Answer:
c. Through light-driven ion pumps
Explanation:
a. Through redox reactions using diatomic oxygen (O2) as the final electron acceptor❌
Oxygen was observed to be lacking on early Earth.
b. Through the controlled oxidation of petroleum❌
Petroleum is categorically a fossil fuel and couldn't be marked as being available or found on early Earth.
c. Through light-driven ion pumps✔✔✔
Light-energy that was derived from the sun was a ready and accessible source of energy that could have directed proton pumps in early or first cells.
d. Through ingesting algae present in the early oceans❌
Algae are known to be eukaryotes and was seen to have been established billions of years after the first cells developed.
Answer:
c. light-driven ion pumps.
Explanation:
According to the records we know that the first cells took energy from the surrounding environment, and lived in the absence of O2, with the shortage of nutrients they began to obtain energy from sunlight and CO2 present in the atmosphere.
Theoretical and experimental measurements show that in many cases, the contributions of ionic and hydrogen-bonding interactions to ΔH for protein folding are close to zero. Provide an explanation for this result. (Hint: Consider the environment in which protein folding occurs.)
The formation of favorable ______ ionic or ________ interactions in a _________ protein replace interactions between solvent (water) and the ionic species (or _________donors and acceptors) in the _________ state. The favorable ΔH obtained by formation of ____________ bonds in the ___________ protein is offset by the energy required to ___________ many interactions, with solvent going from the ________ to the ________ state.
Fill in the blanks above using the words listed below.
H-bonding
H-bond
intermolecular
restore
unfolded
folded
C-bond
break
intramolecular
C-bonding
Answer:
The formation of favorable C-bond ionic or C-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or restore donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of H-bond in the H-bonding protein is offset by the energy required to break many interactions, with solvent going from the inter-molecular to the intramolecular state.
Answer:
The answer is:
The formation of favorable intermolecular ionic or H-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or H-bond donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of intramolecular bonds in the folded protein is offset by the energy required to break many interactions, with solvent going from the unfolded to the folded state.
Explanation:
Suppose that the narrow-sense heritability of wool length in a breed of sheep is 0.92, and the narrow-sense heritability of body size is 0.87. The genetic correlation between wool length and body size is –0.84. If a breeder selects for sheep with longer wool, what will be the most likely effects on wool length and body size?
Answer:
There is an increase in wool length and a little decrease in body size
Explanation:
Given that:
the narrow-sense heritability of wool length in a breed of sheep is 0.92; &
the narrow-sense heritability of body size is 0.87.
As these traits approaches 1, the higher the rate that helps them to provide a positive feedback for selection.
The question proceeds by stating that " The genetic correlation between wool length and body size is –0.84."
You see, the functionality of this negative (-0.84) genetic correlation tends to create an imbalance between these heritable traits. As such, there is a shift in the paradigm in which one traits tends to increase and the other trait to decrease.
Now, if a breeder selects a sheep with longer wool, there is an increase in wool length since he selects it although there will be a little decrease in body size due to the negative genetic correlation effect between the two heritable traits.
In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+) and the gene for this trait is autosomal. The gene for the white-eye trait is on the X chromosome. Suppose a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigial winged male.
a. What will be the genotypes and phenotypes of the F1 flies?
b. what will be the genotypes and phenotypes of the F2 flies?
c. what will be the genotypes and phenotypes of the offspring of a cross of the F1 flies back to each parent?
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
Final answer:
In crossing a homozygous white-eyed, long-winged female Drosophila with a homozygous red-eyed, vestigial-winged male, the F1 generation will have all females with red eyes and long wings and all males with white eyes and vestigial wings. The F2 generation shows a mix of possible phenotypes depending on the sex and inherited genes. Backcrossing the F1 with the original parents yields specific outcomes based on the traits of the parents and the mechanisms of inheritance at play.
Explanation:
Crossing Flies: Drosophila Melanogaster Genetics
In Drosophila melanogaster, the genetics of eye color and wing size are well-studied, with the white-eye trait being X-linked and vestigial wings being autosomal recessive. Considering a cross between a homozygous white-eyed, long-winged female and a homozygous red-eyed, vestigial-winged male, we can predict the outcomes for the F1 and F2 generations, as well as the results of backcrossing F1 individuals with each parent.
a. F1 Generation
b. F2 Generation
c. Backcross to Parents
Cross with white-eyed, long-winged female parent:Produces red-eyed, long-winged females and white-eyed, vestigial-winged males.Cross with red-eyed, vestigial-winged male parent:Results in a similar mix of phenotypes as the F2 generation, depending on sex and inheritance patterns.Dendrites are A. the conduction zone of a nerve cell. B. the input zone of a nerve cell. C. a type of glial cell. D. small interneurons.
Answer:
B. the input zone of a nerve cell.
Explanation:
Dendrites are the small extensions that come out of the soma or cell body of a neuron. The function of dendrites is to receive the nerve signals or information from the axons of the presynaptic neurons and carry them towards the cell body or soma. In a synapse, the signals from the axons of the presynaptic neurons are revived by dendrites of postsynaptic neurons. The plasma membrane of dendrites have receptors to which the chemical messengers from other cells bind. In this way, dendrites serve as input zone of a nerve cell.
Dendrites are the input zone of a nerve cell. They receive information from other neurons and deliver it to the body of the neuron. They are not glial cells, small interneurons, or the conduction zone of a nerve cell.
Explanation:Dendrites are best described as B. the input zone of a nerve cell. These are branch-like structures that protrude from the neuron. They play a crucial role in receiving information from other neurons and transmitting it to the body of the neuron. Unlike the axon, which is the output zone of the nerve cell, the dendrites play an essential role in receiving neural inputs or signals, acting as the 'input zone'. They are not a type of glial cell (which are non-neuronal cells in the nervous system), nor are they small interneurons or the conduction zone of a nerve cell.
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You just bought two black guinea pigs from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black progeny. a) How would you explain this result? b) If the original black guinea pigs produce an average of 10 offspring per mating, how many matings would you have to make with these same parents before you'd expect to see a white guinea pig? c) Indicate the most likely genotype of the white offspring.
Answer:
21 Mating
Explanation:
Ans A)The recessive lethal allele is tightly linked to, and thus co-segregates with, the recessive B allele, which is lethal in the homozygous state (bb)
Ans B)
1% of the gametes will be recombinant (Bl or bL)
0.5% of the gametes will be bL
0.5% of the gametes will be Bl
white animals will have the following genotype
genotype probability
bL/bL(0.005)(0.005) = 0.000025
bL/bl(0.005)(0.495) = 0.002475
bl/bL(0.495)(0.005) = 0.002475
Probability of white animal =0.004975
1/0.004975 = 201 therefore on average you'd have to look at 201 progeny before seeing a white animal so on average you'd have to do 21 mating
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.
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
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.
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.
An enzyme-linked immunosorbent assay requires:
A) a radioactive substrate.
B) a radioactive standard for binding to the antibody.
C) aromatic amino acids.
D) a catalytic antibody.
E) an antibody that binds the protein of interest.
Answer: Option E.
Eliza require an antibody that bind to protein of interest.
Explanation:
Enzyme linked immunosorbent assay is a plate like or immunological technique or assay that Is use to detect and measure peptides, proteins,hormones and antibodies.
Enzyme linked immunosorbent assay require an antibody that can bind to a specific protein of interest.
How does oceanic lithosphere form?
Which of the following techniques is used sparingly because there is a slight but genuine risk of miscarriage or damage to the fetus while having a 99% accuracy in diagnosing genetic problems? amniocentesis CT scan ultrasound chorionic villus sampling
Answer: Amniocentesis
Explanation:
An amniocentesis is performed when a woman is between 14 and 16 weeks gestation. Women who choose to have this test are primarily those at increased risk for genetic and chromosomal problems, in part because the test is invasive and carries a small risk of miscarriage.
Amniocentesis is the technique used sparingly due to the slight risk of miscarriage or damage to the fetus. It offers 99% accuracy in diagnosing genetic problems. Other diagnostic techniques include chorionic villus sampling and prenatal genetic diagnosis.
Explanation:The technique referred to in the question is amniocentesis. This diagnostic procedure is usually done between weeks 15 and 20 of pregnancy, and is used to diagnose genetic disorders in the fetus. A small amount of amniotic fluid, which contains fetal tissues, is sampled from the amniotic sac surrounding the developing fetus. The fetal DNA is then examined for genetic abnormalities. The procedure has a high accuracy rate, but also carries a small risk of miscarriage or damage to the fetus, hence its use is usually reserved for cases wherein the potential benefits outweigh the risks.
Another technique is chorionic villus sampling (CVS) which involves sampling part of the placental tissue, usually performed between weeks 11 and 14 of pregnancy. This technique also carries risk and is used for similar reasons as amniocentesis.
Prenatal genetic diagnosis (PGD), ultrasound, and CT scans are other techniques used to detect conditions in the embryo or fetus, but they differ from amniocentesis and CVS in terms of their procedure, risk, and the type of information they provide. Note that, while ultrasound and CT scans provide images of the fetus, they do not provide the same genetic level of detail as amniocentesis or CVS.
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Blue whales (Balaenoptera musculus) can weigh up to 170 tons and are the largest animals alive today. Their large size is thought to have protected them from large predators, such as Megalodon (Carcharocles megalodon), which would have preferred to eat smaller, less dangerous whales. What process best explains the evolution of the enormous blue whale body size
a. Natural selection.
b. Migration.
c. Genetic drift.
d. Non-random mating.
Answer:
Option c. Genetic Drift is the right answer
Explanation:
Due to random sampling of organisms results in the change of the frequency of an existing gene variant in a population is called as Genetic drift.
example: Baleen whales were moderately large. Baleen whales are the ancestors of the present-day blue whale. However, around 4.5 million year ago a massive evolution occurred in them which was resulted in blue whale origination.
Answer: A. Natural Selection.
Explanation: Blue whales (Balaenoptera musculus) can weigh up to 170 tons and are the largest animals alive today. Their large size is thought to have protected them from large predators, such as Megalodon (Carcharocles megalodon), which would have preferred to eat smaller, less dangerous whales. What process best explains the evolution of the enormous blue whale body size?
One factor that Saruman Enterprises considered in deciding whether to market a line of weight-loss supplements was the increasing rate of obesity in the United States. This trend is considered a(n) _______ factor.
Answer:
This trend is considered a(n) sociocultural factor.
Explanation:
In the field of business studies, sociocultural factors can be described as certain lifestyles or habits that characterize a particular society. The sociocultural environment of a particular area is used in business and marketing to make products which will attract the customers of a particular area. For example, with the increase in obesity of a particular place or city, people of that city will be more attracted to weight loss products.
Because the researcher had access to twenty years of data on the same participants in an education program, the researcher was able to perform a(n) _________ study.
Answer:
Descriptive study
Explanation:
Any research study can be classified into two groups
a) Descriptive and
b) Analytical study
A Descriptive study is also referred as non-analytic study as it involves understanding of something happening in a population as a whole since time frame being referred by doing through literature review, studying case reports, conducting surveys, doing comparisons and referring previous researches. It helps to understand prevalence, incidence, or experience of a group
While analytic study is based on establishing relation between two factors through experiment
Because the researcher had access to twenty years of data on the same participants in an education program, the researcher was able to perform a longitudinal study. This type of research is particularly powerful for tracking changes and developments over time.
Longitudinal studies follow the same group of individuals and assess them repeatedly across a protracted timespan. For instance, psychologist Rich Lucas was able to utilize longitudinal data from over 20,000 Germans across two decades to explore the implications of marriage on happiness. Such studies can reveal impactful insights into the long-term effectiveness of educational programs and behavioral trends but can also be resource-intensive to conduct.
Longitudinal research contrasts with cross-sectional research, which analyzes different segments of a population at a single point in time. While cross-sectional studies can be less expensive and quicker to execute, they do not provide the sequential data over time that longitudinal research can offer.
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%.The body tubes are long, semi-rigid tubes, usually made of _______________________chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end (sticking out of mouth) has a standard ________ mm adaptor.
Answer:
The body tubes are long, semi-rigid tubes, usually made of _polyvinyl_ chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end (sticking out of mouth) has a standard _15_ mm adaptor.
Explanation:
An endotracheal tube (ET) is a flexible plastic tube that is inserted into the trachea through the mouth to help a patient breathe. Then it is connected to a ventilator for oxygen supply to patients's lungs. This process of insertion of tube is refereed as endotracheal intubation.
The body tubes in question are usually part of endotracheal tubes, made of polyvinyl chloride or a similar plastic. The proximal end of these tubes that sticks out of the mouth has a standard 15 mm adaptor.
Explanation:The body tubes referred to in the question are typically part of medical devices known as endotracheal (ET) tubes, used to secure a patient’s airway during operations or critical care. These tubes are long, semi-rigid tubes, typically made of polyvinyl chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end, which sticks out of the patient's mouth, has a standard 15 mm adaptor. This universal size allows it to attach to a variety of equipment like ventilators or bag valve masks.
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Some mutations result in proteins that function well at one temperature but are nonfunctional at a different (usually higher) temperature. Siamese cats have such a "temperature-sensitive" mutation in a gene encoding an enzyme that makes dark pigment in the fur. The mutation results in the breed's distinctive point markings and lighter body color.
Let's complete the question by adding the missing piece of information
The mutation results in the breed's distinctive point markings (ears, mask, tail and legs) and lighter body color. Use this information to explain the pattern of the cat's fur pigmentation.
Answer:
The mutation of the TYR gene results in the enzyme tyrosinase to be heat susceptible. Tyrosinase takes part in the production of melanin to give darker fur in colder areas. The areas like the tail, legs, ears, and face do lack as much body heat and so will get darker.
Explanation:
A unique protein (enzyme), known as tyrosinase, is the major workhorse in the development of the melanin. A research team from the University of California, USA, led by L. A. Lyons, discovered that Siamese cats have tyrosinase that went through mutation due to the changes in the DNA helix and is temperature-sensitive as it's activity reduces with a rise in temperature. This explains why cat’s warm parts of the body are coated with white, melanin-lacking hair since Tyrosinase is deactivated in these regions and melanin is not developed – hair is white-colored. On the other hand, in cooler boundary the enzyme is active and the melanin is formed – hair has dark color.
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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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.
Which of the following statements is/are accurate?
A. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of a hybrid breakdown.
B. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of a hybrid breakdown.
C. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatible. This is a good example of gametic isolation.
D. One species of sea turtle mates during the early spring and a closely related species mates during late spring. This is an example of temporal isolation.
E. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation.
Answer and explanation:
In nature there are different mechanisms that prevent the crossbreeding between different species, what in biology is called reproductive barriers. Some mechanisms that act by preventing hybridization between different species are:
Hybrid breakdown Gametic isolation. Mecanical isolation. Temporary isolation. Ethological aislaminet. Ecological insulation.These mechanisms are responsible for preserving the genetic integrity of each species by preventing hybridization between different species.
A. One species of sea urchins can not fertilize another closely related species because the eggs do not have a receptor for the sperm. This is a good example of a hybrid breakdown.This is not accurate. In the case of sea urchins, the encounter of gametes requires two chemical mechanisms:
The first mechanism is called chemotaxis, which consists of the presence of a chemical signal on the surface of the egg, for which only sperm has a receptor. Another mechanism is that - once the sperm and egg are found - the membrane of the egg releases substances that interact with receptors in the sperm, allowing the sperm to enter it.These two chemical mechanisms ensure that gametes of two different species cannot be joined and fertilized, which is an example of gametic isolation.
B. Horses and Mules can be bred but their offspring is typically sterile. This could be an example of a hybrid breakdown.This is accurate. Horses and donkeys belong to two different species, with a different chromosomal load:
Horses have 32 pairs of chromosomes. Donkeys have 31 pairs of chromosomes.Both species can be bred, but their descendant, mules (Equus africanus x ferus), have an odd number of chromosomes (63) and are infertile. This represents an exact example of hybrid breakdown.
C. Damselflies have sensory receptors that are sensitive to touch (tactile cues). Two related species of damselflies are unable to mate because their touch cues are not compatible. This is a good example of gametic isolation.This is not accurate. The sensitivity to the touch of damselflies is specific to individuals of the same species, preventing mating between male and female of different species.
This, like courtship, is a mechanism that prevents crossbreeding between different species, establishing an example of mechanical isolation.
D. One species of sea turtle mates during the early spring and a closely related species mates during late spring. This is an example of temporal isolation.This is accurate. When two related but different species - such as turtles - have their mating period at different times of the year, there is talk of temporal or seasonal isolation.
Temporal isolation is a reproductive barrier that prevents crossing between different species, due to their mating habits at different times.
E. One sponge species releases its gametes during the night and another species releases its gametes during the day. This is an example of ecological isolation.This is not accurate. In sponges, like some coral species, periods of release of gametes and fertilization vary throughout the day, with some synchrony between individuals of the same species.
The fact that some sponges release their gametes by day and other species do it at night is an example of the reproductive barrier called temporal isolation.
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Two selectively permeable sacs A and B were submerged in a 45% glucose solution that was contained in a beaker. Sac A contained a 15% glucose solution and Sac contained a 15% sucrose solution. In which direction did net movement of water molecules occur? a. Only from Sac A to the beaker b. Only from Sac B to the beaker c. From the beaker to both sacs d. From both sacs to the beaker
Answer:
Option a. Only from Sac A to the beaker is correct.
Explanation:
As beaker contains glucose which is a monosaccharide and Sac A also have glucose in it, So, therefore glucose from sac A will move into beaker through the process of OSMOSIS.
Sac A (15% glucose) is less concentrated as compared to beaker (45% glucose) therefore this phenomenon will occur. (See attached image for more detailed and graphical explanation)
The net movement of water molecules occurred from the beaker to both sacs due to the concentration gradient.
The net movement of water molecules occurred from the beaker to both sacs. This is because the 45% glucose solution in the beaker created a concentration gradient, with a higher concentration of solute outside the sacs compared to inside.
Water molecules will naturally move from an area of lower solute concentration to an area of higher solute concentration through a selectively permeable membrane, in order to balance the concentrations on both sides.
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What are the emergent properties of the nervous system that cannot be predicted by studying individual neurons?
Answer:
Consciousness, human emotion etc, are emergent properties
Explanation:
The properties which are characteristics of an entire system and not its constituting members are called as emergent properties.
Consciousness with in an individual human being can be termed as an emergent property of nervous system. A single neuron cannot generate or holds the sense of consciousness, self-awareness, pride or honor etc. The entire nervous system can also generate complex human emotions such as fear, joy, pride etc. Neuro-biologists have not been able to depict the expression of these functions at micro level such as a single neuron and thus these properties are termed as emergent properties.
Which of the following statements about the evolution of development is false? a. Evolution by natural selection "works" like a tinker, assembling new structures by combining and modifying available materials. b. Nearly all evolutionary innovations are the result of modifications of previously existing mechanisms and structures. c. The genes that control development are highly conserved. d. All of the above are true; none is false
All of the above are true; none is false
Explanation:
Evolution is the phenomena by which an organism gets modified or develops more advanced characters which makes it more fit for the survival in nature.
The first statement is true because evolution refers to the modification of existing characters nothing new develops spontaneously to overcome natural selection. During natural selection organisms that survive gradually surpass the selection pressure. And the evolved character of it gets fixed within the species.
Evolution occurs within the existing mechanisms and structure which find out new ways of surviving in the changing environment.
The genes that control the development are highly conserved and this is the reason we can trace back the similarities with common ancestors. And make a comparitive study of structural anatomy.
Final answer:
The false statement is: c. The genes that control development are highly conserved.
Explanation:
The correct answer is: c. The genes that control development are highly conserved.
Genes that control development can be highly variable between species, allowing for diverse developmental processes and structures to evolve. While certain core developmental genes may be conserved across different species, there is also significant variation that allows for evolutionary innovation.
For example, the Hox genes, which play a crucial role in determining body structure during development, can vary greatly between species, leading to diverse body plans across different organisms.
A bacteria culture starts with 200 bacteria and in 1 hour contains 400 bacteria. How many hours does it take to reach 2000 bacteria?
Answer:
At 0hr it was 200 bacteria
At 1 hr it doubled and was 400
Means: at every 1 hr the population of the bacteria doubles
400 bacteria = 1hr
2000 bacteria = X
X = 2000 × 1 / 400
X = 5 hrs
In wheat plants, the feature of having colored kernels is dominant to having white kernels that lack pigment. A true-breeding plant with colored kernels is crossed to a true-breeding plant with white kernels, resulting in progeny that all harbor colored kernels. The F1 progeny are then crossed, and a few members of the F2 generation have white/colorless kernels. The modified ratio observed is 15 colored: 1 non-colored. Explain these results.
Answer: It occurred a dihybrid cross and epistasis.
Explanation: In dihybrid cross, two different genes controlled two different traits. When they interact with each other is called Epistasis. However, in wheat plants, the genes related to color kernels don't act opposedly to each other. In other words, the genes have the same role in producing protein, so they can substitute for each other.
In the color determination mechanism, a biochemical reaction is necessary to convert a precursor substance into a pigment and that reaction happens with the product of either genes. That's why having a dominant allele makes the wheat colorful. So, crossing colored kernels with white ones will produce a heterozygous F1 generation. Crossing this generation will produce a F2 generation with modified ratio of 15 colored: 1 non colored because, every individual who has dominant alleles will produce the substance and thus the biochemical reaction will happen. Only recessive homozygous ones won't have the substance and so won't have color.
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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"In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was correct. Identify the three proposed models for DNA replication."
Answer:
Three proposed models: 1) Conservative
2) Semiconservative
3) Dispersive
Correct one: 2) Semiconservative
Explanation:
Meselson and Stahl proposed that the process of DNA replication could be conservative, semiconservative or dispersive. Conservative DNA replication will form one DNA duplex with both newly formed strands and the other duplex with both parental strands. The semi conservative mechanism will form two DNA helices each of which would have one parental strand and one newly formed strand (hybrid). The dispersive mechanism will form DNA double helices with patches of the new and parental strand.
However, they found that DNA replication follows a semi-conservative mechanism as after two rounds of replication of heavy chain DNA in the light-medium formed two molecules of hybrid DNA and two DNA molecules with light chains only. There was no DNA molecule with both the strands having the heavy chains.
The three proposed models for DNA replication individually tested by Meselson and Stahl in 1958 were the conservative, semi-conservative, and dispersive models. Their experiment proved the semi-conservative model, where each new DNA molecule consists of one old and one new strand, to be the accurate depiction.
Explanation:In 1958, Matthew Meselson and Franklin Stahl conducted an experiment with the goal of determining the accurate model for DNA replication. The three proposed models that they evaluated were the conservative model, the semi-conservative model, and the dispersive model.
In the conservative model, the parent DNA molecule remains intact and an all-new molecule is formed based on its template. In contrast, the semi-conservative model suggests that each of the two new DNA molecules consists of one original and one new strand, replicating its structure from the parent molecule. Lastly, the dispersive model proposes that each strand of both daughter molecules contains a mixture of old and newly synthesized DNA.
As a result of the experiment, they found that the semi-conservative model was the correct depiction of DNA replication.
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Effect of pH on the Conformation of α-Helical Secondary Structures The unfolding of the α helix of a poly-peptide to a randomly coiled conformation is accompanied by a large decrease in a property called specific rotation, a meas-ure of a solution’s capacity to rotate circularly polarized light. Polyglutamate, a polypeptide made up of only L-Glu residues, has the α-helix conformation at pH 3. When the pH is raised to 7, there is a large decrease in the specific rotation of the solu-tion. Similarly, polylysine (L-Lys residues) is an α helix at pH 10, but when the pH is lowered to 7 the specific
This is because the pH of the solution changes the chemical and stereochemical properties of the solution.
Explanation:All the proteins or polypeptides are formed of polymers of amino acids. These amino acids comprise of an alpha carbon, where a hydrogen, a carboxyl group, an amino group and a variable group R is attached. All these four groups have their specific stereochemistry which gives the polypeptide a particular shape in their coiled form.
Here in case of polyglutamate, the polymer is formed of chains of glutamic acid. This glutamic acid has carboxyl group in the R group which remains free even in the polymerized state. In acidic pH, the carboxyl group has its normal structure - COOH, but as the pH increase to 7,the hydrogen ion dissociates making it - COO⁻. So the stereochemistry changes and the specific rotation also changes.
Similarly in poly lysine, there's amino group in the R group which remains stable in alkaline pH of 10,but in case of neutral or acidic pH, the structure becomes - NH3⁺. So, the specific rotation changes.
Final answer:
The secondary structure of proteins, like alpha-helix, is crucial for protein function and is stabilized by hydrogen bonds. The alpha-helix structure can be altered by pH changes, resulting in different physical properties, such as specific rotation, which indicates structural changes.
Explanation:
Effect of pH on Protein Secondary Structure
The secondary structure of proteins, like the alpha-helix (α-helix) and beta-pleated sheet (β-pleated), plays a critical role in the protein's overall structure and function. These structures are stabilized by hydrogen bonds between the amino acids in the protein chain. The α-helix is particularly stable due to hydrogen bonds that form between the carbonyl oxygen of one amino acid and the amide hydrogen four residues away. Changes in pH can lead to a disruption of these hydrogen bonds, causing the protein to unfold or alter its conformation. This can be observed through changes in physical properties, such as specific rotation, which is a measure of how a substance rotates polarized light. Polyglutamate and polylysine are examples of polypeptides that change from an α-helical structure to a randomly coiled conformation when the pH shifts away from their optimal range, thereby exhibiting a change in specific rotation.
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.
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