Question 1:
a) A: Promoter
B: Splice site
C: 5’ UTR
D: Start codon
E: Stop codon
F: 3’ UTR
b) The sum of the exons and introns (all in kilobases) (1.2+8+0.7+27+0.4+11+3.1) = 51.4 kb.
c) The sum of the exons (all in kilobases) (1.2+0.7+0.4+3.1) = 5.4 kb.
d)
RNA
Protein
Truncation mutation in exon 2
Same length, same amount of RNA produced.
Shorter in length (due to earlier stop codon), same amount of protein produced, usually changes the protein to non-functional, though not always.
3bp in frame deletion in exon 1
Shorter in length by 3pb, same amount of RNA produced.
Shorter in length, same amount of protein produced, usually causes the protein to become non-functional, though not always.
Splice site mutation at the start of exon 3 …show more content…
Questions 2:
a) 45 chromosomes with one sex chromosome, chromosome Y. There would be a spontaneous miscarriage, as monosomy Y is not viable, resulting in no phenotype.
b) 47 chromosomes with an extra chromosome 18 and 2 X sex chromosomes. The resulting phenotype is a female with Edwards Syndrome.
c) 45 chromosomes with one less chromosome 21 and a X and Y sex chromosome. The sex chromosomes are typical of that of a male, though there would be a spontaneous miscarriage, as monosomy 21 is not viable, resulting in no phenotype.
d) 46 chromosomes with one less chromosome 14, a Robesonian translocation between each q arm of chromosome 14 and 21 and a X and Y sex chromosome. The resulting phenotype would be a male with Downs Syndrome, due to the extra q arm of chromosome 21.
e) 46 chromosomes with one less chromosome 21, a Robesonian translocation between each q arm of chromosome 14 and 21 and a X and Y sex chromosome. The resulting phenotype would have been male, though since trisomy 14 is not viable (due to the extra q arm of chromosome 14), there would be no …show more content…
The resultant phenotype would be a wildtype female, as the only genetic material lost (the p arms of chromosome 21 and 14) is redundant.
Question 3:
a) This person has a wildtype phenotype due them still having all required genetic information to be a wildtype even though they are missing the short (p) arms of their chromosome 21. This is due to the p arms of each chromosomes only containing genes for ribosomal RNA and there are multiple copies of those genes on the various other acrocentric chromosomes. So even though they have one less physical chromosome 21, they still have the genetic information of two chromosomes.
b) This individual would produce gametes that are either nullisomic or disomic. This is due to all the genetic information for chromosome 21 being tied to one physical chromosome instead of the usual two. This means that in independent assortment in meiosis, the chromosome will be unable to split and the resultant gamete will either have the equivalent genetic information of two chromosome 21 or no genetic information for chromosome
a) A: Promoter
B: Splice site
C: 5’ UTR
D: Start codon
E: Stop codon
F: 3’ UTR
b) The sum of the exons and introns (all in kilobases) (1.2+8+0.7+27+0.4+11+3.1) = 51.4 kb.
c) The sum of the exons (all in kilobases) (1.2+0.7+0.4+3.1) = 5.4 kb.
d)
RNA
Protein
Truncation mutation in exon 2
Same length, same amount of RNA produced.
Shorter in length (due to earlier stop codon), same amount of protein produced, usually changes the protein to non-functional, though not always.
3bp in frame deletion in exon 1
Shorter in length by 3pb, same amount of RNA produced.
Shorter in length, same amount of protein produced, usually causes the protein to become non-functional, though not always.
Splice site mutation at the start of exon 3 …show more content…
Questions 2:
a) 45 chromosomes with one sex chromosome, chromosome Y. There would be a spontaneous miscarriage, as monosomy Y is not viable, resulting in no phenotype.
b) 47 chromosomes with an extra chromosome 18 and 2 X sex chromosomes. The resulting phenotype is a female with Edwards Syndrome.
c) 45 chromosomes with one less chromosome 21 and a X and Y sex chromosome. The sex chromosomes are typical of that of a male, though there would be a spontaneous miscarriage, as monosomy 21 is not viable, resulting in no phenotype.
d) 46 chromosomes with one less chromosome 14, a Robesonian translocation between each q arm of chromosome 14 and 21 and a X and Y sex chromosome. The resulting phenotype would be a male with Downs Syndrome, due to the extra q arm of chromosome 21.
e) 46 chromosomes with one less chromosome 21, a Robesonian translocation between each q arm of chromosome 14 and 21 and a X and Y sex chromosome. The resulting phenotype would have been male, though since trisomy 14 is not viable (due to the extra q arm of chromosome 14), there would be no …show more content…
The resultant phenotype would be a wildtype female, as the only genetic material lost (the p arms of chromosome 21 and 14) is redundant.
Question 3:
a) This person has a wildtype phenotype due them still having all required genetic information to be a wildtype even though they are missing the short (p) arms of their chromosome 21. This is due to the p arms of each chromosomes only containing genes for ribosomal RNA and there are multiple copies of those genes on the various other acrocentric chromosomes. So even though they have one less physical chromosome 21, they still have the genetic information of two chromosomes.
b) This individual would produce gametes that are either nullisomic or disomic. This is due to all the genetic information for chromosome 21 being tied to one physical chromosome instead of the usual two. This means that in independent assortment in meiosis, the chromosome will be unable to split and the resultant gamete will either have the equivalent genetic information of two chromosome 21 or no genetic information for chromosome