Plant Pathology Feud

(Answers to questions posed in the Powerpoint presentation)

Question 1. DNA

Question 2. A ribosome

Question 3. Click here for more information.

Question 4. Mendel

Question 5. Rudolph Virchov

Question 6. A high number of permutations.....

Question 7. This was an exercise where students had two 3' lengths of thread and had to fit them into an empty gelatin capsule. The successful ones twisted the threads together. This demonstrated the way DNA is packed into the nucleus.
Question 8. Primase

Question 9. The answer was number 1 but this will vary for different classes. This question was asked earlier in the semester and most were indifferent at that time.

Question 10. In this exercise, envelopes contained varying amounts of polymerases, amino acids, mRNA, etc. The whole class was considered a cell and teams had to scramble about to different groups to find what they needed. This demonstration was used to give students a feel for what happens in a cell.

Question 11. Beadle and Tatum

Question 12. Three nucleotides that code for an amino acid.

Question 13. A nucleotide


Question 15. ribosomal, messenger, genomic (viruses)

Question 16. Multiple replication forks

Question 17. Transformed with a beta carotene gene

Question 18. mRNA is made from DNA

Question 19. Transduction

Question 20. Plasmid

Question 3. The Mendel "envelope"

Mendel's cross between the pea genotypes of Ss and Ss yielded the following phenotypes and genotypes.

Phenotype: Smooth Smooth Smooth Shrunken
Genotype: SS  SS SS ss

Obviously, in Mendel's time, molecular probes were not available. Mendel could only observe the phenotype. He saw seeds that were in a three to one ratio of smooth to shrunken. Thus, from the phenotype (physical appearance) he had no way of visually knowing that two of the three smooth seeds had a genotype of Ss.

Devise a non-molecular approach to determining the genotype of the smooth seeds.

Answer: Backcross each phenotype with parent

Diagram your procedure here:

SS X ss = 4 smooth Ss

Ss X ss = 1 smooth Ss, 3 rough ss

ss x ss = 4 rough ss

If cross F1's with SS

SS x SS = 4 smooth SS

Ss x SS = 3 smooth SS, 1 smooth Ss (4 smooth hence no informational gain)

ss x SS = 4 smooth Ss (no informational gain)

Rule: F1 backcross with recessive parent is necessary to delineate genotype.


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Updated: July 7, 2001