a.
To determine:
The reason duuue to which X chromosome will convert into Barr body.
Introduction:
The process of
b.
To determine:
The explanation for interference in the expression of Xist present on the same chromosome as Tsix.
Introduction:
The chromosome from chromatids is formed through the process of supercoiling. The chromatin fibers are condensed into thick structures called the chromosomes. This chromosome in bacteria is circular, whereas in eukaryotes it is linear.
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Genetics: From Genes to Genomes
- A gene encodes a protein with the following amino acid sequence: Met-Trp-His-Arg-Ala-Ser-Phe A mutation occurs in the gene. The mutant protein has the following amino acid sequence: Met-Trp-His-Ser-Ala-Ser-Phe An intragenic suppressor mutation restores the amino acid sequence to that of the original protein: Met-Trp-His-Arg-Ala-Ser-Phe Give at least one example of base changes that could produce the original mutation and the intragenic suppressor.arrow_forwardA gene encodes a protein with the following amino acid sequence: Met-Trp-His-Arg-Ala-Ser-Phe A mutation occurs in the gene. The mutant protein has the following amino acid sequence: Met-Trp-His-Ser-Ala-Ser-Phe An intragenic suppressor mutation restores the amino acid sequence to that of the original protein: Met-Trp-His-Arg-Ala-Ser-Phe Give at least one example of base changes that could produce the original mutation and the intragenic suppressor. (Consult the genetic code in Figure 15.10.)arrow_forwardZidovudine (AZT) is a drug used to treat patients with AIDS. AZT works by blocking the reverse-transcriptase enzyme used by the human immunodeficiency virus (HIV), the causative agent of AIDS. Do you expect that AZT would have any effect on transposable elements? If so, what type of transposable elements would be affected, and what would be the most likely effect?arrow_forward
- A gene encodes a protein with the following amino acid sequence: Met-Lys-Ser-Pro-Ala-Thr-Pro A nonsense mutation caused by a single-base-pair substitution occurs in this gene, resulting in a protein with the amino acid sequence Met-Lys. An intergenic suppressor mutation allows the gene to produce the fulllength protein. With the original mutation and the intergenic suppressor present, the gene now produces a protein with the following amino acid sequence: Met-Lys-Cys-Pro-Ala-Thr-Pro Give the location and nature of the original mutation and of the intergenic suppressor.arrow_forwardSex determination in Drosophila is controlled by a cascade of alternative splicing events. Sex-lethal (Sxl) is involved in regulating splicing of the transformer gene, as outlined below (picture attached). According to this schematic, Sxl functions as a: Splicing enhancer Splicing repressor Splicing activator Splicing silencerarrow_forwardIn humans, the average length of a gene on chromosome 7 is 69,000 base pairs, but the average MRNA length on chromosome 7 is only 2,500 base pairs. Chromosome 7 has 863 protein-coding genes, but many more than 863 proteins are made from the genes on chromosome 7. The observation that the length of the average gene on chromosome 7 is longer than the average mRNA length is explained by the process of while the observation that more proteins are produced from the genes on chromosome 7 than there are genes can be explained by the process of O heterogeneity; complementation splicing of introns; alternative splicing O chromatin remodeling; micro RNA binding acetylation; methylation alternative splicing; splicing of intronsarrow_forward
- Inhibitors for this reverse transcriptase fall in two classes: nucleoside analog inhibitors (NRTIs) or non-nucleoside analog inhibitors (NNRTIs). NRTIs have a similar structure to nucleosides and block the active site of the reverse transcriptase, whereas NNRTIs don't look like nucleosides and bind to an allosteric pocket in the reverse transcriptase. Presently NNRTIs are used more often than NRTIs, as NRTIs have more severe side effects. Why do you think NRTIs would have severe side effects?arrow_forwardThe MAT locus allows yeast to switch mating type through a very complex mechanism. However, it has informed us a great deal about what aspects of gene expression typical to all organisms? Options: higher order changes in chromatin affect transcriptional efficiency that general transcription factors must first bind directly to histone tails and only then can they interact with their cognate binding sites that DNA methylation is involved in this silencing mechanism that SIR2 is required for all types of transcriptional repression that expression of Pol III genes provides a means of identifying active chromatinarrow_forwardBased on the partial diploid sketched shown in Figure 3, where cross-hatched rectangles and the red bar indicate mutations that inactivate that particular genetic element of the operon, state if expression of Lacl (Lac repressor) and LacZ (Beta-Galactosidase) is (select one of the options below for each genetic element and briefly justify your answer): constitutively ON (in the presence/absence of lactose) constitutively OFF (in the presence/absence of lactose) • inducible by lactose • not enough information given to conclude • none of the above FIGURE 3: lacl laco lacz- lacY lacl laco lacz lacYarrow_forward
- Describe the 4 types of suppressor high copy suppression, bypass suppression, nonsense suppression, protein interaction . in your Own words. Often geneticists look for suppressors to find interactive proteins. Which of the type(s) of suppressors you put for part a will help to identify interacting proteins, and which type(s) will not? What are two (or one, if we don’t get a chance to talk about two of them in class) other techniques (not necessarily “genetic” techniques, but at least, lab techniques) that help to identify identifying proteins?arrow_forwardThe length of dogs’ legs varies widely across species, with the Nova Scotia duck tolling retriever representing a short-legged breed. Genome-wide association studies (GWAS) linked this phenotype to a locus on Canine familiaris (CFA) chromosome 12 with strong similarity to the FGF4 gene on chromosome 18. Further work demonstrated that both loci are transcribed. Note: CpG islands (green in diagram) often mark active promoters; we will discuss in a later lecture. Based on the diagram below, what do you think is the relationship between these two loci? The genome sequence shows another locus on chromosome 10 that is almost identical to the one depicted above on chromosome 12, from 5’UTR through 3’UTR. The gene does have one polymorphism, though, in exon 2, that distinguishes it from the Chr. 12 gene. You have developed an assay to easily distinguish mRNA derived these two variants. However, when you assay the Chr. 10-derived mRNA, you don’t detect anything at all.…arrow_forwardThe following double-stranded DNA sequence is part of a hypothetical yeast genome which contains a very small gene. Transcription starts at the Transcription Start Site (TSS), proceeds in the direction of the arrow and stops at the end of the Transcription Terminator (green box). 5' 3' TSS CTATAAAAATGCCATGCATTATCTAGATAGTAGGCTCTGAGAAATTTATCTCACT | | | | | | | | | | GATATTTTTACGGTACGTAATAGATCTATCATCCGAGACTCTTTAAATAGAGTGA - 5' PROMOTER TERMINATOR 3' a) Which strand (top or bottom) is the template strand? Explain why. b) What is the sequence of the mRNA produced from this gene? Label the 5' and 3' ends. c) What is the sequence of the protein produced from the mRNA? d) If a mutation (an insertion) were found where a T/A (top/bottom) base pair were added immediately after the T/A base pair shown in red, what would be the sequence of the mRNA? What would be the sequence of the protein?arrow_forward
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