Gene mutations and what they do to a protein
Grade 9 Science Worksheet
Circle the best answer.
Read this first
Original mRNA (model, not to scale), 5′ to 3′, cut into codons
AUG GGA CUU AAG UAA. Changed mRNA (model), 5′ to 3′, cut into codons: AUG GGA CCU AAG UAA. The marked letter is base 8 of the changed mRNA. Table excerpt of the standard genetic code: AUG, methionine; GGA, glycine; CUU, leucine; AAG, lysine; UAA, stop; CCU, proline.
Ari translates both model mRNAs. How do the two chains compare?
- The changed chain stops after glycine, because CCU ends the chain
- The chains match, because one letter is too small to matter
- Both start the same, but the amino acids after the change are different
- Both have four amino acids; leucine is replaced by proline
Original mRNA (model, not to scale), 5′ to 3′, cut into codons
AUG UCU GCU GGA UAA. Changed mRNA (model), 5′ to 3′, cut into codons: AUG UCA GCU GGA UAA. The marked letter is base 6 of the changed mRNA. Table excerpt of the standard genetic code: AUG, methionine; UCU, serine; GCU, alanine; GGA, glycine; UAA, stop; UCA, serine.
Bo translates both model mRNAs. How many amino acids differ between the chains, and what kind of change is it?
- One amino acid differs; the change is missense
- No amino acid differs, but the protein is harmed anyway
- Two amino acids are lost; the change is nonsense
- No amino acid differs; the change is silent
Original mRNA (model, not to scale), 5′ to 3′, not yet cut
AUGGAUCUGAAACCUGGC. Copy (model), 5′ to 3′, not yet cut: AUGGAUCCUGAAACCUGGC. The marked letter is base 7 of the copy. Read six codons of each strand, starting at AUG. Table excerpt of the standard genetic code: AUG, methionine; GAU, aspartic acid; CUG, leucine; AAA, lysine; CCU, proline; GGC, glycine; GAA, glutamic acid; ACC, threonine; UGG, tryptophan.
Jun compares a model mRNA with a copy that has one extra base (marked). Both are read in threes from AUG. How many of the six amino acids differ?
- 5: the amino acids after methionine
- 1: just the amino acid at the insertion
- 4: the amino acids from the insertion onward
- 0: the extra base is not read as part of a codon
Original mRNA (model, not to scale), 5′ to 3′, not yet cut
AUGCCUGAUAAAUUUGGA. Copy (model), 5′ to 3′, not yet cut: AUGCCUAUAAAUUUGGA. The marked letter is base 7 of the original mRNA. Read five codons of each strand, starting at AUG. Table excerpt of the standard genetic code: AUG, methionine; CCU, proline; GAU, aspartic acid; AAA, lysine; UUU, phenylalanine; GGA, glycine; AUA, isoleucine; AAU, asparagine; UUG, leucine.
Kira’s copy of a model mRNA lacks the marked base. Five codons of the copy are read. How many of its five amino acids differ from the original’s first five?
- 1: just the amino acid at the deletion
- 3: the amino acids from the deletion onward
- 5: the first five amino acids
- 2: the amino acids on each side of the deletion
Original mRNA (model, not to scale), 5′ to 3′, not yet cut
AUGGGAGACCACAACUUC. Copy X (model), 5′ to 3′, not yet cut: AUGCGGAGACCACAACUUC. Copy Y (model), 5′ to 3′, not yet cut: AUGGGAGACCACAACCUUC. The marked letter is base 4 of copy X and base 16 of copy Y. Read six codons of each strand, starting at AUG. Table excerpt of the standard genetic code: AUG, methionine; GGA, glycine; GAC, aspartic acid; CAC, histidine; AAC, asparagine; UUC, phenylalanine; CGG, arginine; AGA, arginine; CCA, proline; CAA, glutamine; CUU, leucine.
Luz compares two copies of a model mRNA, each with one extra base (marked). Which copy changes more amino acids, and why?
- Y: its extra base is nearer the end, so more is lost
- X: more codons come after it and are read in the shifted frame
- Same number: each copy has one extra base, so one amino acid changes
- Neither: an extra base is not read, so the chains are unchanged
Original mRNA (model, not to scale), 5′ to 3′, cut into codons
AUG UGG CAG AAG UAA. Copy (model), 5′ to 3′, not yet cut: AUGUGGCAGACGAAGUAA. The marked letters are bases 9 to 11 of the copy. Table excerpt of the standard genetic code: AUG, methionine; UGG, tryptophan; CAG, glutamine; AAG, lysine; UAA, stop; ACG, threonine.
Milo compares a model mRNA with a copy that has three extra bases (marked). How do the two chains compare?
- 5 amino acids: threonine is new and the others keep their order
- 4 amino acids: just the single amino acid at the insertion site changes
- 7 amino acids: each extra base adds one amino acid
- 5 amino acids: the ones after the insertion are different
Original mRNA (model, not to scale), 5′ to 3′, cut into codons
AUG GAU CUG AAA GGC UAA. Copy (model), 5′ to 3′, not yet cut: AUGGAGAAAGGCUAA. The marked letters are bases 6 to 8 of the original mRNA. Table excerpt of the standard genetic code: AUG, methionine; GAU, aspartic acid; CUG, leucine; AAA, lysine; GGC, glycine; UAA, stop; GAG, glutamic acid.
Noah compares a model mRNA with a copy that lacks three bases (marked in the original). How do the two chains compare?
- 2 amino acids: three bases shorten the chain by three
- 5 amino acids: just the amino acid at the deletion changes
- 4 amino acids: the ones after the deletion are different
- 4 amino acids: two are replaced by one, and lysine and glycine stay
Change card (model data)
change what happened to the gene P 6 bases added Q 1 base added R 2 bases removed S 4 bases removed Pia reads a card of four model changes to one gene. Which change keeps the reading frame of the codons after it?
- No row: any added or removed bases shift the frame
- Q: one base is too few to move the frame
- S: 4 bases removed leaves whole codons behind
- P: 6 bases make two whole codons
Changed DNA template strand (model, not to scale), 3′ to 5′, cut into triplets
TAC CCT ATC GAA TTT ATT. Changed mRNA, made from the template (model), 5′ to 3′, cut into codons: AUG GGA UAG CUU AAA UAA. Original mRNA (model), 5′ to 3′, cut into codons: AUG GGA CAG CUU AAA UAA. The mRNA pairs with the template strand base by base. The marked letter is base 7 of the changed template. Model plant (not a real gene): the original mRNA builds a pigment enzyme of five amino acids. A plant with the enzyme makes purple pigment and has purple flowers. A plant without the enzyme has white flowers. Table excerpt of the standard genetic code: AUG, methionine; GGA, glycine; CAG, glutamine; CUU, leucine; AAA, lysine; UAA, stop; UAG, stop.
Ren reads a model gene card for flower color. A plant has the changed DNA. What color are its flowers, and why?
- White: the enzyme is full length but has a different amino acid
- White: the enzyme ends after two amino acids, so no pigment is made
- Purple: one changed letter is too small to stop the enzyme
- Purple: the changed codon adds a new amino acid to the enzyme
Model plants (model data)
plant change to the pigment gene enzyme flowers A none five amino acids purple B silent substitution same five amino acids purple C stop codon at codon 3 two amino acids white Zee reads cards for three model plants. Which statement do the cards support?
- Plant B has a mutation but the same enzyme; only plant C lost the pigment
- Plant B makes a different enzyme, since its gene has a changed letter
- Plant C makes a longer enzyme, since a stop codon adds amino acids
- Plants B and C both have harmed enzymes, because both have mutations