it inhibits the growth of gram-positive bacteria and highlights the growth of gram-negative bacteria, like Escherichia coli, by a color change. Since TSA agar is non-selective and supports a wider range of bacterial growth, you would generally expect more colonies on the TSA plates compared to the EMB plates, which would select only for specific types of bacteria.
In fruit flies, long wings (W) are dominant over short wings (w), and brown pigments (N) are dominant over yellow pigments (n). Each individual possesses two alleles for each trait.
1. If a fly that is homozygous dominant for both traits is crossed with a fly that is homozygous recessive for both traits, what is the predicted genotype of the offspring?
Answer:
WwNn
Explanation:
Long wings (W) are dominant over short wings (w)Brown pigments (N) are dominant over yellow pigments (n)A fly that is homozygous dominant for both traits (WWNN) is crossed with a fly that is homozygous recessive for both traits (wwnn).
All gametes produced by the WWNN individual will be WN, and all gametes produced by the homozygous recessive individual will be wn.
When the gametes from the two parents combine, the zygote will be heterozygous for both genes, with the genotype WwNn.
The offspring from a homozygous dominant fruit fly (WWNN) and a homozygous recessive fly (wwnn) will all have the heterozygous genotype WwNn, displaying long wings and brown pigments.
Explanation:In classical Mendelian genetics, long wings (W) are dominant over short wings (w), and brown pigments (N) are dominant over yellow pigments (n). When a homozygous dominant fruit fly for both traits (WWNN) is crossed with a homozygous recessive fly (wwnn), the resulting genotypes for all the offspring will be heterozygous for both traits (WwNn), expressing the dominant phenotypes—long wings and brown pigments.
The scenario is straightforward as both traits are being considered separately, and the inheritance pattern follows simple Mendelian inheritance, not taking into account potential linkage or sex linkage as in X-linked crosses. Therefore, all offspring produced from this cross will have the same genotype, with one dominant and one recessive allele for each gene, showcasing the principle of uniformity in the F1 generation.
Primary ciliary dyskinesia (PCD) is a rare genetic disease. Affected individuals exhibit impaired functioning of ciliated cells.
Based on what you know about the role of cilia in eukaryotic cells, why would you expect people with PCD to be particularly susceptible to respiratory infections?
Answer:
Cilia are motile in the lungs responsible for keeping the airways clear of dirt and mucus by their characteristic beating motion and rhythmic waving allowing a person to breathe easily and without irritation. These are present in both the lungs as well as middle ear.
As in Primary ciliary dyskinesia , there are defects in the action of cilia in the lining of the respiratory tract , middle ear , sinuses , eustachian tube etc the cilia cannot peforms its regular role .Infections can lead to an irreversible scarring and obstruction in the bronchi resulting in :
1) Shortness of breath.
2) Recurring chest colds.
3) Sinusitis.
4) Coughing , gagging , choking,
5)Middle ear infections
Answer: The respiratory wall or mucosa is made up of the epithelium and supporting lamina propria. The epithelium of respiratory tract is tall columnar pseudostratified with CILIA and goblet cells The cilia aids in sweeping away dusts and bacteria that adheres to the mucous on the epithelium. Therefore people with Primary ciliary dyskinesia are prone to Respiratory infections.
Explanation: Primary ciliary dyskinesia also called immotile ciliary syndrome is a rare genetic disease that affects the movement of cilia lining the respiratory tract. The major consequences of this dysfunction is reduced or absent mucus clearance from the lungs which subsequently leads to chronic recurrent respiratory infections
I hope this helps. Thanks!
1. If the frequency of two alleles in a gene pool is 90% A and 10% a, what is the frequency of individuals in the population with the genotype Aa?
Answer: 0.18
Explanation:
For the alleles, the percentage distribution of each is 'A' (90% = 0.9)
While 'a' (10% = 0.1)
Hence, 0.9 and 0.1 are the respective frequencies of each allele
Now, apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.
Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 × 0.9 × 0.1 = 0.18
Thus, the frequency of heterozygote is 0.18, while the percentage distribution in the population is 18%
A species of beetle expresses a pigment protein as an adult but not as a juvenile. The pigment is encoded by the gene PIG. A protein called KID, which is only present in juveniles, is a transcription factor that binds to the DNA near the protein-coding sequence for PIG. Do you think KID is a negative regulator of PIG or a positive regulator? Why?
Some individuals in the species never express the pigment (neither as a juvenile or as an adult). Do you think they have a mutation in the KID gene, the KID binding site, or in the protein-coding part of the PIG gene? Why? (the ‘why’ part of the question will only be used to help us give possible credit for answers different than what we think is the clearest answer)
Answer:
The KID protein is responsible for the no pigmentation at the juvenile stage. When the KID protein inhibits in the adult state, the pigmentation occurs in the body. This might occur because the KID protein acts as the repressor molecule and acts as a negative regulator of PIG protein.
The KID protein is responsible for pigmentation an adult stage. Any mutation in the KID gene might result in the loss of pigmentation in the adult. The KID gene is responsible for the binding of the KID protein and mutation in this gene can lead to the arrest of KID protein. The protein is unable to release and PIG continuously repressed in the adults.
6) (1 point) Proteins that span biological membranes often contain -helices. Given that the insides of membranes are highly hydrophobic, predict what type of amino acids would be in such a helix. Why is an helix particularly suited to exist in the hydrophobic environment of the interior of a membrane?
Answer:
The amino acids in such a helix would be hydrophobic in nature.
An α helix is particularly suited to cross a membrane because all of the carbonyl oxygen atoms and the hydrogen atoms of amide of the peptide backbone take part in intrachain hydrogen bonds, thus stabilizing these polar atoms in a hydrophobic environment.
Explanation :
Many transmembrane protiens use several alpha-helices wrapped up together.
It is usually seen as the helical structure can internally satisfy all the hydrogen-bonds , it doesn't leave any polar groups that are exposed to membrane if the sidechains are hydrophobic.
Sometimes, 2 to 3 alpha helices will wrap around each other , forming coiled coil. In an aphipathic alpha helix , the hydrophobic R groups on one side of each helix interact with each other while the hydrophilic R groups on the other side of each helix will interact with water.
Answer:
→alpha-helices
→Non-polar Amino acids
→because they and non -polar and hydrophobic.
Explanation:
Membrane proteins can be intrinsic (integral ) that is embedded in the membrane bilayer or extrinsic(peripheral) attached to the outer membrane layer.
These integral protein transcend the entire phospholipid bilayer, with the alpha- helices. The latter have hydrophobic side chains of non-polar amino acids. They are held to the cell membrane with these side chains which forms hydrophobic interactions with fatty acyl group of the phosphoslipid bilayer, and sometimes ionic bond with the polar head of phospholipid.
These alpha helix are non-polar(uncharged) and hydrophobic.A characteristic feature that make them to interact and fixed into the integral phospholipid hydrophobic medium.
28. Select all accurate statements
A. All chordates will have a muscular post an*l tail
B. All chordates have a dorsal, hollow nerve cord.
C. All chordates have pharyngeal slits or clefts
D. All chordates will have a notochord in their development and/or adulthood
E. All chordates are bilaterally symmetrical animals
Answer:
its A but I am not sure I just tried
Answer:
All the options A,B,C,D and E are correct.
Chordates possess a muscular posterior tail, a dorsal, hollow nerve cord, pharyngeal slits or clefts, a notochord in their adulthood and are bilaterally symmetrical animals.
Which of the following is/are true?
A. The fungal pathogen Mycosphaerealla graminicola is found worldwide with its host, cultivated wheat. Mycopharealla graminicola is host specific and does not occur on other host species such as Barley.
The closest known relative of M. graminicola is a barely-adopted pathogen Septoria passerinii.
You are researching these fungi and have the following hypothesis: If M. graminicola and 5 passerinii both had a common ancestor that lived in one geographic area where wheat and barley grew, it may be possible that the ancestor gave rise to these two species in one geographic location. This would be classified as sympatric speciation.
Your hypothesis and definition of sympatric speciation is correct.
B. A flood causes the loss of all red-headed males ducks in a population. This is an example of sympatric speciation.
C. A plant species obtain an extra set of homologous chromosomes. This would be an example of sympatric speciation.
D. Sympatric speciation may be due to sexual (mate) selection.
E. Sympatric speciation can occur when a single species occupies the same geographic location.
Answer: Option C, D, E and A
Explanation:
Sympatric speciation is the evolution or isolation of new species from the original population of species occupying the same geographic area. It is also due to sexual selection of mates leading to reproductive barriers. A plant with extra set of homologous chromosomes is an example. Sympatric speciation is due to isolation of new species from the population of species who arise from common anscetor.
True or False, it's likely that organisms other than LUCA existed alongside LUCA before the divergence of bacteria from archaea and eukaryotes.
Answer:
False
Explanation:
Which of the following is/are true?
A. Sympatric speciation most commonly occurs due to sexual (mate) selection.
B. Sympatric speciation can only occur when a single species occupies the same geographic location.
C. A plant species obtain an extra set of homologous chromosomes. This would be an example of sympatric speciation.
D. A flood causes the loss of all red-headed males ducks in a population. As a result, the red-headed female ducks must breed with yellow-headed male ducks, which are not their preferred mates. This is an example of sympatric speciation.
E. The fungal pathogen Mycosphaerella graminicola is found worldwide with its host, cultivated wheat. Mycosphaerella graminicola is host specific and does not occur on other host species such as Barley.
The closest known relative of M. graminicola is a barely-adapted pathogen Septoria passerinii.
You are researching these fungi and have the following hypothesis: If M. graminicola and S passerinii do not have a common ancestor that lived in one geographic area where wheat and barley grew, it may be possible that a common ancestor gave rise to these two species. This would be classified as sympatric speciation.
Your hypothesis and definition of sympatric speciation is logical.
Answer: A, C, and E are correct
Explanation:
Sympatric speciation is a random or naturally occurring event whereby organisms of the same species:
- live in the same territory or nearby territories ( i.e no single specie occupy
an area in isolation)
- DO NOT interbreed, but select a sexual mate from a much diverse territory and practice non-random mating, which favors some genes results in an uneven gene flow or disruption of alleles previously common among the population.
- produce offspring with extra sets of chromosomes known as polyploidy, leading to show genetic variations
Finally, M. graminicola and S passerinii are Sympatric species based on the already given explanation.
Many properties of living things involve the transfer and transformation of energy and matter. For example, plant chloroplasts convert energy from sunlight to what forms of energy or matter?
Answer:all of
Explanation:
List the sources of experimental uncertainty. List steps you will take to minimize the uncertainties.
Answer:
Sources of experimental uncertainties
1. Environment- change in environment can bring about experimental error e.g change Temperature can affect crop yield.
2. Wrong caibration of equipment- some equipment need to be calibrated before use. Wrong calibration brings error
Explanation: steps to uncertainties
1.Calibrate equipment when necessary.
2. Ensure fomulars are rightly imputed for electronic devices
3. Experience and competency is needed to avoid Esperanza uncertainties. Expert can be employed
4. Replication for field work reduces uncertainties. E.g maize of the same varieties can be planted in three places on the field to avoid uncertainties.
Crenation and hemolysis A cell placed in a hypertonic solution will shrink in a process called crenation. A cell placed in a hypotonic solution will swell in a process called hemolysis. To prevent crenation or hemolysis, a cell must be placed in an isotonic solution such as 0.9% (m/v) NaCl or 5.0% (m/v) glucose. This does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes. Part D
Question options:
A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur.
Solution A: 3.21% (m/v) NaCl
Solution B: 1.65% (m/v) glucose
Solution C: distilled H2O
Solution D: 6.97% (m/v) glucose
Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl
Answer:
Crenation: A, D, E
Hemolysis: B, C
Explanation:
Crenation is an osmotic process in which blood cells shrink while placing hypertonic or alkaline solutions.
Crention caused by these hypertonic solutions.
A: 3.21% (m/v) NaCl (more solutes)
D: 6.97% (m/v) glucose (more solutes)
D: 5.0% (m/v) glucose and 0.9%(m/v) NaCl (more solutes)
Hemolysis is the destruction of red blood cells in which cells bloat up and may explode while placing in a hypotonic solution.
Hemolysis caused by these hypotonic solutions.
B: 1.65% (m/v) glucose
C: distilled H2O
Crenation (cell shrinkage) occurs when cells are placed in hypertonic solutions, while hemolysis (cell swelling) occurs when they are put in hypotonic solutions. To maintain cell size and form, cells should be in isotonic solutions, which have the same solute concentration as the cells. Examples of isotonic solutions are 0.9% m/v NaCl or 5.0% m/v glucose.
Explanation:Crenation and hemolysis are processes that pertain to the behavior of cells in different types of solutions. When a cell is placed in a hypertonic solution - a solution with a greater concentration of solutes compared to the cell's cytoplasm - the water molecules inside the cell will move outside to balance the solute concentration, causing the cell to shrink. This process is known as creation.
On the other hand, if a cell is placed in a hypotonic solution - a solution with a lower solute concentration than the cell's cytoplasm - the water molecules will move into the cell, causing it to swell and potentially burst in a process called hemolysis.
To prevent either crenation or hemolysis from occurring, the cell should be placed in an isotonic solution. An isotonic solution has the same solute concentration as the cell's cytoplasm, ensuring equal water movement in and out of the cell, and thus, maintaining the cell's size and shape. Examples of isotonic solutions are 0.9% (m/v) NaCl or 5.0% (m/v) glucose. However, these percentages do not imply that a cell has a 5.0% (m/v) glucose concentration. It means that these solutions exert the same osmotic pressure as that inside the cell, which has multiple solutes.
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The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent and deep seated earthquakes is characteristics of ___?
Answer:
ocean-continent plate convergenceExplanation:
The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent and deep seated earthquakes is characteristics of ocean-continent plate convergence
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The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent and deep-seated earthquakes is characteristic of a convergent plate boundary, which typically forms when an oceanic plate is subducted under a continental plate.
Explanation:The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent, and deep-seated earthquakes, are indications of a convergent plate boundary. This typically happens when an oceanic plate is subducted or pushed beneath a continental plate, forming a deep oceanic trench. The intense heat and pressure cause the subducted plate to partially melt, and this molten material can rise to form a chain of volcanic mountains along the edge of the continent, often known as a volcanic arc. The process of subduction also leads to seismic activity or earthquakes that are deep-seated.
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Sickle cell anemia and albinism are both recessive traits in humans. Imagine that a couple, already pregnant with twins, has just learned that they are both heterozygous for both of these traits.As the couple's genetic counselor, the couple asks you the following questions about how their carrier status will affect their offspring.Part AIf the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?Part BIf the couple has fraternal twins, what is the probability that both of the couple's children will have both sickle cell anemia and albinism?Part CWhat is the probability that one of the fraternal twins is a carrier of either, but not both, of the conditions?(Hint: You will need to use both the product law and the sum law to answer this question.)Part DIf the couple has fraternal twins, what is the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism?(Hint: You will need to use both the product law and the sum law to answer this question.)
A. 81/256 B.1/256 C 4/9 D 1/4
Explanation:
A. Both the parents are heterozygous for sickle cell anaemia, the probability of gene passing to one child is
1/2A*1/2= 1/4 A
The probability of albinism in one of the child of heterozygous parents:
1/2*1/2= 1/4 A
Now from the above data the probability of a child not having albinism the diseases will be known by:
1-1/4 A
= 3/4 A
Similarly for sickle cell anemia the probability of not occurring the disease is also 3/4 S
Now applying the product rule for getting the probability of both the diseases not occurring.
3/4 A*3/4 S = 9/16
Here the two fraternal twins are in question, so the probability of both the children to be unaffected will be 9/16*9/16
=81/256
B. The chances for each child to inherit defective gene from each parent in case of Albinism .is A 1/2*1/2 = 1/4 A
Similarly same with sickle cell anaemia
1/4 S
applying the product law, we can determine the probability of the occurrence of both the diseases in one child
1/4*1/4
1/16
Hence in two children that are fraternal twins, the probability is
1/16*1/16
= 1/256
C. From the data, we can see that the probability of twin to be a carrier is 2/3
the chance of occurrence of both the diseases in one child is
2/3*2/3
=4/9
The chances of not carrying the gene of either disease is1/3*1/3
=1/9
Thus, we can know the probability that if it happens to be fraternal twins the chances of diseases
1-4/9+1/9
=4/9
D. we know that the probability of one fraternal twin carrying a gene for either disease:
chances of carrying the gene is 3/4
chances of not occurrence of gene 1/3
So, in case of fraternal twins 3/4*3/4= 9/16
1/3*1/3 = 1/9
So,1- 9/16+1/9
= 1/4
Using the principles of natural selection, __________ studies how behavior and the mind have evolved.
Answer:
Evolutionary psychology
Explanation:
Evolutionary psychology is the theoretical branch of psychology that describes that how the human behavior have evolved by the effect of evolution through a lens
Only body is not effected by evolution but the brain is also sculpted; the behavior it produce and psychological procedure.There were many psychological mechanisms that were involved in survival of species known as psychological adaptation.The psychological adaptations and their byproducts that are activated in modern environment are different from their ancestral environment in some features.Example;
Our brain insructs us to behave in an adaptive way.
Was the rate of increase of sucrase activity greater when sucrose concentration went from 2.5 to 7.5 g/l or when it went from 22.5 to 27.5 g/l?
Answer: It is greater when sucrose concentration went from 2.5 to 7.5g/l.
Explanation: The rate of reaction of an enzyme is known to be affected by the rate of concentration of its substrate, which in this case is the sucrose Solution.
If the rate of increase of concentration is high,the activities of the enzyme SUCRASE will increase accordingly, in order to breakdown the substrate.
The rate of increase of Sucrose from 2.5 to 7.5g/l is higher(300%) than the rate of Increase of Sucrose from 22.5 to 27.5g/l (1.22%). It is expected under circumstances that the action of SUCRASE will increase at a rate higher in the first Solution than in the second Solution.
The rate of increase in sucrase activity depends on the concentration of sucrose and whether or not the enzyme is saturated. The increase could be greater at lower concentrations (2.5 to 7.5 g/l) if sucrase is not yet saturated. The increase might be less at higher concentrations (22.5 to 27.5 g/l) if sucrase is near or at saturation point.
Explanation:The increase in sucrase activity is generally considered to be a response to the concentration of substrate present, in this case, sucrose. The increase in activity happens because more substrate (sucrose) is available for the enzyme (sucrase) to act upon. However, there is a limit to this increase. Once the enzyme is saturated with substrate, further increases in substrate concentration do not increase the enzyme's activity. This is known as the saturation point.
To determine whether sucrase activity increased more when sucrose concentration increased from 2.5 to 7.5 g/l or from 22.5 to 27.5 g/l, we would need specific data on the rate of sucrase activity at these different concentrations. It's possible that the increase from 2.5 to 7.5 g/l was greater if this is in the ascending portion of the enzyme activity curve and the sucrase was not yet saturated with sucrose. Conversely, the increase from 22.5 to 27.5 could be lesser if the sucrase is near or at saturation point.
Learn more about sucrase activitySickle cell anemia and albinism are both recessive traits in humans. Imagine that a couple, already pregnant with twins, has just learned that they are both heterozygous for both of these traits.
As the couple's genetic counselor, the couple asks you the following questions about how their carrier status will affect their offspring.
Part A
If the couple has fraternal twins, what is the probability that both children will be unaffected by both conditions?
Part B
If the couple has fraternal twins, what is the probability that both of the couple's children will have both sickle cell anemia and albinism?
Part C
What is the probability that one of the fraternal twins is a carrier of either, but not both, of the conditions?
(Hint: You will need to use both the product law and the sum law to answer this question.)
Part D
If the couple has fraternal twins, what is the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism?
(Hint: You will need to use both the product law and the sum law to answer this question.)
Answer:
A. 81/256
B. 1/256
C. 4/9
D. 1/4
Explanation:
A. Given,
Both parents to be heterozygous.
Therefore, the chances for one child to inherit a defective gene from each parent
= Â ½ * ½ =Â ¼.
In the same vein, the probability of being affected by albinism
=Probability of possessing a defective gene from each parent
= Â ½ * Â ½
= Â ¼
Now, The probability of not being affected is represented using the formulae
= 1- the probability ofbeing affected
= 1-¼ Â
= Â ¾.
Thus, the probability of not being affected by both albinism and sickle cell anemia
= Â ¾ * ¾
= Â (3 * 3)/(4 * 4)
= Â 9/16.
So,
The probability of neither twins to be affected
= 9/16 * 9/16
= (9 * 9)/(16 * 16)
=81/256.
B. We obtain the chances for a child to possess one defective gene from each parent
= Â ½ * ½
= Â ¼.
In the same vein, the probability of being affected by albinism
=Probability of inheriting one defective gene from each parent
= Â ½ * Â ½
= Â ¼
Now, the probability of being affected by both albinism and sickle cellanemia
= Â ¼ * Â ¼
= 1/16
Therefore, the probability of both twins to be affected
= 1/16 * 1/16
= 1/256.
C. Take, the probability of twin 1 to be a carrier for albinism to be = 2/3 (note, the homozygous recessive is affected, not the carrier).
We can say that the chances of twin 1 to be a carrier for both the diseases
= 2/3 * 2/3
= 4/9.
In like vein, the chances of carrying neither recessive allele
= 1/3 * 1/3
= 1/9.
Therefore, the probability that one of the fraternal twins is a carrier of either, but not both,
= 1- (4/9 + 1/9)
= 1 - (5/9)
= 4/9
D. Given, the probability of one fraternal twin carrying a gene for either disease:
Take;
The provability of carrying the gene to be 3/4
= 3/4 * 3/4
= 9/16
Thw chances of non occurrence of gene = 1/3
And for fraternal twins
= 1/3 * 1/3
= 1/9
Therefore, the probability of having two phenotypically normal children, one being a carrier of only the sickle cell anemia recessive allele, and the other being a carrier of only the recessive allele for albinism
= 1 - (9/16+1/9)
= 1/4.
Suppose two parents, a father with the genotype AaBbCcDdee and a mother with the genotype aabbCcDDEe, want to have children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of the offspring will have each of the specified characteristics?
Answer:
A. same genotype as the father (AaBbCcDdee) = (½ x ½ x ½ x ½ x ½) = 0.03
B. phenotypically resemble the father (A_B_C_D_ee) = (½ x ½ x ¾ x 1 x ½) = 0.09
C. same genotype as the mother (aabbCcDDEe) = (½ x ½ x ½ x ½ x ½) = 0.03
D. phenotypically resemble the mother (aabbC_D_E_) = (½ x ½ x ¾ x 1 x ½) = 0.09
E. phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1) = 1 – (0.09375 + 0.09375) = 0.81
Explanation:
The computation of the total probability would simply be the product of individual chances for each particular gene.
This can be better gotten with the punnette sqaure
The study of genes and inheritance is called genetics.
The correct answer is 0.81
What are genes?The structural and functional unit of the DNA is genes and able to code the sequence of the DNA.The answer is as follows:-
Same genotype as the father (AaBbCcDdee) =[tex]\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} [/tex] = 0.03Phenotypically resemble the father (A_B_C_D_ee) = [tex]\frac{1}{2}*\frac{1}{2}*1*\frac{3}{4}*\frac{1}{2} = 0.09[/tex]Same genotype as the mother (aabbCcDDEe) = [tex]\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2} [/tex]= 0.03phenotypically resemble the mother (aabbC_D_E_) =[tex]\frac{1}{2}*\frac{1}{2}* \frac{3}{4} *1*\frac{1}{2} = 0.09[/tex]Phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1)[tex] = 1 - (0.09375 + 0.09375) = 0.81[/tex]
Hence, the correct answer is 0.81
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g A chunk of tissue is treated so that each cell's membrane is broken open to release the contents inside, and then subjected to differential centrifugation. What is the end of the centrifugation
Answer: The end of the centrifugation is microsomal fraction
Explanation:
A chunk of tissue being treated for its cell's membrane to break and release the contents inside describes the process of HOMOGENIZATION.
After homogenization, the various cell components (nuclei, mitochondria, Microsomes etc) can be separated by step-by-step fashion of differential centrifugation based on their size
At low speed, nuclei fraction is collected due to its larger size
At high speed, mitochondrial fraction is collected
At the much higher speed (usually the end) microsomal fraction is collected due to its microscopic size
Angela wants to start a company developing apps. She needs access to R&D to be able to use the newest technologies. Of the conditions that need to be put in place for the Entrepreneurial Ecosystem, she needs ______
Answer:
Research and Development Transfer
Explanation:
she needs Research and Development Transfer, which involves the transferring of technology/skills from the individual or company that owns or holds it to another individual or company.
Explain why evolutionary biologists monitor selec- tively neutral polymorphisms as molecular clocks.
Answer:
Because of their property of preciseness and rhythmic repetitions like a normal clock sequence...
Explanation:
Evolutionary biologist use these neutral polymorphisms as molecular clock because of their gradual accumulation of mutations that remain on changing gradually in a uniform sequence overlarge number of generations. These changing sequences are highly precise over and over, that's why these are also termed as neutral mutations.
Select the true statements about protein secondary structure. a. The α‑helix is held together by hydrogen bonds between the amide N − H N−H and C = O C=O groups. b. In a β‑pleated sheet, the side chains are located between adjacent segments. c. In an α‑helix, the side chains are located on the outside of the helix. d. The secondary level of protein structure refers to the spatial arrangements of short segments of the protein. e. Peptide bonds stabilize secondary structure.
Answer:.
→The α‑helix is held together by hydrogen bonds between the amide N − H N−H and C = O C=O groups
→ In a β‑pleated sheet, the side chains are located between adjacent segments.
→ In an α‑helix, the side chains are located on the outside of the helix. .
→ The secondary level of protein structure refers to the spatial arrangements of short segments of the protein
Explanation:
This is the level of protein which results from spatial arrangement produced by the formation of hydrogen bonds between the oxygen atom of one carboxyl group(c=0) group, and hydrogen of the NH group of amino acids four places ahead of it .( The resulting structure is coiled and are therefore called alpha-helices)
AND
Hydrogen bonds between adjacent amino acids that join them side by side so that the bonds appear straight rather than coil, and the chains form upwards-downwards-upward- downwards format to form flat shaped structure called beta-pleated sheet.
The hydrogen bonding is due to strong polarities of the –NH- CO- groups of amino acids.
The two structures account for the spatial arrangement of secondary protein structure. Secondary structure is stabilized by the orientation and aggregation of these hydrogen bonds. . The outwards distributions of the side chains, the non-polar nature (hydrophobic) of alpha-helix makes some secondary proteins ideal as integral membrane proteins.
Note- peptide bond stabilizes primary protein structure.
The following statements are true for protein secondary structure:
The amide N-H and C=O groups form hydrogen bonds that hold the -helix together.In a helix, the side chains are found outside the helix.The secondary structure is stabilized by peptide bonds.Therefore, the correct options are A, C and E.
The polypeptide chain is regularly coiled to form a -helix, and hydrogen bonds between the amide NH and C=O groups help to stabilize the structure. Because they occur outside the helix and extend outward in a -helix, amino acid side chains enable interactions with the environment. The covalent bonds that link amino acids together, known as peptides, are essential for maintaining the secondary structure. Rotation is hindered by the planar structure of the peptide bond, which helps to form regular patterns such as the -helix.
Therefore, the correct options are A, C and E.
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Unconditioned negative reinforcers must be related to our inherited capacity to respond to them (for example, aversive, painful stimuli), and conditioned negative reinforcers must be stimuli that were originally neutral events that acquired their effects through previous pairing with existing negative reinforcers.
O True
O False
Answer: False
Explanation:
Negative reinforcers strengthen a behavior that avoids or removes a negative outcome.
Negative reinforcers are stimuli factors that influence behavior because individuals have capacity we inherited to respond to them and effects of these reinforcers have been established through records and histories of learning. Negative reinforcers are at the mercy of type of response acquired. Negative reinforcers are sources of negative reinforcements.
Unconditioned negative reinforcers
stimuli is one whose removal strengthens the choice behavior in absence of prior learning. Example includes the place of shock, loud noise, intense light.
Other times pain will occasion behavior and other response that eliminates discomfort will be reinforced.
Final answer:
True, unconditioned negative reinforcers are connected to our inherent responses to stimuli like pain, whereas conditioned negative reinforcers are learned through association with existing negative reinforcers. In operant conditioning, negative reinforcement increases the likelihood of a behavior's occurrence by removing an unpleasant stimulus after the behavior happens.
Explanation:
True, unconditioned negative reinforcers are related to our inherent responses to stimuli such as pain, and these unconditioned responses do not require learning. On the other hand, conditioned negative reinforcers are originally neutral but become effective through association with existing negative reinforcers. Consider the scratch response to an itchy bug bite: scratching (the response) leads to the removal of the itch (the aversive stimulus), thus negatively reinforcing the behavior of scratching. Punishment, however, whether positive (adding an undesirable stimulus) or negative (removing a desirable stimulus), always serves to decrease a behavior.
Operant conditioning is the learning process through which the strength of a behavior is modified by reinforcement or punishment. B.F. Skinner, a renowned psychologist, distinguished between reinforcement and punishment, and further between the positive and negative forms of these phenomena. A stimulus that serves as a negative reinforcer for someone may not serve the same function for another, highlighting the subjectivity of these responses.
Four of the five factors below support the mechanism of natural selection. In contrast, the fifth factor blocks natural selection from occurring. Which one of the factors listed below does NOT lead to natural selection, but instead blocks it from occurring?
a. Characteristics are inherited passed from parent to offspring.
b. Organisms in a population are identical to each other.
c. There is competition among organisms in a population.
d. More offspring are born than can survive.
e. Organisms in a population vary in characteristics that affect how well they survive and reproduce.
Answer:
Four of the five factors below support the mechanism of natural selection. In contrast, the fifth factor blocks natural selection from occurring. Which one of the factors listed below does NOT lead to natural selection, but instead blocks it from occurring
Organisms in a population are identical to each other
Explanation:
If organisms are identical then all of them should be able to survive harsh condition but since not all organisms has this identity it makes the law of natural selection not to be supported by the claim. The best organism that thrives survives and pass such trait to the next generation of offspring
Tadpoles raised in water with atrazine levels of 0.1 ppb should produce a higher percentage of male frogs with gonadal abnormalities than those raised in pure water.
This statement is an example of:
A. a question leading to a hypothesis
B. a hypothesis
C. a testable prediction leading to design of an experiment
D. data from an experiment
E. an interpretation of data
Answer:
Option C, a testable prediction leading to design of an experiment
Explanation:
The statement here predicts the outcomes of raising tadpoples is water with atrazine levels of 0.1 ppb as compared to the ones raised in pure water. Hence, it cannot be question. Now since this prediction can be tested, an experiment can be designed where a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with 0.1ppb of atrazine level. The difference in two populations can be then compared to either support the prediction or contradict it.
Hence, option C is correct
Non-segmentation allows for evolutionary innovation in body form.
a. True
b. False
Answer:
True.
Explanation:
Answer: a. True
Explanation: Through evolutionary time, animals have developed more-complex body plans, including true tissues, non-segmentation and bilateral symmetry which is possible as a result of evolutionary innovation--the introduction and progression of novel traits compared to what exists before leading to a more advanced or complex form. Non-segmentation indeed allows for evolutionary innovation in body form.