Question

17. A diploid plant is a dihybrid for flower color (gene “R”) and leaf size (gene “L”). Its
phenotype is red flowers and large leaves. This plant is crossed to a tester plant (which has
white flowers and small leaves). The progeny is as follows:
23 Red, large
25 White, small
230 Red, small
235 White, large
What can be concluded about the linkage relationship of the loci in the dihybrid parent?
A) Genes R and L are unlinked
B) The dihybrid parent’s genotype is RL/rl
C) The dihybrid parent’s genotype is Rl/rL
D) Genes R and L are linked and 9 map units apart
E) C and D

18. Over the years, a person’s skin cells are exposed to UV damage leading to pyrimidine dimers.
Which DNA repair mechanism repairs this damage?
A) Double-strand break repair
B) Non-homologous end joining
C) Mismatch repair
D) Base-excision repair
E) Nucleotide-excision repair

17. A diploid plant is a dihybrid for flower color (gene "R") and leaf size (gene "L"). Its phenotype is red flowers and large leaves. This plant is crossed to a tester plant (which has white flowers and small leaves). The progeny is as follows: 23 Red, large 25 White, small 230 Red, small 235 White, large What can be concluded about the linkage relationship of the loci in the dihybrid parent? A) Genes R and L are unlinked B) The dihybrid parent's genotype is RL/rl C) The dihybrid parent's genotype is RI/L D) Genes R and L are linked and 9 map units apart E) C and D 18. Over the years, a person's skin cells are exposed to UV damage leading to pyrimidine dimers. Which DNA repair mechanism repairs this damage? A) Double-strand break repair B) Non-homologous end joining C) Mismatch repair D) Base-excision repair E) Nucleotide-excision repair

Answer 1

Hello,

18.Exposure to UV can lead to mutations in DNA.UV exposure mainly leads to the formation of pyrimidine dimers especially thymine dimers.This can be repaired by Nucleotide-excision repair mechanism.In this the thymine dimers are removed and correct bases are added.Helicase enzyme unwind the DNA at the given site,incorrect bases are removed,then,DNA polymerase adds the bases and finally the nick is ligated by DNA ligase.So,option E is correct.

Hope this helps.