Answer:
Isocitrate dehydrogenase and α‑ketoglutarate dehydrogenase are the two enzymes that produces carbon dioxide as a by-product.
Explanation:
The release of carbon dioxide from any compound is known as decarboxylation. The first step where carbon dioxide is released as a by product is when Isocitrate is oxidized and decarboxylated to α‑ketoglutarate and this reaction is catlyzed by isocitrate dehydrogenase. The second step where carbon dioxide is released as a by product is from decarboxylation of α‑ketoglutarate into succinyl-Co A and this reaction is catalyzed by α‑ketoglutarate dehydrogenase.
What Kingdom do euglena belong to? What Phylum?
Answer:
Euglena are classified into the Kingdom Protista, and the Phylum Euglenophyta.
Explanation:
In Arabidopsis thaliana, the flowering locus C (FLC) gene codes for a regulatory protein that suppresses flowering. FLC is expressed in seedlings to prevent premature flowering. In mature plants, FLC expression decreases with cooler temperatures, and flowering occurs once sufficiently cool temperatures are reached.
If small interfering RNA (siRNA) that is complementary to FLC mRNA is introduced, how would RNA interference (RNAi) affect flowering?
a. RNAi is found in prokaryotes and would not affect Arabidopsis.
b. RNAi would bind irreversibly to FLC mRNA and stimulate flowering.
c. RNAi would degrade FLC mRNA and repress flowering.
d. RNAi would degrade FLC mRNA and stimulate flowering.
e. RNAi would induce methylation of chromatin and repress flowering.
Answer:
In Arabidopsis thaliana, the flowering locus C (FLC) gene codes for a regulatory protein that suppresses flowering. FLC is expressed in seedlings to prevent premature flowering. In mature plants, FLC expression decreases with cooler temperatures, and flowering occurs once sufficiently cool temperatures are reached.
If small interfering RNA (siRNA) that is complementary to FLC mRNA is introduced, how would RNA interference (RNAi) affect flowering?
D- RNAi would degrade FLC mRNA and stimulate flowering.
Explanation:
Arabidopsis thaliana is a small flowering plant native to Eurasia (the old continent) and Africa. This plant has a small genome that is diploid. This is why it is useful and easier for genetic mapping and sequencing. It is commonly used for understanding the molecular biology of some plant traits, maybe due to the fact that it was the first plant to have its genome sequenced.
Plant A and Plant B are heterozygous for a gene that expresses their tall trait (Tt). Upon crossing, what is expected in the offspring, assuming random chromosome segregation?
A) They will always have the tall phenotype.
B) They will always have the short phenotype.
C) They are more likely to be tall plants than short plants.
D) They are more likely to be short plants than tall plants.
Answer:
C
Explanation:
They are more likely to be tall plants than short plants
Which two positions shows the location of the moon in its crescent city phases as seen from earth?
Answer:
waxing and waning crescent
Explanation:
Which scientist developed the idea that micro organisms can cause disease
Answer: Paul Ehrlich, Anton van Leeuwenhoek and Louis Pasteur
Explanation:
What is a function of the backbone in animals?
Answer:
It has a lot of functions, but its mainly to protect the spinal cord
It surrounds the spinal cord.
Explanation: it said it on my quiz
1. The organs of sexual reproduction in plants
are the
Answer:
As a plant's reproductive part, a flower contains a stamen (male flower part) or pistil (female flower part), or both, plus accessory parts like sepals, petals, and nectar glands (Figure 19). The stamen is that the male sex organ. It consists of a pollen sac (anther) and a protracted supporting filament.
Explanation:
In silkmoths, red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these 2 genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wings. The F1 are crossed with moths that have red eyes and white banded wings in a test cross. The progeny of this test cross are:
wild-type eyes, wild-type wings 418
red eyes, wild-type wings 19
wild-type eyes, white-banded wings 16
red eyes, white-banded wings 426
a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes?b. What is the rate of recombination between the genes for red eyes and those for white-banded wings?
Answer:
a. The phenotypic proportions that would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes is 1:1:1:1
b. The rate of recombination between the genes for red eyes and those for white-banded wings is 0.04
Explanation:
To know that two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.
If the genes for red eyes and white-banded wings were located on different chromosomes, they would assort independently, and hence the expected phenotypic proportions would be 1/4:1/4:1/4:1/4
4/16=1/4 re+re wb+wb
4/16=1/4 rere wb+wb
4/16=1/4 re+re wbwb
4/16=1/4 rere wbwb
In the present example, the genotype, in linked gene format, of the double heterozygote is re+wb+/rewb.
In this way, we might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny.
To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.
In the present example:
Parental) re+ wb+/re wb x re wb/re wb
Gametes) re+wb+ parental type
re wb parental type
re+wb recombinant type
re wb+ recombinante type
Phenotypic class Number of offspring
wild-type eyes, wild-type wings, re+ wb+, 418red eyes, wild-type wings, re wb+, 19wild-type eyes, white-banded wings, re+ wb 16red eyes, white-banded wings, re wb, 426The recombination frequency is:
P = Recombinant number / Total of individuals
P = 19 + 16 / 418 + 19 + 16 + 426
P = 35/ 879
P = 0.0398=0.04
The genetic distance between genes is 0.04 x 100= 4 MU.
If the genes for red eyes and white-banded wings in silkmoths were on different chromosomes, a 1:1:1:1 phenotypic ratio would be expected. In the test cross given, the rate of recombination is approximately 7.7%, which suggests linkage with some degree of crossing over between the two genes.
In silkmoths, where red eyes (re) and white-banded wings (wb) are encoded by two recessive alleles located on the same chromosome, the expected outcome of a test cross can reveal the rate of recombination and indicate a linkage between the two genes.
If the genes for red eyes and white-banded wings were located on different chromosomes, we would expect to see a 1:1:1:1 phenotypic ratio due to independent assortment.Based on the given progeny of the test cross, the rate of recombination between the genes for red eyes and white-banded wings can be calculated by adding the number of recombinant offspring (19 wild-type eyes, red wings + 16 red eyes, white-banded wings) and dividing it by the total number of offspring, then multiplying by 100 to get a percentage. This calculation yields a recombination frequency of approximately 7.7%, indicating that the two genes are linked but not completely linked as crossing over does occur.A microbiologist is in the process of classifying a newly discovered organism. Its characteristics include:
· eukaryotic
· unicellular
· has a cell wall
· autotrophic
Which type of organism is it most likely to be?
Answer:
a diatom
Explanation:
Answer:
c
Explanation:
The Champaign-Urbana area has long been suffering from the heinous pathogenic bacterium, Michiganious wolverinous, which causes rabid wolverine fever. During 2019, the Champaign Public Health Department reported an incidence of 1000 cases of rabid wolverine fever. If the number of pre-existing cases carried over from 2018 to 2019 was 2000 cases, then the prevalence of rabid wolverine fever in 2020 was __________________.
a. 3000 cases
b. 4000 cases
c. 1000 cases
d. 0 cases
e. 2000 cases
Answer:
The correct answer is option a, that is, 3000 cases.
Explanation:
The measurement of all the individuals getting influenced by the disease at a specific time is known as the prevalence. On the other hand, the measurement of the number of novel individuals that came into contact with a disease during a specific time duration is known as the incidence.
Based on the given question, the number of prevailing cases carried from 2018 to 2019 is 2000, and the new diseases recorded in the year 2019 is 1000 (incidence). Therefore, the prevalence of the disease in 2020 will be 3000 cases.
Recent research has identified a gene (SLC45A2) that codes for a protein that affects melanin production in humans and other animals. A mutation in this gene has given rise to two variant alleles. The L374 allele correlates with darker pigmentation, and the F374 allele correlates with lighter pigmentation. Researchers collected DNA samples from people in 14 European, Asian, and African populations and identified the frequency of the F374 allele in these groups. Review the data in the chart on p. 242 of your lab manual and answer the questions that follow. In what populations do we see the highest frequencies of the F374 allele for lighter pigmentation?
Complete Question:
Recent research has identified a gene (SLC45A2) that codes for a protein that affects melanin production in humans and other animals. A mutation in this gene has given rise to two variant alleles. The L374 allele correlates with darker pigmentation, and the F374 allele correlates with lighter pigmentation. Researchers collected DNA samples from people in 14 European, Asian, and African populations and identified the frequency of the F374 allele in these groups. Review the data in the chart below (see the attached first image) and answer the questions that follow.
1. In what populations do we see the highest frequencies of the F374 allele for lighter pigmentation?
2. In what populations do we see the lowest frequencies of the F374 allele for lighter pigmentation?
3. The map below(see the attached second image) shows the ancestral locations of the populations studied, superimposed on Figure 8.2, a map of given regional skin color variation. Are these results what you would expect based on the skin color information in the map? Why or why not? Figure Credit: Map George Chaplin. First published in Skin: A Natural History by Nina G. Jablonski, UCP 2013. Used with permission.
4. Why might some light-skinned populations, such as the Japanese, be missing the F374 allele for lighter skin pigmentation? (Hint: Consider the various forces of evolution that may be at play.)
Answer and explanation:
Going by the given data
1.
The highest frequencies of the populations of the F374 Alleles for lighter pigmentations is said to be as, going by the data in the above table is (german)GERMANY where as frequency of F374 of 0.965(96.5%)
2.
The lowest frequency of the population of the F374 alleles for lighter pigmentations going by the data in above table is
10. (CHINA)
12. Japanese (JAPAN)
14. African (GERMANY, JAPAN)
where the frequency of the F374 is 0.000(0%)
3.
Ancestral location of population which is shown on map is similar to the data which is provided in the table.
=> when parental genome has L374 allele at high rate then the offspring will have darker pigmentation in them
=> when parental genome has F374 Allele at low rate then the offspring will have lighter pigmentation in them
=> when parental genome is having both alleles at same rate then the offspring will have moderate color pigmentation ( that is not too dark or light)
4.
The lighter skin pigmentation in Japanese of F374 Alleles is high in rate than the genome compared to L374 Allele
Due to loss of L374 Allele in their genetic material leading to lighter pigmentation in their skin for adaptation of L374 allele may be lighted up with various reason said to be genomic changes, environmental, climatic changes and style of living with their food habits
The F374 allele of the SLC45A2 gene, associated with lighter skin pigmentation, would likely be more prevalent in populations where lighter skin is beneficial, for example, in regions with less sunlight. To definitively answer the question, one would need to refer to the specific data in their lab manual.
Explanation:The measurements of the F374 allele, the variant of the SLC45A2 gene that is associated with lighter pigmentation, are likely higher in populations where lighter skin would be advantageous due to ecological factors. This might include areas with lower levels of sunlight, such as Northern Europe, where darker skin could lead to deficiencies in vitamin D production. From looking at your lab manual, you would need to determine which of the 14 studied populations show the highest frequencies of this allele. This question is basically looking at how mutations in a specific gene can affect skin color and how the frequencies of these mutations can vary in different human populations.
Learn more about Human Genetics here:https://brainly.com/question/29408544
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The main function of organelles is
translation
protect DNA
move proteins throughout the cell
Answer:
translation
Explanation: USATestprep
Why does some members of the military oppose the idea of a “surgical strike” against Cuba
Answer:
the answer is A, could not be limited
Explanation:
Answer:
A) Could not be limited
Explanation:
I did it on edgenu!ty.
Fragile X syndrome is an X-linked dominantly inherited condition that contributes to autism and intellectual inabilities. Two individuals mate and produce three children. The female is heterozygous for the condition, and the male also displays the condition. What is the probability that the first two children will have the disorder and the last child will not?
Answer:
The probability for the given scenario will be 9/64
Explanation:
As indicated by the above expressed inquiry,
Two people (i.e) male (hetero) + female (hetero)
Prevailing + Latent = First Child
Latent + Dominan = Second Child
Latent + Passive = Third Child
The probability of gene communicating will be low than initial two children.The coding succession is called codons..There are 64 codons for each 3 grouping one amino-acid will be produced.The likelihood of prevailing gene will be 9.
So the appropriate response is 9/64.
substances that have a physiological effect when introduced into the body are?
Answer:
Drugs
Explanation:
Drugs are substances that alter the body function physically or psychologically when introduced into the body.
Drugs could be legal e.g alcohol or illegally e.g cannabis.
Which statement describes a Disadvantage of non renewable resources?
Answer:
A disadvantage of a non renewable resource is that it runs out.
Explanation:
If people become reliant on a non-renewable resource then when it runs out they are screwed.
Final answer:
The main disadvantage of nonrenewable resources like fossil fuels and nuclear power is that they are being used up much faster than they can naturally replenish, leading to potential depletion and environmental problems like global warming and hazardous waste.
Explanation:
A disadvantage of nonrenewable resources is that they exist in fixed amounts and are being consumed at a much faster rate than they can be naturally replaced. This means that as we continue to extract and use these resources, particularly fossil fuels and nuclear power, we are depleting the Earth's reserves, possibly leading to a future where these critical energy sources are no longer available.
Fossil fuels such as petroleum, coal, and natural gas, formed over millions of years, are expected to run out in the foreseeable future due to the current rates of consumption. With oil potentially running out by 2055, natural gas by 2085, and coal by around 2250, reliance on these resources is unsustainable long-term. Additionally, the combustion of fossil fuels contributes significantly to global warming, while nuclear power presents challenges with the safety and disposal of hazardous waste.
The environmental and availability concerns of nonrenewable resources provide a compelling case for the urgent development of sustainable alternative energy sources. These alternative sources, which are typically renewable, seek to replace the energy produced by nonrenewable resources and mitigate their negative impact on the environment.
Long-chain fatty acids have an alkyl chain of more than 10 carbon atoms. Fatty acids with alkyl chains of this length are characterized as ____ in terms of polarity and ___ in terms of size. This length decreases their ability to cross the lipid bilayer of the mitochondrial membrane. As a result, such fatty acids are activated in the cytoplasm by ___ . However, the resulting molecule, fatty acyl-CoA, is still too ____ to cross the membrane. Therefore, a protein system involving carnitine is required for this transport to occur.
This system consists of enzymes ___ (located in the outer mitochondrial membrane) and ____ (located in the inner mitochondrial membrane), which are responsible for the conjugation of the fatty acyl-CoA with carnitine and the delivery of the resulting molecule into the ____ .
The purpose of such a mechanism is that the carnitine molecule is ____ in size compared to coenzyme A but is classified as ____ , which gives it the ability to pass the lipid bilayer. When a fatty acyl group is transferred to carnitine, the conjugate molecule passes into the intermembrane space via the pores in the outer mitochondrial membrane and the inner membrane is crossed with the help of _____ .
Answer:
The answers of blanks are:
---Non-polar
---Large
---Acyl-coA synthetase
---Large
---CPT I
---CPT II
---Mitochondrial matrix
---Small
---polar
---Translocase
Explanation:
Filling up he blanks, we have;
Long-chain fatty acids have an alkyl chain of more than 10 carbon atoms. Fatty acids with alkyl chains of this length are characterized as nonpolar in terms of polarity and large in terms of size. This length decreases their ability to cross the lipid bilayer of the mitochondrial membrane. As a result, such fatty acids are activated in the cytoplasm by acyl CoA synthetase. However, the resulting molecule, fatty acyl-CoA, is still too large to cross the membrane. Therefore, a protein system involving carnitine is required for this transport to occur.
This system consists of enzymes CPT I (located in the outer mitochondrial membrane) and CPT II (located in the inner mitochondrial membrane), which are responsible for the conjugation of the fatty acyl-CoA with carnitine and the delivery of the resulting molecule into the mitochondrial matrix .
The purpose of such a mechanism is that the carnitine molecule is small in size compared to coenzyme A but is classified as polar , which gives it the ability to pass the lipid bilayer. When a fatty acyl group is transferred to carnitine, the conjugate molecule passes into the intermembrane space via the pores in the outer mitochondrial membrane and the inner membrane is crossed with the help of translocase .
Answer: simultaneously the answers includes: nonpolar, large, acyl coa synthetase, large,CPT1,CPT2,mitochondrial matrix,small, polar,translocase.
Explanation:
Long-chain fatty acids have an alkyl chain of more than 10 carbon atoms. Fatty acids with alkyl chains of this length are characterized as NONPOLAR in terms of polarity and LARGE in terms of size. This length decreases their ability to cross the lipid bilayer of the mitochondrial membrane. As a result, such fatty acids are activated in the cytoplasm by ACYL COA SYNTHETASE . However, the resulting molecule, fatty acyl-CoA, is still too LARGE to cross the membrane. Therefore, a protein system involving carnitine is required for this transport to occur.
This system consists of enzymes CARNITINE PALMITOYLTRANSFERASE 1 (located in the outer mitochondrial membrane) and CARNITINE PALMITOYLTRANSFERASE 2 (located in the inner mitochondrial membrane), which are responsible for the conjugation of the fatty acyl-CoA with carnitine and the delivery of the resulting molecule into the MITOCHONDRIAL MATRIX.
The purpose of such a mechanism is that the carnitine molecule is SMALL in size compared to coenzyme A but is classified as POLAR , which gives it the ability to pass the lipid bilayer. When a fatty acyl group is transferred to carnitine, the conjugate molecule passes into the intermembrane space via the pores in the outer mitochondrial membrane and the inner membrane is crossed with the help of TRANSLOCASE .
What are the parts of a seed? Check all that apply.
Answer:
A PINE SEED: Embryo, Cotyledons, Endosperm , and seed coat
A DICOT (BEAN) : embryo, 2 Cotyledons, , and seed coat
A MONOCOT (CORN): Seed coat, single Cotyledons, Embryo, Endosperm
If the population of rabbits in this ecosystem decreased dramatically because of certain environmental changes, which organisms’ food supply would decrease?
Answer:
The organisms' that eat the rabbits
Explanation:
Their food supply will decrease because the rabbit populations are decreasing causing the organisms' that eat the rabbits to look for something else to eat.
The malate/aspartate shuttle: if inhibited, could lead to the accumulation of cytosolic glyceraldehyde-3-phosphate within the cell (assume the glycerol-3-phosphate shuttle, LDH, and GNG are not found in this particular cell). if inhibited, would have little to no effect on the flux of GNG (gluconeogenesis). if inhibited, would result in the accumulation of cytosolic NAD (assume the glycerol-3-phosphate shuttle and GNG are not found in this particular cell). D) if inhibited, would result in the accumulation of cytosolic lactate (assume the glycerol-3-phosphate shuttle and GNG are not found in this particular cell). A) and D). A), B), and D). A), B), C), and D).
Malate aspartate shuttle is a biochemical process for translocating electrons which are produced during glycolysis. Its inhibition can result in accumulation of cytosolic glyceraldehyde-3-phosphate and cytosolic lactate within the cell.
What is Malate aspartate shuttle?Malate aspartate shuttle is an important mechanism to regulate glycolysis and lactate metabolism in heart. It transfers reducing equivalents into mitochondria from cytosol.
The electrons from this enters electron transport chain of mitochondria and generate ATP. Process can be seen in the image attached.
Its inhibition can leads to accumulation of cytosolic glyceraldehyde-3-phosphate within the cell and accumulation of cytosolic lactate
Thus the correct option is A and D
For more details about malate aspartate shuttle, visit:
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Final answer:
Inhibiting the malate/aspartate shuttle can result in the accumulation of cytosolic glyceraldehyde-3-phosphate, cytosolic NAD, and cytosolic lactate in the cell.
Explanation:
The malate/aspartate shuttle is a mechanism that transports NADH molecules produced in the cytosol to the mitochondria for oxidation through the electron transport chain. If this shuttle is inhibited, it can lead to the accumulation of cytosolic glyceraldehyde-3-phosphate within the cell. This is because glyceraldehyde-3-phosphate cannot be further metabolized or transported to the mitochondria without the malate/aspartate shuttle. Therefore, inhibiting the shuttle would limit the cell's ability to generate ATP through glycolysis.
Additionally, inhibiting the malate/aspartate shuttle would not have a significant effect on the flux of gluconeogenesis (GNG) because gluconeogenesis primarily occurs in the mitochondria, while the malate/aspartate shuttle operates in the cytosol. Gluconeogenesis can still proceed using substrates derived from other metabolic pathways.
Inhibiting the malate/aspartate shuttle would also result in the accumulation of cytosolic NAD. This is because NADH produced in the cytosol cannot be transported to the mitochondria for oxidation, leading to an increased ratio of NADH to NAD in the cytosol.
Considering these points, the correct answer is option D) if inhibited, would result in the accumulation of cytosolic lactate (assumed glycerol-3-phosphate shuttle and GNG are not found in this particular cell) and also option A) and D) if we consider the accumulation of cytosolic glyceraldehyde-3-phosphate as well.
The diagram shows what enters and exits a plant during photosynthesis. Which labels belong in the areas marked X and Y? X: Glucose Y: Chlorophyll X: Chlorophyll Y: Glucose X: Carbon dioxide Y: Oxygen X: Oxygen Y: Carbon dioxide
Answer:
X: Carbon Dioxide
Y: Oxygen
Explanation:
Answer:
X: carbon dioxide
Y: oxygen
In the living world there is a great deal of genetic variation The genetic information of dois differs from the genetic information of cats The genetic
information of plant differs from the genetic information of bactors Your genetic informations differs from your father's gonetic
On
formation, and so
Which of the following is least ilkely to be a source of genetic variation among living organisins?
A asexual reproduction
B. mutations
C.meiosis
D
Sexual
reproduction
Answer:
A
Explanation:
Asexual reproduction involves a simple exact copying of the genetic material. An exact copying is FAR less likely to result in mutations, which cause variation. The other three answers lead to mutations which cause variation.
what does kelp eat. this is for marine science but there is no option so I just put biology.
Answer:
Kelp, like other plants photosynthesize. This means they get their "food" or energy rather, from the sun.
Explanation:
Tropical leafcutter ants collect leaf cuttings, which they transport to special underground chambers. There, they chew the leaves to create nursery beds on which they grow a species of fungus they use for food. When ant queens disperse to establish new colonies, they carry the fungus with them, dispersing it as well (this benefits the fungus). In the ants’ nest, the fungus is at risk of being replaced by another fungal species that is able to grow in the same habitat, using the same limited resources. On their bodies, the ants carry and provide a home for bacteria that produce antibiotics the ants use to kill this second fungus and thereby protect their food supply. In this system, the relationship between the two fungal species is:
a. parasitism.
b. predation.
c. mutualism.
d. commensalism.
e. interspecific competition.
Answer:
option E
Explanation:
In the system described above, the relationship that exists between this two different species of fungus is interspecific competition. Interspecific competition is the type of competition that occurs when individuals of different species compete for the same resources available. The two different fungal species is able to grow and utilize the same limited resources in the above description.
Answer:
The correct answer is option e. "interspecific competition".
Explanation:
Interspecific competition is defined as an ecological interaction when different species of the same ecological area compete for resources. This type of interaction is commonly seen in nature, and the relationship of the two fungal species herein described belongs to this interaction. The species of fungus that grows in the nursery beds that ants create compete with the second fungal species that is able to growth the same habitat. Since both species are members of the same ecological area and both compete for the resources that ants offer they have an interspecific competition relationship.
Mltochondria are the
of the cell.
Answer:
powerhouse
Explanation:
The mitochondria are organelles located in the cell and play a crucial role in cell respiration where they produce ATP. ATP is basically the cell's energy source, so mitochondria is commonly known as the powerhouses of the cell because they produce this energy.
Hope this helps!
Explanation:
Powerhouse is your answer :)
What occurs when a peptide bond forms between two amino acids?
Answer:
Water molecule
Explanation:
The carboxylic group of one molecule reacts with the amino group of the other molecule, linking the two molecules and releasing a water molecule.
You are interested in mapping the location of two loci in crested geckos that control different aspects of the crest phenotype. Gene D controls the extension of the crest along the back. 'd codes for spikes that continue along the back past the neck and is recessive to 'D' which codes for spikes that stop at the neck. Gene e controls the length of spikes; short spikes (ee) are recessive to long spikes (E_). (For you crestie fans, yes, this is totally made up).
1. You testcross double heterozygote geckos and observe the phenotypes listed below in the offspring
Phenotypic class Number of offspring Space for notes
Spikes along back and long 79
Spikes along back and short 12
Spikes only on neck and long 10
Spikes only on neck and short 76
a. What was the genotype, in linked gene format, of the double heterozygote parent? Explain how you know.
b. What is the recombination frequency between these two genes? Show your work.
Answer:
a. Ed/eD
b. RF=0.12
Explanation:
The alleles for genes D/d and E/e are:
D_: only on neck spikesdd: along back spikesE_: long spikesee: short spikesAfter testcrossing a double heterozygote (DdEe x ddee) there are 4 types of offspring, two of them much more abundant than the other two. The homozygous recessive parent can only produce ed gametes, so the phenotypes of the offspring depend on the gametes that the double heterozygous parent produced.
The offspring was:
Ed/ed 79 ed/ed 12 ED/ed 10 eD/ed 76Total: 177
a) This result suggests that the genes are linked. Since recombination is a rare event, the most abundant phenotypes always come from the parental gametes, and the least abundant come from the recombinant gametes.
Therefore, the genotype of the doube heterozygote parent was Ed/eD.
b) Recombination frequency (RF) = Recombinants / Total
RF = (12+10)/177
RF = 0.12
Answer:
a. The genotype, in linked gene format, of the double heterozygote parent is De/dE
b. The recombination frequency between these two genes is 0.124
Explanation:
To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, and that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.
In the present example, the genotype of the double heterozygote parent is De/dE.
In this way, we might recognize which are the recombinant gametes produced by the di-hybrid, by looking at the phenotypes with lower frequencies in the progeny.
To calculate the recombination frequency we will make use of the next formula: P = Number of Recombinant individuals/ Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.
In the present example:
Parental) De/dE x de/de
Gametes) De parental type
dE parental type
DE recombinant type
de recombinante type
Phenotypic class Number of offspring
Spikes along back and long, dE, 79--> parentalSpikes along back and short, de, 12 --> recombinantSpikes only on neck and long, DE, 10--> recombinantSpikes only on neck and short, De, 76-->parentalThe recombination frequency is:
P = Recombinant number / Total of individuals
P = 12 + 10 / 79 + 12 + 10 + 76
P = 22 / 177
P = 0.124
The genetic distance between genes is 0.124 x 100= 12.4 MU.
A trait is a characteristic. Think about your physical traits (eye color, skin tone, height, hair, face, allergies, etc.) What traits do you think you inherited or received from your parents?
Answer:
Explanation:
Genetic inheritance occurs due to genetic material in the form of DNA being passed from parents to their offspring, we inherit physical traits from our parents when parents pass copies of their genes to their children and such traits are observable characteristics that children inherit from their parents which can be dominant or recessive inheritance of a single gene.
For example chrildren can inherite : Free or attached ear lobes ,
Eye Colour. Dominant and recessive genes play a role in determining eye colour of the child,brown eyes,blue eye color,green and hazel.
Hair color
Height
Dimples
Handedness
Fingerprints
Hair color
Teeth structure
Freckles
Final answer:
Physical traits like eye color and skin tone are often inherited from our parents, while traits like intelligence and disease susceptibility are influenced by both genetic and environmental factors. Examples of inherited traits include curly hair and tall stature, but personality characteristics can also have a genetic component and be shaped by the environment.
Explanation:
Understanding Inherited Traits
Genetics is the study of how traits are inherited from one generation to the next. Many physical traits, such as eye color, skin tone, and height, are inherited from our parents due to genetic factors. These traits, defined by our genes, make up our genotype, which manifests externally as our phenotype, the observable characteristics.
However, traits are not solely determined by genetics. The environment also plays a significant role in shaping who we are, which is why even traits with a genetic component, such as intelligence or susceptibility to certain diseases, can be influenced by environmental factors. For instance, while a parent might pass down genes that predispose a child to high intellectual capacity, the child's environment, such as access to education, can greatly affect how this potential is realized.
As an example of inherited traits, one might have inherited curly hair and a tall stature from their parents. Conversely, personality characteristics could be influenced not only by genetic predisposition but also by modeling from parents, social interactions, and cultural context. Thus, understanding that a trait can be a feature like bipedalism or a specific genetic sequence, it's clear that traits encompass a wide range of inherited and environmentally influenced characteristics.
Inherited Traits and the Environment
When considering traits we have inherited from our parents, examples might include physical attributes like hair color or facial features. Personality characteristics often have a genetic component, with research suggesting that attributes such as extraversion or introversion can be inherited. Yet, these same traits are also shaped by environmental factors, or modeling, as we grow and interact with the world around us. An individual's tendency towards kindness, for instance, may be influenced by being raised in a nurturing environment, regardless of their genetic predispositions
Based on the information in the article, which title
would be most appropriate?
In 2012, astronomers reported the discovery of a planet
that revolved around two Sun-like stars. Kepler-34b, a
gaseous planet, revolves around the stars every 289
days, and the stars revolve around each other every 28
days. This information was presented in the January 11,
2012. issue of the journal Nature.
O New Planet Found in Open Cluster System
O New Planet Discovered in Binary Star System
Answer: New planet discovered in binary star system
Explanation:
Answer:
B)New planet discovered in binary star system
Explanation:
I just got it right on the edge
. What is the main purpose of mRNA is to? *
1 point
act as a readable copy of DNA which takes the message to produce proteins to the ribosomes
act as DNA and goes to the golgi body to make proteins
it is a type of RNA with mutations
describes genetic traits and characteristics
Answer:
act as a readable copy of DNA which takes the message to produce proteins to the ribosomes
Explanation:
mRNA is a copy of DNA that is small enough to leave the nucleus and deliver instructions to make proteins to the ribosome.