CHAPTER 21
EXTENDING YOUR IDEAS
1.
1h
2i
3d
4g
5a
6
7
8
9
c
e
b
f
2.
a.
2x
b.
x
c.
x
d.
½x
chromatids.
3.
Since 38% were adenine, a further 38% would be the complimentary base thymine. Guanine
and its complimentary base cytosine would therefore make up 24% of the DNA molecule.
Thus guanine would make up 12%.
4.
The labelled nitrogen was used by the bacterium to produce nucleotides which were
incorporated into nucleotides of the DNA strands Each daughter cell was fed on normal
nitrogen thus any new nucleotides formed would contain normal nitrogen. On replication,
therefore, each replicated strand was normal. With cell division, a normal chromosome goes
to one daughter cell and a labelled chromosome goes to the other.
5
1:
2:
3:
4:
5:
6:
6
It would not account for the same number of A and T or C and G. It would not be able to
explain replication.
7
a.
Replication of DNA has occurred in preparation for division.
Separation of the chromatids has occurred.
Homologous chromosomes, but not chromatids have separated.
There has been separation of both the homologous chromosomes and
DNA helix unwinding
unzipping
formation mRNA
mRNA to cytoplasm for translation
energy required for linking amino acids etc. together
formation new proteins which will control the cell function.
Mutant X will grow if provided with cystathionine, homocystine or cysteine – thus lacks
an enzyme which will convert a precursor into one of these three chemicals.
Mutant R will grow only on homocystine but accumulates cystathionine – thus
homocystine is the end product and R lacks an enzyme somewhere in the reactions which
will convert cystathionine to homocystine.
Mutant W will grow if provided with cystathionine or homocystine but accumulates
cysteine – thus W lacks an enzyme which will convert cysteine to cystathionine.
Mutant X
precursor
homocystine
a
Mutant W
cysteine
Mutant R
cystathionine
b
c
b. i. X crossed with R:
X homozygous recessive for enzyme a, and dominant for B and C.
R homozygous recessive for enzyme c. Shows at least one dominant allele for enzymes A
and B.
If R
AA BB cc
x
aa BB CC
then all offspring (AaBbCc) of cross will be able to grow if provided with basic precursor
substances.
If R
If X
ABc
Abc
aBc
abc
Aa Bb cc
aa Bb Cc then
aBC
AaBBCc
AaBbCc
aaBBCc
aaBbCc
abC
AaBbCc
AabbCc
aaBbCc
aabbCc
aBc
AaBBcc
AaBbcc
aaBBcc
aaBbcc
abc
AaBbcc
Aabbcc
aaBbcc
aabbcc
Phenotype Mutant X aaBBCc 1/16
aaBbCc 1/16
Phenotype Mutant W AabbCc 1/16
AabbCc 1/16
aabbCc 1/16
Phenotype Mutant R AaBBcc 1/16
AaBbcc 2/16
Aabbcc 1/16
aaBBcc 1/16
aaBbcc 2/16
aabbcc 1/16
Normal
AaBBCc 1/16
AaBbCc 2/16
(other combinations could also be X aaBBCc / aaBbCC
Y AaBBcc / AABbcc etc)
ii. X crossed with W:
If X aaBBCC and W AAbbCC, all offspring normal.
If X aaBbCc
W
AabbCc
AbC
Abc
abC
abc
aBC
AaBbCC
AaBbCc
aaBbCC
aaBbCc
abC
AabbCC
AabbCc
aabbCC
aabbCc
Phenotype Mutant X aaBBCC 1/16
aaBbCc 1/16
aaBbCC 1/16
Phenotype Mutant W AabbCC 1/16
AabbCc 2/16
aabbCC 1/16
aabbCc 2/16
Phenotype Mutant R AaBbcc 1/16
aaBbcc 1/16
Aabbcc 1/16
aabbcc 1/16
Normal Phenotype
AaBbCC 1/16
AaBbCc 2/16
Other possible combinations X
W
iii. W crossed with R:
Crosses could be made with
W
R
aBc
AaBbCc
AaBbcc
aaBbCc
aaBbcc
abc
AabbCc
Aabbcc
aabbCc
aabbcc
aaBBCc or aaBbCC
AAbbCc or AabbCC)
AAbbCC; AabbCC; AabbCc; AAbbCc
AABBcc; AaBBcc; AABbcc; AaBbcc
Results of offspring give an indication of genotypes provided:
a. genes not linked and
b. if linked, no crossing over.
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The enzyme will be closest to normal if it is at the end of the gene since the deletion will
result only in one amino acid (the last one) being different or absent. If the deletion occurs at
the beginning of the gene the message will be still read in triplet codons and thus there is a
possibility that all subsequent codons will be read 'incorrectly' for that gene.