Answers
Answer:
According to biologists the coelacanth is NOT the closest living relative to amphibians,
Explanation:
In this modern era of molecular biology there is evidence that the coelacanth and tetrapods are not closely related. While in the other hand, evidences indicated a close relationship between lungfishes and tetrapods. The molecular analysis was based on mitochondrial DNA sequences. I have attached picture of evolutionary relationship.
FIG. 1. Alternative hypotheses of sister group relationships between sarcopterygii and tetrapods.
(A) Lungfish as the sister group of
tetrapods.
(B) Coelacanth as the closest living relative of tetrapods.
(C) Coelacanth and lungfish equally closely related as sister groups of tetrapods.
Reference: Zardoya, R., & Meyer, A. (1996). Evolutionary relationships of the coelacanth, lungfishes, and tetrapods based on the 28S ribosomal RNA gene. Proceedings of the National Academy of Sciences, 93(11), 5449-5454.
Answer:
No, it doesn't support
Explanation:
The DNA based alternative hypothesis shows that the lung fish is actually the closest relative to amphibians rather than the coelacanth.
Related Questions
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 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.)
Answers
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.
which of the followings are true about V0 and Vmax? A. The unit for each of them is M B. Vmax is a special V0 when all enzymes bind substrate C. Vmax is independent of enzyme concentration D. V0 can be determined using a linear correlation of product and time in the beginning of a reaction
Answers
Answer:
The correct answer is option B.
Explanation:
The unit of Vmax and Vo is moles per second. Km refers to the concentration of the substrate, which is needed to attain the maximum reaction velocity. In case, when [S] is far greater in comparison to Km then Vo will be close to Vmax.
Vmax relies upon the concentration of the enzyme, Vmax enhances with the concentration of the enzyme and becomes steady when all the active sites of enzymes get occupied. Vo increases with the enhancement in the concentration of the substrate with time. Thus, the correct answer is option B, that is, Vmax is a special Vo when all the enzymes combine with the substrate.
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
Answers
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.
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
Answers
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
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.
Answers
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
during the course of development, the phenotype interacts with a environment to produce the genotype true of false
Answers
Answer:
True
Explanation:
Development in general involves the interaction between a genotype and its environment. For example, a organism born and raised in a cold climate likely will exhibit phenotipic features closest associated with such climatic conditions compared than other from the same species born and raised in a temperate region. It means that genes differentially expressed in different conditions can influence organismal development, although both organisms have potential to express the same genes
The statement is false. Genotype, a set of genes in an organism's DNA, interacts with the environment to influence phenotype, the physical expression of those genes. Environmental factors can modify phenotypic expression, demonstrating a reciprocal interaction between genotype and environment.
Explanation:The statement provided is false. During the course of development, the genotype - the set of genes in an organism's DNA - influences the organism's traits or phenotype. However, the environment can interact with the genotype, modifying the way these traits are expressed.
For instance, in Mendel's hybridization experiments, true-breeding plants with yellow pods and green pods produced hybrid offspring with yellow pods - the same color as one parent (phenotypically identical) but with a different genetic makeup (genotypically different).
Phenotypes and genotypes are interconnected, with the genotype serving as a blueprint, subject to environmental factors that can influence phenotypic expression. An example of environmental influence is sun exposure affecting skin color or temperature affecting sex determination in certain reptilian species. In essence, there is a reciprocal interaction between the genotype and the environment to determine the phenotype - not the other way around.
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List the sources of experimental uncertainty. List steps you will take to minimize the uncertainties.
Answers
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.
Generally, membrane filters are a used to remove bacteria from liquids. b are made of microscopic pores c used to remove spoilage agents from alcoholic beverages. d all the above
Answers
Answer:All of the above
Explanation:
Membrane filters are very thin, highly porous media composed of foamed and/or stretched polymeric compounds. Because of their homogeneous structure they cannot contaminate the filtrate with fibres or particles from the membrane matrix.
Microfiltration membranes retain particles according to the size of the pore, and the affinity of the filter materials for the solute. While materials are held largely on the surface of the membrane, they can also retain within the matrix itself, as in depth filtration. Unlike depth filters, however, which bind solutes of nominal size ranges only, membrane filters retain particles with absolute accuracy due to their controlled, predetermined pore size. Some membranes also demonstrate retention of specific substances as for example in the case nitrocellulose membranes which bind proteins and nucleic acids in high concentration. This feature may be an advantage for some applications where exclusion based on chemical properties rather than porosity alone is required.
Membrane filters for microfiltration are used primarily for separating particulate materials or micro-organisms, larger than the rated pore size, from gases or liquids. On filtration of gases particles are also retained which are smaller than the pore size rated. Also they show a low particle capacity per unit area as compared to a depth filter. For some filtration applications it may be advisable to prefilter through a depth filter, eg, a glass fiber filter, prior to microporous filtration to prevent clogging. This may be necessary, because the solution is highly viscous, contain molecules which may precipitate in the membrane filter or are heavily contaminated by microorganisms and particulates.
This separation is carried out either for the purpose of cleaning to obtain highly pure or sterile products or recovering (concentrating) these materials in order to carry out further chemical, microscopic, microbiological or other analyses on the separated sample.
Typical applications include Sterile filtration
–Dialysis
–Fluid clarification/purification
–Gas filtration/particle control
–Microbiological investigations
–HPLC solvent filtration and sample preparation
In this exercise, you will research the effects of excess greenhouse gases in the atmosphere. You will apply the scienbific method and m ow greenhouse gases warm the planet by covering a jar with plastic wrap. You will then relate the findings of your experiment to global warming Procedure
In Exercise 1, we observed that gases have different properties In this section we will further investigate gases. How are greenhouse gases related to global warming?
Answers
Answer:
By heat trap mechanisms...
Explanation:
The main green house gases in our atmosphere are Carbon dioxide, Nitrous oxide, water vapors and ozone. All of these gases have a property of heat trap that trap the heat waves coming from the sun and heat up our atmosphere as a result. This is the reason why Venus is farther from sun but is more hotter than mercury because of it's dense Carbon Dioxide atmosphere.
How do you plant a lavender? What are the best conditions to grow it
Answers
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.
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.)
Answers
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
The house fly, Musca domestica, has a haploid chromosome number of 6. How many chromatids should be present in a diploid, somatic, metaphase cell?
Answers
Answer:
12
12
24
Explanation:
Musca domestica is the scientific name of an animal popularly known as Housefly. From the question, it is known that this organism has a haploid chromosome number of 6.
In a diploid chromosomes, the number of chromatids that will be present will be twice of that in haploid.
Since haploid (n) = 6, diploid (2n) will be; 2 × 6 = 12
Somatic cells on the other hand also exhibit diploid number of chromosomes, again it means we will have (2n) = 2 × 6 = 12
In metaphase cell, cells do make sure they complete the S'phase of the cell cycle before cellular division.Therefore, the DNA present in the chromatids in the S'phase actively engage in chromatids doubling, as a result, 24 chromatids (i.e 12 × 2= 24) exists in the metaphase cell.
The chromatids should be present in a diploid, somatic, metaphase cell-
diploid - 12somatic - 12metaphase cell- 24Chromatids at different stageThe scientific name of the fly Housefly is Musca domestica. It is known that this organism has a haploid chromosome number of 6. In the diploid cells, the number of chromatids that will be present will be twice that in haploid.
Diploid genome (2n)
= [tex]2\times6=12[/tex].
Somatic cells also exhibit diploidy (2n)
= 12
Metaphase stage cells contain 24 chromatids as cells before entering cell division complete S-phase of the cell cycle in which DNA present in chromatids become double due to replication.
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What phenotypic ratio would you expect as a result of a test cross between two individuals where 9) one that is homozygous recessive for alleles at two independent loci?
Answers
What phenotypic ratio would you expect as a result of a test cross between a dihybrid organism and one that is homozygous recessive for alleles at two independent loci?
a. 3:1
b. 9:3:3:1
c. 1:1:1:1
d. 1:2:1
e. 9:4:2:1
Answer:
c. 1:1:1:1
Explanation:
When a heterozygous individual for two genes is test crossed with a double homozygous recessive individual, the progeny is obtained in 1:1:1:1 phenotypic ratio. This occurs as the heterozygous dominant individual forms four types of gametes in 1:1:1:1 ratio while the homozygous recessive individual would form only one type of gamete having one recessive allele for each gene.
For example, a test cross between TtRr (tall and red) and ttrr (short and white) would produce a progeny in following ratio=
1 tall, red: 1 tall, white: 1 short, red: 1 short, white
Here, T= tall, t= short, R= red, r= white
Final answer:
The genotypic ratio expected from the mating of two individuals heterozygous for a recessive lethal allele expressed in utero is 2:1 (homozygous dominant: heterozygous) among surviving offspring because the homozygous recessive genotype results in death before birth.
Explanation:
When two individuals that are heterozygous for a recessive lethal allele expressed in utero are mated, the expected genotypic ratio (homozygous dominant:heterozygous: homozygous recessive) among the offspring would be 2:1. This is because the homozygous recessive genotype, which would normally make up one-quarter of the offspring, results in death before birth due to the lethal allele. Therefore, the live offspring genotypic ratio becomes 1:2:0, as the homozygous recessive individuals do not survive. This contrasts with a typical Mendelian 3:1 phenotypic ratio observed in monohybrid crosses of non-lethal traits. The Punnett square for such a cross shows that out of the four possible genotypes, one (homozygous recessive) does not survive, leaving a genotypic ratio of homozygous dominant:heterozygous at 1:2 among surviving offspring.
You decide to designate the twist allele as FT to distinguish it from the forked allele F. Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels can be used once, more than once, or not at all.
Answers
Image attached
Answer:
FTFT, F, FFT (in order left to right)
Explanation:
The twist allele is FT, the forked allele is F. We are told there are pure lines, so this means they are homozygous. That means the parents are FF x FTFT.
The F1 generation is both twisted and forked (as can be seen from the image), suggesting the alleles are codominant (both are expressed).
In the F2, there are three different types of flowers, 2 matching the parental and 1 matching the F1 twisted, forked, and both.
The order from left to right is twisted, forked both. We know twisted is the genotype FTFT, and forked is the genotype FF. The both phenotype would have a copy of each allele, so would be FFT
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.
Answers
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.
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
Answers
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.
A researcher conducts an experiment on the secretion of a particular hormone in mice. Scientists inject mice with a substance that stimulates the production of the hormone. The scientists then test the levels of hormones produced by the mice. The tool used to measure the hormones consistently detects the levels at 10 points lower than the actual hormone levels in the mice. This tool makes______ measurements, but the measurements aren’t _____.
first blank options:
1.)qualitative
2.)Reliable
3.)Valid
second blank options
1.)quantitative
2.)reliable
3.)valid
Answers
The first blank should be filled with Reliable, and the second blank should be filled with Valid. Therefore option 2 and 3 is correct.
The tool used to measure the hormones is reliable because it consistently produces consistent results, albeit with a consistent offset of 10 points lower. The reliability aspect pertains to the consistency and stability of the measurements.
However, the measurements from the tool are not valid because they don't accurately reflect the true hormone levels in the mice. Validity refers to the extent to which a measurement accurately measures what it's intended to measure.
In this case, the tool's measurements are consistently inaccurate by a fixed amount, which means they lack validity as they don't represent the actual hormone levels in the mice.
Therefore option 2 and 3 is correct.
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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)
Answers
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.
is rectus femoris NOT a muscle of the the medial forelimb of the fetal pig that you will be dissecting?
Answers
Answer:
Yes it in not a muscle of the medial forelimb of the fetal pig
Explanation:
Rectus femoris is not the part of forelimb of the fetal pig.
However, rectus femoris in fetal pig originates from the hip and that to through the ilium and extends towards the knee and insert into its tendon. These are responsible for flexing the hip and extension of lower knee and legs. This muscle is a part of muscles of thigh and lies at its top. In quadriceps these muscles are the most anterior muscles.
Answer: Rectus femoris is not a muscle of medial forelimb of the fetal pig.
Explanation:
Pigs are mammals like humans. Pigs have similar structures with humans but there are slight differences in these structures because pigs are quipedal and humans are bipedal.
Rectus femoris is not found in the medial forelimb of fetal pigs instead it is found in the medial hindlimbs of fetal pigs and is one of the quadriceps muscles. It starts from the hip through the ilium and extend to the knee.Rectus femoris function in knee extension, its extend the lower leg and it flexes the hip.
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?
Answers
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!
Alleles of the gene that determines seed coat patterns in lentils can be organized in a dominance series:Marbled > Spotted = Dotted > ClearA lentil plant homozygous for the marbled seed coat pattern allele was crossed to one homozygous for the spotted pattern allele. In another cross, a homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross was then mated to an F1 plant from the second cross.What are the expected phenotypes of the F1 plants from the two original crosses?
Answers
Answer:
first original cross = 100 % marbled seed coat lentils
second original cross = 100% dotted seed coat lentils
Explanation:
Given,
Marbled > Spotted = Dotted > Clear ( Spotted and dotted alleles are codominant )
Let, marble = m
spotted = s
dotted = d
clear = c
First original cross: mm * ss :
m m
s ms ms
s ms ms
Entire progeny has ms genotype. Since m is dominant over s, result will be 100% marbled seed coat lentils.
Second original cross : dd * cc :
d d
c dc dc
c dc dc
Entire progeny has dc genotype. Since d is dominant over c, result will be 100 % dotted seed coat lentils.
You are given a metaphase chromosome preparation (a slide) from an unknown organism that contains 12 chromosomes. Two that are clearly smaller than the rest appear identical in length and centromere placement.
What would most likely be true of these two chromosomes? Select all that apply.
They have similar banding patterns.
They contain identical genetic information.
They would replicate synchronously during the S phase of the cell cycle.
They are homologous chromosomes.
Answers
Answer:
True answers are as follow:
a. They have similar banding patterns.
c. They would replicate synchronously during the S phase of the cell cycle.
d. They are homologous chromosomes.
Explanation:
In Mitosis, chromosomes condense and align in the center before moving to each opposite pole is called meta-phase. (See attached picture)
In this stage you will observe similar banding pattern of homologous chromosomes that were replicated during S phase of cell cycle.
The two smaller, identical chromosomes most likely are homologous chromosomes, containing the same gene order and potentially identical genetic information. They would replicate together during the S phase of cell cycle.
Explanation:The two chromosomes that are smaller and identical in length and centromere placement are likely to be homologous chromosomes. As such, they would have similar banding patterns because they have the same order of genes. They would contain identical genetic information only if there hasn't been any genetic recombination during meiosis. The replication of these chromosomes would occur synchronously during the S phase of the cell cycle as DNA replication happens to all chromosomes of a cell at the same time.
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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 ______
Answers
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.
Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 18 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities. (a) No samples are mutated. (b) At most one sample is mutated. (c) More than half the samples are mutated.
Answers
Answer:
a) [tex]0.58[/tex]
b) [tex]0.598[/tex]
c) [tex]0[/tex]
Explanation:
Given -
Total sample i.e n [tex]= 18[/tex]
Probability (p) [tex]= 3[/tex] % [tex]= 0.03[/tex]
We will use binomial distribution theory for determining the probability of mutated sample
Let X be the number of mutated sample
a) No samples are mutated i.e [tex]X = 0[/tex]
[tex]P(X=0) = 0.03^0 * 0.97^{18}\\= 0.5779 = 0.58\\[/tex]
[tex]0.58[/tex]
b) At most one sample is mutated
[tex]P(X=0) = 0.58 + 0.03^1 * 0.97^{17}\\= 0.598[/tex]
c) More than half the samples are mutated.
[tex]P(X = 10) + ........+ P(X = 18) = 0[/tex]
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.
Answers
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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Using the principles of natural selection, __________ studies how behavior and the mind have evolved.
Answers
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.
. Given that 3 of the 64 possible codons are stop codons, what is the chance of having a stop codon at any given position, assuming that the sequence is random?
Answers
Answer:
[tex]1 - (\frac{61}{64})^n[/tex]
Explanation:
Given
There are three stop codon
TAA, TAG, TGA.
Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.
So the probability of choosing a non stop codon is
[tex]\frac{61}{64}[/tex]
Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is
[tex](\frac{61}{64})^x\\[/tex]
Probability of choosing a stop codon is equal to
[tex]1 -[/tex] probability of choosing a non stop codon
[tex]1 - (\frac{61}{64})^n[/tex]
The chance of having a stop codon at any given position in a random sequence of codons can be calculated by dividing the number of stop codons by the total number of possible codons.
Explanation:In a given sequence of codons, the chance of having a stop codon at any given position can be calculated by dividing the number of stop codons by the total number of possible codons. The number of stop codons is given as 3, and the total number of possible codons is 64. Therefore, the chance of having a stop codon at any given position is 3/64, or about 0.046875, assuming that the sequence is random.
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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.
Answers
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.
Why do bacteria usually contain an even number of replisomes? Choose one: A. Bacteria need only one replisome; the other serves as a backup copy. B. Because replication is bidirectional, there are two replication forks emanating from the origin of replication. C. There are two copies of the gene that codes for DNA polymerase, an important enzymatic protein component of replisomes. D. DNA is double-stranded and each single strand requires its own replisome.
Answers
Answer: Option B.
Because replication is bidirectional, there are two replication forks emanating from the origin of replication.
Explanation:
Replisomes are complexes that involve in DNA replication. Replisomes consist of many proteins like ligase, helicase,topoisomerase,DNA polymerase III, e.t.c. The replisomes first unwind double stranded DNA to single strands DNA.
In prokaryotes,the nucleoid divides and requires two replisomes for bidirectional replication. Bacteria normally have even number of chromosome because the replication is bidirectional, and there are two forks which emanates from the origin of replication. The replication forks duplicate both the leading and the lagging strands.
The major function of replisomes is to carryout DNA replication.
Bacteria usually contain an even number of replisomes because DNA replication in bacteria is bidirectional, emanating from a single origin but forming two replication forks, each with its replisome for DNA synthesis (Option B),
Explanation:The correct answer to your question, 'Why do bacteria usually contain an even number of replisomes?' is B. Because replication is bidirectional, there are two replication forks emanating from the origin of replication. For a deeper understanding, let's explain it. In bacteria, the DNA replication starts at a single origin of replication but proceeds bidirectionally, creating two replication forks.
The enzyme helicase unzips the double helix at the origin of replication forming a replication fork. Each of these forks has a complex of proteins called replisome that is required for DNA synthesis. The replisome is responsible for copying the entire bacterial chromosome from the origin of replication to the terminus.
The synthesis is semiconservative where one strand—called the leading strand—is synthesized continuously in the direction of the replication fork, and the second strand, called the lagging strand, is synthesized discontinuously, away from the replication fork, creating Okazaki fragments. Thus, because of this bidirectional nature, there are always two replisomes in prokaryotic DNA Replication, hence an even number of replisomes.
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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
Answers
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)