In the dalmatians, there is a single gene that encodes for the protein that make
ID: 50854 • Letter: I
Question
In the dalmatians, there is a single gene that encodes for the protein that makes the coat pigments for the black fur spots. You are studying how this gene is regulated and isolated the DNA that encodes the gene as well as the processed mRNA molecule transcribed from the gene.
With the isolated DNA and mRNA molecules, you mix them in a test tube to hybridize them to one another, meaning that the DNA and mRNA molecules pair and align with one another based on sequence complementarity. Then, using a high-powered specialized microscope you take a picture of the mRNA paired with the DNA, which is shown below. In the image, the DNA is black and the mRNA is blue.
(A) You notice that there are loops in the DNA strand that are not paired with the mRNA molecule. Based on what you know about the similarities and differences between the DNA sequence and processed mRNA sequence for a given gene, explain what the loops in the DNA strand could be.
(B) Use what you know about transcription and mRNA processing to show on the diagram (label) which side of the mRNA is its 5´ and which is its 3´ end and explain how you can identify which end is which.
(C) A mutant spot gene was isolated that does not make pigment, and thus the animals have no spots. This mutant gene does make mRNA encoding the protein at similar levels as the wild type gene, but the mRNA molecules are slightly longer than usual. You repeat an experiment mixing and hybridizing DNA of the mutant gene with mRNA transcribed from the mutant gene and take a picture of it, shown below.
Based on your answers above and changes in the pattern of pairing between the wildtype DNA/mRNA molecules and the mutant DNA/mRNA molecule, suggest a possible explanation for (1) the nature of the mutation, (2) the effect it would have on the mRNA sequence, and (3) the effect it likely will have on the protein product.
Explanation / Answer
transcriptional process where the DNA is transcripbed into the m RNA with the help of polymerase enzyme . it is the first step in transcription .both DNA and RNA are complemetary ,DNA sequence is read by the RNA polymerase which produces the complementary antiparalell strand called primary transcript . A DNA transcription unit encoding for the protein may contain both coding sequence which is translated into protein . the regulatory sequence before the coding seqence is five prime untranslated region and seqence for the coding sequenc eis 3 prime untranslated region . direction of reading transcritpiton is 3 prime to 5 prime ,so mRNA produced will be in 5 prime to 3 prime . RNA polymerase can add nucleotides in only in three prime directions. the m RNA produced will have uracil instead of the thymine .
Answer for the option a ;
the reulation of transcription occurs through the positive control and in some it s through the negative control DNA looping is involved in regulation of transcription in prokaryotes and in eukaryotes the effect of enhancer appear to form the loop .loop formation may have diffrent functions . loop itself can act as requisite for the regulation while in others it increases the local concentartion of the proteins .formation of loops may result in transfer of protein form one segment to another there by facilitating in search of target site within the DNA .so the DNA looping is exactly the protein -protein complexes binding to different regions in the DNA structure . the proteins interact with each other forms the loop .DNA looping is used to increase the specificity and affinity of the interactions .so to conclude the DNA loops are formed where the proteins away from the genescan be bought togather by the formation of the DNA loop which there by they control the gene expression.
Answer for b ;
mRNA are the molecules that covey the genetic information from DNA to the ribosome . these are stored in the form of codons consists of 3 bases .each code for specific aminoacid except for the stop codon . the RNA produced form the DNA contains the coding and non coding regions ,this is called pre m RNA which undergoes processing to form the matured m RNA . The five prime end of them RNA contain the five methy gaunine cap which will protect {starting portion is the five prime end } , and the polyadneylation ,where addition of adenine molecules to mRNA is the three prime end { bend portion towards the right side in the above diagram}.
answer for C;
the type of mutation will be point mutation . the muation occured in the splicing site of the intron ,so thi s will inetrfere with the correct splicing of mRNA so the mRNA is slightly longer than the wild type.