1) For each of the following partial diploids, give the cellular phenotypes for A and B. Explain.
A B
a. for a negatively regulated gene:
F’pR- poA-B/ pR po-AB-
F’p-R p-oA-B/ pR- poAB-
b. for a positively regulated gene:
F’pR- poA-B/ pR po-AB-
F’p-R p-oA-B/ pR- poAB-
2) Would attenuation of the type seen in the tryp operon of E. coli be likely to occur in humans? Why or why not?
3) In the yeast gal gene system, in which genes should you be able to isolate mutations which can not turn on the gene?
4) In the developing Drosophila embryo system, what kind of mutation would give rise to:
a. an embryo with 3 abdominal segments missing
b. a fly with the correct number of segments, but had two additional thoracic segments and two fewer abdominal segments.
5) How is it possible for an organism with 20,000 genes to produce millions of different (protein) antibodies?
I was wondering if you could help out with any of these questions to help me study for a test. Thank You
Of course! I'll be happy to help you with your questions. Let's go through each question one by one and explain how to get the answers:
1) For each of the following partial diploids, give the cellular phenotypes for A and B. Explain.
In order to determine cellular phenotypes for A and B, we need to analyze the genotypes of the given partial diploids.
a. For a negatively regulated gene:
- Genotype: F’pR- poA-B/ pR po-AB-
Explanation: In this case, A is negatively regulated. The presence of the repressor protein encoded by the pR gene (pR-) prevents the activation of the A gene, resulting in the A- phenotype. However, the lack of the repressor protein encoded by the pR gene (pR) allows the A gene to be activated, resulting in the A+ phenotype. For B, in both cases, the presence of the repressor protein encoded by the poB gene (poB-) prevents the activation of the B gene, resulting in the B- phenotype.
b. For a positively regulated gene:
- Genotype: F’pR- poA-B/ pR po-AB-
Explanation: In this case, A is positively regulated. The absence of the repressor protein encoded by the poA gene (poA) allows the A gene to be activated, resulting in the A+ phenotype. However, the presence of the repressor protein encoded by the poA gene (poA-) prevents the activation of the A gene, resulting in the A- phenotype. For B, in both cases, the absence of the repressor protein encoded by the pR gene (pR) allows the B gene to be activated, resulting in the B+ phenotype.
2) Would attenuation of the type seen in the tryp operon of E. coli be likely to occur in humans? Why or why not?
To determine if attenuation of the type seen in the tryp operon of E. coli is likely to occur in humans, we need to understand the concept of attenuation.
Attenuation is a regulatory mechanism that allows bacteria to rapidly respond to changing nutrient conditions. It involves the premature termination of transcription depending on the availability of certain amino acids.
In E. coli, the tryp operon is regulated by attenuation, which controls the production of enzymes involved in tryptophan synthesis based on the levels of tryptophan in the environment. However, in humans, tryptophan metabolism and regulation are controlled in a different way, primarily through feedback inhibition and direct regulation of gene expression rather than attenuation.
Therefore, attenuation of the type seen in the tryp operon of E. coli is unlikely to occur in humans due to differences in metabolic regulation mechanisms.
3) In the yeast gal gene system, in which genes should you be able to isolate mutations which can not turn on the gene?
In the yeast gal gene system, the genes that are responsible for turning on the expression of the gene are called regulatory genes. To isolate mutations that cannot turn on the gene, we need to identify the specific regulatory genes involved.
In the case of the yeast gal gene system, the gene GAL4 is a key regulatory gene that activates the expression of other galactose metabolism genes. Therefore, if there are mutations in the GAL4 gene or its regulatory elements, it can result in the inability to turn on the gene.
Therefore, to isolate mutations that cannot turn on the gene in the yeast gal gene system, you should focus on mutations in the GAL4 gene or its regulatory elements.
4) In the developing Drosophila embryo system, what kind of mutation would give rise to:
a. an embryo with three abdominal segments missing?
Explanation: In the developing Drosophila embryo, the number of segments and their identity are determined by the expression of specific genes called homeotic genes. Mutations in these homeotic genes can result in changes in segment identity or missing segments. In this case, a mutation in a homeotic gene responsible for abdominal segment development would result in an embryo with three abdominal segments missing.
b. a fly with the correct number of segments, but had two additional thoracic segments and two fewer abdominal segments?
Explanation: Similar to the previous question, this scenario also involves mutations in homeotic genes that control segment identity. In this case, a mutation in a homeotic gene responsible for specifying the identity of thoracic and abdominal segments can result in a fly with two additional thoracic segments and two fewer abdominal segments.
5) How is it possible for an organism with 20,000 genes to produce millions of different (protein) antibodies?
In organisms with a relatively small number of genes, such as humans with approximately 20,000 genes, the ability to produce a diverse range of antibodies lies in a process called V(D)J recombination.
V(D)J recombination is a DNA rearrangement process that occurs during the development of immune cells, specifically B cells and T cells. It involves the recombination of variable (V), diversity (D), and joining (J) gene segments to generate a unique epitope recognition site on immunoglobulins or antibodies.
By rearranging the V, D, and J segments in different combinations, the immune system is able to produce a vast repertoire of antibodies, each with a unique antigen-binding site. The combinatorial possibilities of V, D, and J segments result in the potential for millions of different antibody sequences from a limited set of genes.
This process allows the immune system of organisms with a limited number of genes to recognize and respond to a wide variety of foreign antigens effectively.
I hope this helps you with your studies! If you have any more questions, feel free to ask.