50%. The segregation at anaphase I will result in 50% of the gametes carrying two 21 chromosomes and 50% carrying one 21 chromosome.
P: ffMM X FFmm
F1: Ff Mm
F1 cross: Ff Mm X Ff Mm
¾ floppy ears ¾ normal teeth
¼ normal ears ¼ overgrown teeth
Probability of normal ears and overgrown teeth: 1/16, hence 64/16=4
P: BBaa X bbAA
Cross: BbAaMm X bbAaMm
bb= 1/2 Aa=1/2 mm=1/4
+ t h 385 y t h 85
y + h 18 + t + 20
+ + + 90 + + h 1 y + + 401
No other phenotypes were observed.
+ t h 385 NO crossover y + + 401 NO crossover y t h 85 Single CO + + + 90 Single CO y + h 18 Single CO + t + 20 Single CO
+ + h 1 Double CO
yt = (85+90+1)/1000=0.176=17.6cM
p=1/2 (boy) *2/4 (het) =1/4
A nuclear RNAprotein complex responsible for removing introns from premRNA
A tRNA attached to its amino acid
iii) ShineDalgarno sequence (1 mark)
Ribosome binding site in prokaryotic mRNAs
Translation; Transcription initiation; RNA export from the nucleus; Splicing (1 mark)
Transcription initiation; Splicing; RNA export from the nucleus; Translation
Yes. Exons do not represent individual open reading frames and so shifting the reading frame in one exon will affect everything from that point in the mature mRNA.
|Compound X Absent||Compound X Absent||Compound X Present||Compound X Present|
|Mutation in A||||+|||||
|Mutation in B||+||+||+||+|
|Mutation in C||+|||||||
|Mutation in D|||||||||
Vertical/parental, Horizontal: Transformation, Transduction, Conjugation (0.5 mark each)
Conjugation (1 mark) – F factor/plasmid in donor strain (1 mark), producing pili on cell surface (1 mark). F factor can exist as plasmid, or integrated into genome in Hfr strains (1 mark).
Transformation (1 mark) – bacterial cells can be made chemically competent or electroporated to uptake recombinant plasmid DNA for gene cloning (1 mark)
It has an origin of replication (1 mark), a selectable marker (1 mark), multiplecloning site with unique restriction site (MCS) (1 mark)
Fragment amplified by PCR (with restriction sites in primers) OR cut from other plasmid with restriction enzymes (either alternative OK, 1 mark). Cloning vector is cut with restriction enzymes to produce compatible ends (1 mark). DNA ligase is used to insert gene fragment into MCS of cloning vector (1 mark).
DNA would need to be (shotgun) sequenced and annotated (1 mark). Any two of the following in the sequences: higher density of genes in prokaryote vs eukaryote; introns in coding regions in eukaryotes; large amount of repetitive DNA in eukaryotes; total genome size (small prokaryote, large eukaryote), evidence of mitochondrial or chloroplast DNA in eukaryotes (2 marks)
Could not know structural difference eg. circular or linear chromosomes, presence of nucleosomes, chromatin etc. (0 marks)
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