A mutation is found in a tRNA-encoding gene. The wild type allele produces a tRNA that recognizes the codon GAA, and is charged with the amino acid Glutamic acid. The mutant tRNA is still charged with Glu, but the anticodon is mutated such that it recognizes the codon UAA. What effect will this have on translation in these cells
It will bring about a stop to the translation process
The mutant tRNA despite still being charged with Glu, since it's anticodon is mutated and then recognizes and reads the codon UAA which is one of the stop codons (UAA, UAG and UGA) on the mRNA strand causes the translation process of that particular mRNA strand to stop. And the growing polypeptide chain to be released if any from the ribosomes.
The anticodon will be unable to recognize the mRNA codon that is GAA, and the translation of this protein/polypeptide will be abruptly stopped. This may result in a truncated protein which is defected and hence, will be degraded by the relevant mechanisms in place. Since, UAA is actually one of the stop codons, the ribosome will not continue onward with the translation and fall off the mRNA.
A researcher is studying gene expression in different types of cells. For a gene of interest, he notes its expression no longer appears to be regulated—it occurs in all cells, rather than in just a few cell types. The gene’s corresponding transcript and protein, however, appear normal. This may be the result of a mutation in which of the following regions associated with the gene? (5 POINTS)Silencer
9What role does the MutL protein play in DNA repair? (5 POINTS)
It performs a proofreading function, similar to DNA polymerase, and removes incorrect bases from the template DNA strand.
It recognizes and binds to a pair of "mismatched" nucleotides, preventing their translation.
It recognizes the newly-synthesized DNA strand (e.g., by detecting nicks), ensuring only it—and not the template strand—is repaired.
It helps to link together regions of DNA, after an error has been removed via exonuclease.
10If a mutation occurs that affects the large subunit of a ribosome, rendering it inactive, how would this affect translation? (5 POINTS)
The small subunit of the ribosome would fail to associate with a mRNA.
Methionine tRNA would not recognize the start codon of a mRNA.
New amino acids would not be added after the initial methionine, so translation would be effectively inhibited.
Arginine tRNA, rather than methionine tRNA, would associate with a mRNA’s stop codon.
8) The region associated with the gene is : ( A ) Silencer
9) The role played by Mutl protein in DNA repair is : ( C ) It recognizes the newly-synthesized DNA strand (e.g., by detecting nicks), ensuring only it—and not the template strand—is repaired.
10) The effect on translation is : ( C ) New amino acids would not be added after the initial methionine, so translation would be effectively inhibited.
Function of the silencer
The silencer is the region of the DNA that regulates the expression of DNA in RNA and protein, since the mutation cannot regulate the expression of genes the region associated with this is the silencer.
Mutl proteins helps in identifying template DNA strand and newly synthesized DNA strand. they are also used in the treatment of methyl mismatch. also
Translation is inhibited when a large unit of ribosomes is affected by a mutation.
Hence we can conclude that The region associated with the gene is : Silencer. The role played by Mutl protein in DNA repair is :It recognizes the newly-synthesized DNA strand (e.g., by detecting nicks), ensuring only it—and not the template strand—is repaired.The effect on translation is New amino acids would not be added after the initial methionine, so translation would be effectively inhibited.
9- It recognizes the newly-synthesized DNA strand (e.g., by detecting nicks), ensuring only it—and not the template strand—is repaired
10- New amino acids would not be added after the initial methionine, so translation would be effectively inhibited.
As the question is, mutation is unable to regulate gene expression and occurs in all cells. Thus, silencer are the DNA control region and regulate the expression of DNA into RNA and protein. Hence, option a is correct.
This protein is used to differentiate between tempelate DNA strand and newerly synthesized DNA strand to rectify the foundation incorporated during DNA replication; it helps to identify new synthesizes of DNA. MutL Proteines are a proteins used to treat a methyl mismatch repair system. Hence, option c is correct.
If the large unit of ribosome is affected by a mutation, new amino acid is not introduced after inhibition of the original methionine and translation. Hence, option c is correct.
Which is the answer for it? I will appreciate the help
What do U need help with
In this activity, you will use presentation software to create a presentation of 15 to 20 slides explaining your arguments in support of or against the following statements:AIDS continues to be a global health crisis. AIDS still meets the definition of a pandemic.
Your claims should be supported by scientifically complete evidence. Follow the steps provided to research, plan, and create your presentation. This guide about the research process can help.
Time to complete: 2 to 3 hours
Part A: Ask Questions Begin by creating a list of questions to guide your research and help you form a convincing argument. Your presentation should answer some of these questions: What is the definition of a pandemic? How does a pandemic differ from an endemic or epidemic? What is the life cycle of the HIV virus once it enters the body? How does the genetic code of the virus change? How does HIV affect the body? How do people become infected with HIV? How does AIDS develop from an HIV infection? What are the current infection rates of HIV across the globe? Where is HIV/AIDS most prevalent? What treatments are currently available for HIV/AIDS? Do some regions of the world have better access to treatments than others? What is the average life expectancy for someone with HIV? Does life expectancy differ around the globe? Now write four additional questions about HIV/AIDS that could help strengthen your argument in the presentation.
Based on the information given, a pandemic simply means the spread of an infectiousdisease.
What is HIV?
It should be noted that the difference between an epidemic and a pandemic is the degree if its spread.
HIV is simply a virus that attacks the immune system. It can be gotten through unprotected intercourse, sharing sharp objects, etc. It should be noted that in the long run, HIV leads to AIDS.
It should be noted that HIV is most prevalent in Sub Saharan Africa. There are presently no cure for HIV but there are medications to control the complications.
The average life expectancy for someone with HIV is 56 years.
A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?
The best way of solving this is to draw a Punnett square.
You know the F0 had one parent with singed bristles (s), and normal wings (L), and the other parent is normal bristles (S) with vestigial wings (l).
If you do the cross ssLL x SSll you'll find 100% of the offspring is F1: SsLl, this means, all of them show the dominant traits: normal wings and normal bristles.
If you cross two parents from F1 to have F2, you'll find:
SsLl x SsLl = SSLL + SslL + sSlL+ ssll = 25% SSLL,all dominant traits. 50% SsLl is a recessive trait carrier but shows dominant traits. 25% ssll this one has all recessive alleles, which means, it will show vestigial wings and signed bristles.
The expected ratio of phenotypes among the progeny of a test cross is 1:1:1:1. Out of 200 total resulting progeny, 48 occur in one of the four phenotypic classes. Given this information, which of the following must also be true?1) Since 48 is so close to the expected value, there is no need to calculate chi square before drawing a conclusion about the ratio. 2) At least one additional cell must also contain a count of 48. 3) The value of observed - expected for this cell = -2. 4) The progeny of this cross do not conform to a 1:1:1:1ratio.
Answer: The progeny of this cross do not conform to a 1:1:1:1 ratio
Explanation: This is because out of the 200 total resulting progeny, we must have 50 in each phenotype class to conform with the 1:1:1:1 which is not so as we have 48 in one of the phenotypic class already. Therefore, it did not conform to the ratio.
The value of observed - expected for this cell = -2
"With a total of 200, the expected number in each cell when the predicted ratio is 1:1:1:1 = 50. The observed number is 48." - Mastering Genetics
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Which of the following can be true about invasive species?B Invasive species are not a a threat to new environments. А Invasive species increase the biodiversity of an ecosystem с Invasive species can help destroy other invasive species in an ecosystem