answer question 27,28,29,30 please thank you so much. 27. A homozygous tomato pl
ID: 133630 • Letter: A
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answer question 27,28,29,30 please thank you so much.
27. A homozygous tomato plant with red fruit and yellow fowers was crossel with homozygous tomato plant with golden fruit and white fowers The FI al fruit and yellow flowers. The FI were testcrossed and the offspring table below. How many map units separate these genes? had red are shown in the Phenotype Red fruit, yellow flowers Red fruit,white flowers Golden fruit, yellow flowers Golden fruit, white flowers a. b. 8 Number of plants 41 15 d. 44 28) In the lab, you are recreating Griffith's experiment. The table below shows your experimental groups and results Experimental Bacteria Mouse Result Group Living S LivingR Dead Healthy Heat-killed S+ Living R Heat-killed R+Living S What results to you expect for experimental groups 3 and 47 a.) Both "healthy" Both "dead" 3 "healthy": 4 "dead 3 "dead: 4 "healthy 29,In the Hershey Chase experiment, what was labeled with radioactive phosphorus (32p)? a. Viral DNA b. Viral proteins d. Bacterial DNA- re is a mutation in a gene called dnaB that alters the helicase enzyme that normally acts at the origin of replication. expect as a result of this mutation? 30. In E. coli, the Single-strand binding proteins will expose the DNA template Replication will occur by the conservative model instead of the semiconservative model. a. formed o replication bubble will be A replication bubble will form but no DNA polyme begin DNA replication. rase will be recruited to d.Explanation / Answer
Please find the answers below:
Answer 27: Choice c (the total recombination frequency accounts for the distance between the genes, i.e. 8+7=15 map units apart)
Answer 28: Choice b (the rough strain is avirulent, hence the mouse lives whether its alive or heat-killed. However, transformation of the rough strain prior to heat-killing of smooth strain is still able to kill the mouse as the smooth strain is virulent in nature)
Answer 29: Choice b (since the DNA contains phosphate groups, it was radiolabelled with active phosphorus. The proteins were labelled with radioactive sulphur)
Answer 30: Choice c (since the normal activity of helicase is essentially required for initiation of DNA replication by opening the helix, failure of helix opening secondary to enzymatic mutation results in failure of replication of the DNA)
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