IDNLearn.com: Your go-to resource for finding precise and accurate answers. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
In yeast, mating requires the activation of two protein kinases encoded by the STE11 and STE7 genes. Ste 11 phosphorylation of Ste7 activates Ste7 and the activated Ste7 then phosphorylates downstream proteins. Deletion of either STE11 (ste 11A) or STE? (ste7A) makes cells sterile, that is unable to mate. Constitutively active mutant versions of STE11 (STE11-c) and STE7 (STE7-c) exist (that is, they are active under all conditions) and are able to mate. Based on this information which of the predictions is MOST LIKELY to be TRUE? A. A haploid containing ste11A ste 7A will be able to mate. B. A haploid containing ste11A STE7-c will be able to mate. C. A haploid containing STE11-c ste 7A will be able to mate. D. A haploid containing STE11-с STE7-c will be unable to mate. E. A haploid containing STE11 STE7 will be unable to mate.
Sagot :
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
