In lizards, chromosome I can be represented by the sequence ABC.DEFG and chromos
ID: 62636 • Letter: I
Question
In lizards, chromosome I can be represented by the sequence ABC.DEFG and chromosome II can be represented by the sequence STUV.WXYZ Normal diploids have two of each of these chromosomes. A reciprocal translocation is known in which FG and YZ are exchanged with each other, giving rise to the translocation chromosomes ABC.DEYZ and STUV.WXFG When a translocation heterozygote undergoes meiosis, which of the following gametes will most likely be viable?
A gamete containing the chromosomes ABC.DEFG and ABC.DEFG
A gamete containing the chromosomes ABC.DEFG and STUV.WXYZ
A gamete containing the chromosomes STUV.WXYZ and ABC.DEYZ
A gamete containing the chromosomes ABC.DEYZ and STUV.WXFG
b and d
a.A gamete containing the chromosomes ABC.DEFG and ABC.DEFG
b.A gamete containing the chromosomes ABC.DEFG and STUV.WXYZ
c.A gamete containing the chromosomes STUV.WXYZ and ABC.DEYZ
d.A gamete containing the chromosomes ABC.DEYZ and STUV.WXFG
e.b and d
Explanation / Answer
Chromosome 1 sequence: ABC. DEFG
Chromosome 2 sequence: STUV.WXYZ
Normal diploids will have 2 copies of both the sequences.
Given, the reciprocal translocation cross is between heterozygotes for the sequences, which means one normal and one mutant copy:
ABC.DEFG and ABC. DEYZ
And
STUV.WXYZ and STUV.WXFG
So the possible outcomes for gametes are as follows:
1) ABC.DEYZ and STUV.WXFG
2) ABC.DEYZ and STUV.WXYZ
3) ABC.DEFG and STUV.WXFG
4) ABC.DEFG and STUV. WXYZ
Out of these 4 results only the first (ABC.DEYZ and STUV.WXFG ) and fourth (ABC.DEFG and STUV. WXYZ) are viable since they contain all the nucleotides. Hence, the correct answer is e (both b and d).
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