Answer/Explanation:
An important way of selecting for bacteria carrying specific recombinant DNA is to add a resistance gene (such as antibiotic resistance) to your gene of interest. That means that any bacteria that carry your DNA will be resistant to a specific antibiotic (in this case streptomycin). Therefore, you can grow your bacteria on plates with streptomycin, and in theory, the bacteria will be unable to grow if they don't have the resistance gene and your gene of interest.
Here, strain A does not grow on streptomycin plates, but strain B now has 100 colonies. This suggests strain B can grow on streptomycin, meaning it is resistant.
Strain B is therefore the streptomycin resistant master plate strain.
Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Which of the following is not true regarding the denaturation and reannealing of double-stranded DNA molecules? Decreasing the salt concentration of the solution lowers DNA's melting point (Tm). Increasing the G-C content of DNA raises its melting point (Tm). Single-stranded DNAs can only anneal to one another if they are 100% identical in nucleotide sequence. Melting point (Tm) is the temperature at which the DNA is one-half double-stranded and one-half single-stranded. DNA is more likely to denature when exposed to a high pH.
Answer: "Decreasing the salt concentration of the solution lowers DNA's melting point (Tm)" is not a true statement
Explanation:
Increasing salt concentration would lower the DNA's melting point (Tm), not otherwise.
For instance:
- In 8M urea (8M means 8 Moles per dm3), Tm is decreased by nearly 20°C.
- 95% formamide at room temperature would completely denature the double stranded DNA.
Thus, higher concentration of salts like urea or formamide lowers Tm, not otherwise
For which one of the following observations were both Lamarck's hypothesis and Darwin's hypothesis in complete agreement?
a. More complex species are descended from less complex species.
b. Gradual evolutionary change explains why organisms are well-suited to their environments.
c. Acquired characteristics are inherited.
d. Use and disuse of organs determines their size in progeny.
Answer:
b. Gradual evolutionary change explains why organisms are well-suited to their environments.
Explanation:
Darwin and Larmack are both evolutionists who believed that the present organisms or offspring are descendants of previous organisms. And they are able to survive till today from previous generation because they have certain features that make them resistant to selective pressure in their environment. Lamarck called concluded that changes in the physical features of the present organisms from using a certain body part more than others in order to survive is the major reason for their existence and resistant to selective pressure. He called his theory Law of use and disuse .
Darwin believed that, the ability to survive by theses organisms was due to certain factor or trait(gene) which the descendants must have inherited from the previous generations,. And these trait(gene) made them to resist selection pressures,and therefore have higher competition for survival than other organisms which lack these traits. Thus he concluded that nature must have selected these organisms with certain traits(genes) than others, He tagged his findings theory if evolution by natural selection.
However both scientists believed that present organisms are products of Gradual evolutionary changes of million of years ago.
Answer: Option B.
Gradual evolutionary chnges explain why organisms are well suited for their environment.
Explanation:
Lamarck and Darwin are both naturalist that formulate theories. Lamarck theory was called theory of acquired characteristics and Darwin theory was called theory of evolution by natural selection. The two scientist agree that gradual evolutionary changes I.e change in trait and characteristics explain why organisms are well suited in their environment. Lamarck theory describe changes that happen to organisms in their life time and how these changes are passed to their offsprings and Darwin theory supported these that organisms arise from natural selection and inherited variations which help them survive and reproduce in their environment.
Principles of Ecology: Part 4 FANR/MARS 1100 Law of Tolerance Survival depends on _________________________ of many factors, which can vary greatly. For each factor a given species has a ______________ of __________________________. If environmental conditions exceed upper or lower limitof tolerance, deathcan result.
Answer:
Survival depends on interaction of many factors, which can vary greatly. For each factor a given species has a range of conditions. If environmental conditions exceed upper or lower limit of tolerance, death can result.
Explanation:
This is the Shelford's law of Tolerance.
This law states that the success of distribution of an organism can be controlled by some factors like topographic, climate, and biological requirements of plants and animals, in which their levels exceed the higher or lower limits of tolerance of the organism.
This was a law proposed by V. E. Shelford in 1911.
An organism is classified as a heterotroph or autotroph based on the type of ________ it utilizes.. A. respiration source (e.g., oxygen or other) B. nitrogen source C. carbon source D. none of the above
Answer: Option C) carbon source
Explanation:
The type of Carbon source is the distinction between heterotroph or autotroph. As autotrophs like green plants generate their food (glucose) from simple inorganic molecules like carbon dioxide (CO2) in the atmosphere, while heterotroph like man utilize carbon in carbohydrates such as starch that have been stored in the tissues of autotrophs.
Thus, the type of carbon source utilized distinguished heterotroph and autotrophs
Fungi can be identified at the genus and species level via visualization of their nucleus under a transmission electron microscope. Group of answer choices True False
Final answer:
The statement is false because fungi are not typically identified at the genus and species level by visualization of their nucleus under a transmission electron microscope, but rather by morphological characteristics and other methods.
Explanation:
The correct answer to the question is False. While fungi are eukaryotic organisms that do contain a nucleus, transmission electron microscopy of a nucleus is not typically used for identifying fungi down to the genus and species level. In practice, fungi are more commonly identified by their morphological characteristics, which can be observed under lower magnification microscopy, along with genetic sequencing and biochemical tests. For instance, the morphology of hyphae and the structure of mycelium, along with other characteristics like reproductive structures and spore types, are key attributes used in the identification of fungi. Transmission electron microscopy might be used for detailed cellular studies but not typically for standard identification at the genus and species level.
A population is made up of individuals where 104 have the A1A1 genotype, 55 have the A1A2 genotype, and 37 have the A2A2 genotype. What is the allele frequency of A1? Answer to 2 decimal places.
Answer:
0.67
Explanation:
In this population there are:
104 A1A1 individuals55 A1A2 individuals37 A2A2 individualsN: 196
The frequency of the A1 is the amount of times that allele appears divided by the total number of alleles in the population.
The A1 allele is present twice in the A1A1 individuals and once in the A1A2. Each individual has 2 alleles, so the total number of alleles is N × 2.The frequency of the A1 allele can be thus calculated using the following formula:
[tex]freq(A1)=\frac{2*A1A1\ + A1A2}{2*N} \\\\freq(A1)=\frac{2*104\ + 55}{2*196} \\\\\\freq(A1)= 0.67[/tex]
A dehydration reaction starting with 3.1 g cyclohexanol produces 2.2 g cyclohexene. Calculate the theoretical yield for this reaction. Report your answer with two significant figures.
Answer:
2.54 g
Explanation:
Equation for the reaction was followed by the dehydration of:
Cyclohexanol ----> Cyclohexene
Given that:
mass of cyclohexanol = 3.1 g
mass of cyclohexene = 2.2 g &
the standard molar mass of cyclohexanol is known to be = 100.16 g/mol
Also, the standard molar mass for cyclohexene = 82.14 g/mol
From what we have above, we can calculate for their respective numbers of moles;
By the formula which say;
number of moles = [tex]\frac{mass(g)}{molar mass}[/tex]
number of moles of cyclohexanol = [tex]\frac{3.1(g)}{100.16g/mol}[/tex]
= 0.03095 moles
From the equation for the dehydration reaction above;
one mole of cyclohexanol gives one mole of cyclohexene
∴ 0.03095 moles of cyclohexanol is said to be equal to 0.03095 moles of cyclohexene
Thus, having gotten that; we can proceed to calculate for our theoretical yield of cyclohexene.
Theoretical yield of cyclohexene = (number of moles × molar mass) of cyclohexene
= (0.03095 g × 82.14 g/mol)
= 2.542233 g of cyclohexene
≅ 2.54 g to two significant figures.
Following the balanced equation and converting g to mol and back, we find that the theoretical yield for the dehydration reaction of cyclohexanol to cyclohexene is 2.54 g.
Explanation:Firstly, we need to calculate the molar masses of cyclohexanol and cyclohexene. The molar mass of cyclohexanol (C6H12O) is approximately 100.16 g/mol while the molar mass of cyclohexene (C6H10) is about 82.14 g/mol. We then convert the masses of cyclohexanol and cyclohexene into moles. This is done by dividing the given mass of each compound by its respective molar mass. So, for cy Convert this theoretical yield in moles back to grams by multiplying by the molar mass of cyclohexene. So, the theoretical yield is 0.0309 mol x 82.14 g/mol = 2.54 g.
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An elephant had a long, powerful trunk. According to the ideas of Lamarck, how did the trait of long, powerful trunks develop develop in elephants
Answer:
D
Explanation:
Lamark believed in the inheritance of acquired characteristics. That is, that if an individual's body or organs changed in some way during the course of their life time (due to the environment or some other pressure), that these characteristics could be passed on to the next generation.
This was his theory of evolution. Therefore, if, slowly elephants started using their trunks to an advantage more, they would start to develop and grow. Lamark believed these acquired traits would be passed on to the next generation. We now know this not to be true, and random, selectively advantageous mutations in DNA are what cause evolution over time.
Cells of a normally rod-shaped bacterium (e.g., Bacillus subtilis) that have completely lost the ability to produce the MreB protein would mostly likely be Choose one: A. nonflagellated. B. stalked (like Caulobacter). C. coccoid-shaped. D. filamentous in form. E. unable to divide symmetrically.
Answer:
C. coccoid-shaped.
Explanation:
MreB protein serves as function as actin proteins do in the eukaryotic cells. MreB protein is involved in maintaining the cell shape. It is observed in many rod-shaped bacteria and archaea such as Bacillus subtilis. MreB polymerizes to form a spiral around the inside periphery of the cell and maintain the elongated shape of the cell. A bacterium that is not able to produce functional MreB protein cannot maintain the rod shape of its cell and is coccoid-shaped.
What would you expect the F1 generation flies to look like if two scarletmutants were crossed with each other?
Answer:
complementation test
Explanation:
The easiest examination to differentiate between the two possibilities is the complementation test. The test is simple to perform: two mutants cross and F1 is analyzed. If F1 expresses the wild-type phenotype, we conclude that each mutation is in one of two possible genes necessary for the wild-type phenotype. When it is shown that it is genetically shown that two (or more) genes control a phenotype, the genes are said to form a complementation group. Otherwise, if F1 does not express the wild type phenotype, but rather a mutant phenotype, we conclude that both mutations occur in the same gene.
These two results can be explained considering the importance of genes for phenotypic function. If two separate genes are involved, each mutant will have an injury to one gene while maintaining a wild-type copy of the second gene. When F1 occurs, it will express the mutant allele of gene A and the wild-type allele of gene B (each contributed by one of the mutant parents). F1 will also express the wild type allele for gene A and the mutant allele for gene B (contributed by the other mutant parent). Because F1 is expressing the two necessary wild-type alleles, the wild-type phenotype is observed.
Which of the following statements are accurate? (MORE THAN ONE ANSWER)
A. Invertebrates are animals that lack a backbone
B. Ecdysozoans typically have an exoskeleton
C. Bilaterians include three clades:
1) Deuterstomia
2) Lophotrochozoa
3) Ecdysozoa
D. Givens about animal "A'
I. the animal has three prominent germ layers
II. an animal has bilateral symmetry
III. the animal lacks a backbone
You would be able to classify this animal, animal A, as a bilaterian
E. All animals share a common ancestor
Answer:
The statements in options A,B,C,D and E are all correct.
A. Invertebrates lack a back bone
B. Ecdysozoans typically posses a flexible exoskeleton that protects these animals from harmful external factors like water loss.
C. Deuterstomia, Lophotrochozoa and Ecdysozoa are classes of Bilaterians.
D. Animal A is a bilateral because bilateral embryos are triploblastic possessing three germ layers, are bilaterally symmetrical and lack a back bone.
E. All animal from studies are found to have a common ancestry.
Early twentieth century Russia was in great turmoil as the Bolshevik Revolution ended 300 years of monarchical rule. Tsar Nicholas Romanov, his wife, Tsarina Alexandra, and their five children were placed under house arrest, but disappeared in the summer of 1918. Rumors abounded about their possible execution or escape from Russia.
Two years later, a young woman named Anna Anderson claimed that she was Anastasia, the youngest daughter of Tsar Nicholas II. There was much controversy surrounding Anna's claim. Some believed she was a fraud, others that she was indeed the lost Princess Anastasia. Anna remained steadfast in her claim until her death in 1984.
In 1991, the remains of nine skeletons, five male and four female, were exhumed from a shallow grave east of Moscow. Evidence from nuclear DNA showed that three of the young women were related and that one of the men and one of the women were their parents.
Further evidence from mitochondrial DNA (mtDNA) showed that one of the women could be positively identified as Tsarina Alexandra and that one of the men was indeed Tsar Nicholas II. The other three women had mtDNA that matched that of the Tsarina's and were identified as three of the Tsar's children. Anastasia and her younger brother, Alexei, were not among those found in the grave. This led to further speculation about Anastasia's possible escape from Russia after her parents were killed.
In 2007, an additional grave was found near that exhumed in 1991. It contained the remains of a teenage girl and boy. These remains were mtDNA tested, as well as samples from Anna Anderson, uncovered from hospital storage long after her death.
Analysis of the mtDNA supported the hypothesis that Princess Anastasia had been killed with her family in 1918. Which of the following statements support this hypothesis? Select all that apply.
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra.
b. The mtDNA from the remains of the teenage girl matched that of Tsar Nicholas II.
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
d. The mtDNA from Anna Anderson matched that of Tsarina Alexandra.
e. The mtDNA of the teenage girl and boy matched each other.
Answer:
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
Explanation:
Mitochondria contain their own DNA known as mtDNA, a strand which is inherited maternally. mtDNA can be used to find maternal ancestry and cannot be passed down by a paternal parent.
This information provides evidence given that the remains of the teenage girl had mtDNA that was identical to the mother, Tsarina Alexandra. It also gave evidence as to why Anna Anderson's claim of being Anastasia were false as the mtDNA of her and Tsarina Alexandra were not similar.
As to why the other answers cannot support the hypothesis, mtDNA is a maternally passed-down strand making the father, Tsar Nicholas Romanov, from having identical mtDNA to his children. On the other hand, if the teenage girl were to not have matching mtDNA to that of Tsuarina's mtDNA, this would disprove the theory of the girl being related to her.
While the teenage boy also had matching mtDNA with the girl, this only proves that the two were related to one another but not necessarily related to Tsarina unless the mtDNA of all three were proven identical.
The statements supporting the hypothesis concerning the killing of Princess Anastasia with her family in 1918 are stated below:
a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra.
c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.
e. The mtDNA of the teenage girl and boy matched each other.
Clearly, these statements negated Anna Anderson's claim of being the youngest daughter of Tsar Nicholas II.
The matching of Anastasia's mtDNA with Alexei's mtDNA in 2007 confirmed that Tsarina Alexandria was indeed their mother. Recall that teenagers were not among the first nine skeletons exhumed in 1991.
Thus, the 2007 discovery unequivocally confirmed the hypothesis that Princess Anastasia was killed alongside her family in 1918 during the Russian revolution.
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Two linked genes, A and B, are separated by 18 cM. A man with genotype AB/ab marries a woman who is ab/ab. The man’s father was AB/AB.a. What is the probability that their first child will be Ab/ab?b. What is the probability that their first two children will both be ab/ab?
Answer:
a) 9%.
b) 16.8%.
Explanation:
a).
We are provided with the information that Two linked genes, A and B, are separated by 18 cM (centiMorgan). i.e the recombinant frequency is 18%
Also , the man's genotype is AB/ab... This only result to one explanation, that The man will definitely produce 18% of recombinant gametes which entails
9% Ab & 9% aB
i.e 0.09 Ab & 0.09 aB
On the other-hand, The mother ab/ab have tendency to produce just one single type of gamete which is ab
∴
The probability that their first child will be Ab/ab will be
Pr ( Ab/ab) = (0.09) x (1)
= 0.09
= 9%.
b).
If the father produces 18% of recombinant gametes which entails
9% Ab & 9% aB , this typically implies that the number of the non-recombinant gametes will be;
100%-18% = 82% ( non-recombinant gametes)
i.e genotype AB/ab = 82%
AB =41%; ab = 41%
AB = 0.41 ; ab = 0.41
Now, the probability that their first two children will both be ab/ab:
Using Multiplication Rule to calculate the probability that their first two children (ab/ab); we have:
(0.41)(1) ×(0.41)(1)
= 0.1681
= 16.8%.
The probability of the first child being Ab/ab as 25% and the probability of the first two children being ab/ab as 6.25%.
Probability of first child being Ab/ab:
Since the man is AB/ab and woman is ab/ab, the possible gametes are AB, ab, and Ab.The probability of their first child being Ab/ab is 1/4 or 25%.Probability of first two children both being ab/ab:
The probability of having a child with ab/ab is 1/4.For both children to be ab/ab, the probability is (1/4) x (1/4) = 1/16 or 6.25%.Micah has straight hair. If he is homozygous for the gene that determines straight hair, what is true of his parents?
Answer:
They have contributed same allele for straight hair.
Explanation:
There are two conditions homozygous and heterozygous. In homozygous conditions both the allele of a particular gene is same in an offspring which means the offspring got two same alleles from its parent for a particular gene and if an offspring is heterozygous for a gene that suggests that the parent has contributed two different alleles for that gene in the offspring.
So If Micah is homozygous for straight hair then it can be concluded that his parents have contributed two same alleles for straight hair genes.
If Micah is homozygous for the straight hair gene, both of his parents must have contributed an allele for straight hair. They could be homozygous or heterozygous for this trait.
Explanation:If Micah is homozygous for the gene that determines straight hair, it means he has two of the same type of allele for that gene - in this case, the allele for straight hair. This indicates that both of his parents must have contributed an allele for straight hair. The simplest scenario is that both of his parents are also homozygous for the straight hair allele. However, it's also possible that one or both parents are heterozygous, possessing one straight hair allele and one potentially recessive curly hair allele. In this scenario, they would have each passed the straight hair allele to Micah.
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Each of two parents has the genotype , which consists of the pair of alleles that determine , and each parent contributes one of those alleles to a child. Assume that if the child has at least one allele, that color will dominate and the child'swill be .
Answer:
Here is the full question:
Each of two parents has the genotype blond divided by red, which consists of the pair of alleles that determine hair color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one blond allele, that color will dominate and the child's hair color will be blond.
a. List the different possible outcomes. Assume that these outcomes are equally likely.
b. What is the probability that a child of these parents will have the red divided by red genotype?
c. What is the probability that the child will have blond hair color?
Explanation:
a.) Remember that the pair of allele determines the hair color. So, each of the two parents have the genotype red/blond. Therefore, the possible outcomes are blond/blond, blond/red, red/blond, and red/red.
b.) Since there are four outcomes, We calculate this as P(red/red) = (1/1)/(1/4) = 0.25
c.) There are three outcomes of the child having a blond hair which are blond/blond, blond/red, and red/blond. The probability is therefore P(outcome of blond/total outcome) = 3/4 =0.75
. A normal vision male marries a female who is a carrier of the color-blind gene and she becomes pregnant. What will a genetic counselor tell the parents regarding the a) percentage of having a daughter who is color-blind; b) what is the percentage of having a son who is color-blind? Consider a different couple with a female with normal vision and a male with normal vision. Would it be possible for their daughter to be a carrier for color-blindness? Explain your answer.
Answer:
Explanation:
a, The genetic counselor should explain that color-blindness is a sex-linked disorder, that is its expression is related to a particular sex-cells.
a) The percentage is 0% . This is because your husband is normal, so no sex-linked alleleis attached to his only X-chromosomes.However either of your X-chromosomes can bear the allele for color blind. Since a girl child sex-chromosomes combination is XX, and your husband X chromosomes is normal,hence your daughters can only be carries (XCX, XCX)like you if they inherited any of your X -chromosomes color blind allele.But none will be colorblind.
b)Because you are carrier for this alllele, and your husband is normal; chances are all your sons be colourblind is 50% proportion(YXC/YXC). This is because your husband Y chromosomes determines a male child, and you can bear the defective(carrier allele) for colourblind on either of your X-chromosmes to give a male child as XY. Thus either of the male child assuming with Mendelian fashion of inheritance, will be colour blind in 50% proportion.
No. A girl child can only be a carrier for color blindness if the inherited the color blind gene from her carrier mother or the father is color-blind. None of this is present in this scenario. The two parents are normal Therefore there is no chance of their daughter emerging a carrier.
A genetic counselor would tell the parents that (a) the percentage of having a daughter who is color-blind is 0% and (b) the percentage of having a son who is color-blind is 50%.
When considering the genetic possibilities for a couple where the female is a carrier of the color-blindness gene and the male has normal vision, genetic counselors would provide the following insights:
The percentage of having a daughter who is color-blind is 0%. This is because for a female to be color-blind, she must inherit two color-blindness genes – one from each parent. In this scenario, the father can only pass on a normal vision gene.
The percentage of having a son who is color-blind is 50%. Since males have only one X chromosome, inherited from their mother, they will be color-blind if that X chromosome carries the color-blindness gene.
For a different couple, both with normal vision, their daughter can still be a carrier for color-blindness if the mother is a carrier or if the father's mother is a carrier. This is because the mother can pass a carrier X chromosome to the daughter, or the father can pass an X chromosome that he inherited from his color-blindness carrier mother.
An RFLP can be created:_________. a) only by a mutation creating a new restriction enzyme recognition site b) only by a basepair change within the restriction enzyme recognition site c) only by a deletion removing a restriction enzyme reognition site d) by any mutational event that cretes or deletes a restriction enzyme recognition site e) only in RNA
Answer:
Option B, only by a basepair change within the restriction enzyme recognition site
Explanation:
A restriction fragment length polymorphism (RFLP) consists of alternative alleles with varying sizes of restriction fragments. In the traditional RFLPs, base pairs were changes at the restriction sites. These restriction sites comprises of nucleotide sequences that are identified by restriction enzymes. In RFLP analysis, the restriction fragments are created when restriction enzyme divide DNA into fragments. These restriction fragments are then separated while gel electrophoresis.
Before the advent of PCR, RFLPs (which were predominant form of DNA variation) were used to analyze linkages.
Hence, option B is correct
Which of these lists reflects the order in which cell organelles participate in the overall process that produces and transports many kinds of protein though the cell?
O Golgi apparatus, vesicles, ribosomes bound to rough endoplasmic reticulum, vesicles
O ribosomes bound to rough endoplasmic reticulum, vesicles, Golgi apparatus, vesicles
O Golgi apparatus, vesicles, ribosomes bound to rough endoplasmic reticulum
O vesicles, Golgi apparatus, ribosomes bound to rough endoplasmic reticulum
Answer:
O ribosomes bound to rough endoplasmic reticulum, vesicles, Golgi apparatus, vesicles
Explanation:
→Ribosomes bounded to R.ER are responsible for the mediation of the process of translation and elongation of the amino acids chains during protein synthesis for gebe expression. These R.ER bounded ribosomes are concerned for the synthesis of proteins for excretion out of the cells.
→As the protein chains were eleongated on the ribosomes units they are pushed into the lumen of R.ER into the intracistern space.Therefore R.ER is concerned with synthesis and packaging of protein synthesised by the attached ribosomes,
→These proteins are transported in vesicles,( which move through the cytoskeleton of the Cystosol) which budded off from the R.ER membranes.These vessicles transport the synthesized proteins to the Golgi complex.
→The Golgi apparatus receive the protein from the (R.ER), process it, and sorted it out into its vesicles(Golgi vesicles) for onward secretion out of the cells through excytosis from the plasma membranes. Example of this sorting and processing by the Golgi apparatus is the formation of Glycoprotein by addition of sugars molecules to proteins .
The idea that the bacterial genome is ""loose"" in the cytoplasm is incorrect because Choose one: A. the DNA is attached to the cell envelope and organized into domains through supercoiling and DNA-binding proteins. BY. attached ribosomes prevent tangling of the DNA. C. bacterial cells have a nucleus. D. the DNA is usually condensed into a chromosome.
Answer:D. the DNA is usually condensed into a chromosome
Explanation:
The bacterial genome is found loose in the cytoplasm but with a form. This unique feature is called the chromosomal DNA and it is not contained within a nucleus in the bacterial cell.
Answer: Option D.
DNA is usually condensed into a chromosome.
Explanation:
Bacterial genome refers to the complete set of DNA or complete set of genetic material of bacteria. Bacteria chromosomes are circular DNA. The chromosomes are packed by histone proteins into a condensed structure called chromatin. The condensed DNA is supercooled.
would cephalization be an advantage to jellyfish? why or why not?
Answer:
Cephalization does not confer an advantage to free-floating or sea organisms. Many aquatic species display radial symmetry like the jelly fish for instance.
Aquatic animals are not cephalized because they must be able to find food, take caution and defend themselves from harm from all directions.
Many kangaroo rats live in the Sonoran desert of the southwestern United States. They have a variety of adaptations for living in the desert. Under which circumstances would would the kangaroo rats of the Sonoran desert be most likely to develop new adaptations by natural selection?
Kangaroo rats in the Sonoran Desert would be most likely to develop new adaptations through natural selection if their environment undergoes significant changes that challenge their survival, leading to the evolution of advantageous traits that are passed on to offspring.
Explanation:The kangaroo rats of the Sonoran desert would be most likely to develop new adaptations by natural selection under circumstances where there are significant changes in their environment that create new challenges or pressures for survival. The key to this process is variation within the kangaroo rat population and the selective pressures that favor certain traits over others. For example, if the desert climate were to become even drier or if a new predator were introduced, those individuals with traits that improve their chances of survival, like even more efficient water conservation or better evasion tactics, would be more likely to survive and reproduce. These advantageous traits would be passed on to the next generation, leading to a population that is better adapted to the new conditions. This process over many generations results in evolution and the development of new adaptations.
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A woman with Turner syndrome is found to be color-blind (an X-linked recessive phenotype). Both her mother and her father have normal vision. a. Explain the simultaneous origin of Turner syndrome and color blindness by the abnormal behavior of chromosomes at meiosis. b. Can your explanation distinguish whether the abnormal chromosome behavior occurred in the father or the mother? c. Can your explanation distinguish whether the abnormal chromosome behavior occurred at the first or second division of meiosis? d. Now assume that a color-blind Klinefelter man has parents with normal vision, and answer parts a, b, and c
Answer:
A) The turner syndrome is characterized by lacking half chromosome numbers which are also termed as the monosomy as there is only half the number of X linked chromosomes that are responsible for the colorblindness will be greater. The chances are 50% of having this disorder in such a case.
B) The correct answer would be no, as the error will be shown in only one of the due to n number of chromosomes present in both parents.
C) Anaphase II of meiosis II it takes places as here only the strands are pulled to opposite poles of the cell
.
D) Assuming it as Klinefelter man with colorblindness has normal parents then,
a)double XX +Y , the colorblind gene is recessive in XX and Y. or recessive in XX but dominating Y only rules out the other recessive.
b) Explanation is similar as B)
c) Due to a nondisjunction event during meiosis, I or meiosis II in the female X chromosome is still present in it.
Which of the following specialized structures/inclusions would aquatic photoautotrophic bacteria likely possess? 1. Thylakoids 2. PHB Granules 3. Carboxysomes 4. Gas vacuoles 5. Chloroplasts
Answer:
1. Thylakoids 2. PHB Granules 3. Carboxysomes
Explanation:
1. Thylakoids are membrane bounded compartments present in bacteria for light dependednt reactions.
2. PHB Granules help in carbon fixation
3. Carboxysomes help to retain carbon when enough supply isn't available.
Answer: Option 1,2 and 3.
Thykaloids, PHB granules and carboxysomes.
Explanation:
Carboxysomes consist of polyhedral protein and enzymes Rubisco which is important to supply carbon and fix it.
Thykaloids are membrane bound part of chloroplast which is the site of light dependent reaction during photosynthesis.
PHB granules are important component which help to fix carbon .
How much ATP (in solid form) do you need to weigh, in order to make 1 mL of stock solution with a concentration of 100 mM ATP? The molecular weight of ATP is 551.14
Answer:
mass(g) = 0.55114g
Explanation:
Given that;
Molarity (M) 100mM = 0.1M
Volume (V) = 1mL
= 1.0 × 10⁻³ Litre
numbers of moles = Molarity × Volume
= 0.1M × 1.0 × 10⁻³ L
= 0.001
= 1.0 × 10⁻³ mol
Now, to determine the mass of ATP (in solid form) that we need to weigh;
we have our numbers of moles to be =[tex]\frac{mass(g)}{molarmass(g)}[/tex]
where;
numbers of moles = 1.0 × 10⁻³ mol
molar mass = 551.14
mass(g) = numbers of moles × molar mass
mass(g) = 1.0 × 10⁻³ mol × 551.14
mass(g) = 0.55114g
∴ 0.55114g ATP(in solid form) is required to be weighed in order to make 1 mL of stock solution with a concentration of 100 mM ATP.
In proteins, amino acids are linked by peptide bonds; in polynucleotides, nucleotides are linked by ________.
Answer: Phosphodiester bond
Explanation:
The backbone of DNA consists of deoxyribose nucleotides linked by phosphodiester bridges.
The 3'-hydroxyl of the adjacent sugar of one deoxyribonucleotide is joined to the 5'-hydroxyl of the adjacent sugar by an internucleotide linkage called a phosphodiester bond.
Thus, phosphodiester bond is the answer
Answer:
Phosphodiester bond
Explanation:
Nucleotides are linked by phosphodiester bonds that are formed between the 5' phosphate group of one nucleotide and the 3' hydroxyl group of another nucleotide in the DNA.
What are the differences between the G0, G1, and G2 phases of the cell cycle? Do all cells proceed through each of these phases? g
Answer:
The cell division is divided into two main phase - interphase and M (mitosis) phase. The interphase is further divided into G1, S phase and G2 phase.
G1 (gap 1 phase): The G1 phase is marked by the complete growth of the cell. The cells accquire different proteins and factors that are necessary for the cells to undergo the process of DNA rep[lication of the synthesis phase.
G2 phase (gap 2 phase) : During the G2 phase the cells prepares itself to undergo the M phase and underwent through the different phases of the cell cycle. The nuclear envelope forms around DNA and centrosomes are fully developed.
G0 phase: The G0 phase is also known as the extended phase of the G1 cycle. The G0 phase occurs when the cells are completely ready but do not undergo G1 phase, this might occur when the tissue is under the generation process.
The cells will go through the G1, S, G2 and M phase of the cell cycle. This is not necessary that all cells undergo the G0 phase of the cell cycle.
In all of these phylogenetic trees, the gain or loss of a trait is represented by a red trait mark. These marks are a way of showing change---that is, evolution. Which tree does the best job of showing which branches are evolving?
The tree with the most red trait marks best shows which branches are evolving.
Explanation:In phylogenetic trees, the gain or loss of a trait is represented by a red trait mark, which shows evolution. The tree that best shows which branches are evolving is the one with the most red trait marks. This indicates that those branches have undergone the most changes in traits over time.
Learn more about Phylogenetic trees here:https://brainly.com/question/31419981
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State whether each of the following statements is true or false, and explain your choice: a. The chromosomes in a somatic cell of any organism are all morphologically alike. b. During mitosis, the chromosomes divide and the resulting sister chromatids separate at anaphase, ending up in two nuclei, each of which has the same number of chromosomes as the parental cell.
Final answer:
The chromosomes in a somatic cell of any organism are not all morphologically alike. During mitosis, chromosomes divide and sister chromatids separate, resulting in two nuclei with the same number of chromosomes as the parent cell.
Explanation:
a. False. The chromosomes in a somatic cell of any organism are not all morphologically alike. Somatic cells contain pairs of homologous chromosomes, which are matched pairs containing the same genes in identical locations along their lengths. This means that each homologous chromosome may have different variants called traits caused by differences in the DNA sequence for a gene. So, the chromosomes in a somatic cell can show morphological variation.
b. True. During mitosis, the chromosomes divide and the resulting sister chromatids separate at anaphase, ending up in two nuclei, each of which has the same number of chromosomes as the parental cell. This ensures that each daughter cell receives an identical set of chromosomes to the parent cell.
Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where T is tall and t is short, and for color, where R is red and r is tan. If a short, tan male goat mates with a tall, red female goat of an unknown genotype, what is the probability that they would produce short, tan offspring?
and....
One of a pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of green and round offspring.
Answer:
(A) 0.0625
(B) 25%
Explanation:
A
Given that ;
T is tall
t is short
R is red
r is tan.
If a short, tan male goat mates with a tall, red female goat of an unknown genotype;
Let's take it one after the other;
A short tan male goat = ttrr
The female goat is an unknown genotype, we will have to determine that but we were told that it is tall and red
∴ For tall it is T and red is R;
Definitely, the female genotype should be T_ R_;
the probability genotype of the female for each allele is either:
- TT or Tt
- RR or Rr
However, from above;
the probability that the female will be Tt = [tex]\frac{1}{2}[/tex]
the probability that the female will be Rr = [tex]\frac{1}{2}[/tex]
the probability that the recessive allele 't' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]
the probability that the recessive allele 'r' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]
∴ the probability that they would produce short, tan offspring (ttrr) will be;
= [tex](\frac{1}{2})^{4}[/tex]
= [tex](\frac{1}{16})[/tex]
= 0.0625
(B)
One of a pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of green and round offspring.
For Color, Let;
Y = Yellow
y = Green
For shape, Let;
R = Round
r = wrinkled
If a cross occur between Yy Rr × yy rr. (i.e yellowround and green wrinkled)
the percentage of green and round offspring will be;
If Yy Rr self crossed; we have the following traits (YR, Yr, yR, yr)
yy rr will result to (yr, yr)
YR Yr yR yr
yr YyRr Yyrr yyRr Yyrr
yr YyRr Yyrr yyRr Yyrr
Yellow Yellow Green Green
Round Wrinkled Round Wrinkled
Therefore, the percentage of Green and Round offsprings from above is:
= [tex]\frac{2}{8}*100%[/tex]%
=[tex]\frac{1}{4}*100[/tex]%
= 25%
In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2.
(a) What will be the phenotypic ratio in the F2?
(b) If an F1 plant is backcrossed to the bitter, yellow-spotted parent, what phenotypes and
proportions are expected in the offspring?
(c) If an F1 plant is backcrossed to the sweet, non-spotted parent, what phenotypes and proportions
are expected in the offspring?
(d) If you testcross the F1 plant, what phenotypes and proportions are expected in the offspring?
Answer:
(a) 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)
(b) 100%, bitter, yellow spotted.
(c) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.
(d) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.
Explanation:
(a) The parental cross is BBSS x bbss.
We know that the F1 will be all BbSs (see attached punnet square). The only possible gametes the parental generation can pass on to the F1 are BS and bs, respectively.
The F1 intercross is therefore BbSs x BbSs. the alleles that each can pass on are BS, Bs, bS, or bs. The punnett square attached shows the phenotype ratio is 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)
(b) An F1 plant is BbSs. If it is backcrossed with a bitter, yellow spotted parent (BBSS), the cross is BbSs x BBSS (see attached punnett square). The resulting genotypes all give a bitter, yellow spotted phenotypes, so the phenotype of the offspring is 100% bitter, yellow spotted
(c) An F1 plant is BbSs, if it is backcrossed with a sweet, non spotted parent (bbss), the cross is BbSs x bbss. (see attached punnett square). The resulting phenotypes are 1:1:1:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. I.e. 25% of each genotype
(d) A test cross is a cross between an individual with a known genotype and another plant. The plants showing the recessive trait have a known genotype (they must be homozygous). So in this case, the F1 plant would be crossed with the homozygous recessive plants (BbSs x bbss). Therefore, the resulting phenotypes would be as in option c, 1:1:1:1