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From the pedigree it is reasonable to conclude that A - VCE - SSCE Biology - Question 15 - 2010 - Paper 1
Question 15
From the pedigree it is reasonable to conclude that A. II 4 is homozygous normal at the DMD locus. B. I 1 is heterozygous with respect to the DMD allele. C. I 1 a... show full transcript
Worked Solution & Example Answer:From the pedigree it is reasonable to conclude that A - VCE - SSCE Biology - Question 15 - 2010 - Paper 1
Step 1
A. II 4 is homozygous normal at the DMD locus.
Answer
To determine whether II 4 is homozygous normal at the DMD locus, we need to analyze the family pedigree. If the individual does not show symptoms of DMD and neither do their offspring, it is reasonable to conclude that they are likely homozygous normal. However, this cannot be definitively stated without additional information regarding the carrier status of their parents.
Step 2
B. I 1 is heterozygous with respect to the DMD allele.
Answer
In cases of X-linked recessive disorders like DMD, males (XY) who are affected must have the DMD allele on their X chromosome, while females (XX) can be carriers. If there is evidence showing that these individuals produced affected offspring, it is likely that I 1 is heterozygous (carrying one normal and one affected allele), making this statement reasonable.
Step 3
C. I 1 and 2 have a one-in-four chance of producing a daughter with DMD.
Answer
The chance of producing a daughter with DMD can be calculated based on the genetic makeup of the parents. If I 1 is heterozygous and I 2 is not a carrier, the chance of having a daughter with DMD would not be one-in-four, but rather zero, as daughters inherit X chromosomes from both parents. This statement is therefore incorrect.
Step 4
D. II 4 and 5 have a one-in-three chance of producing a daughter who is a carrier of DMD.
Answer
If II 4 is homozygous normal, then there are no chances of producing a carrier offspring. However, if II 5 is a carrier, the probability of their union resulting in a carrier daughter depends on the locus of alleles. Considering multiple scenarios could yield a different probability, but generally, with one carrier and one non-carrier, we often assume a chance of 0.5 (not one-in-three). Therefore, this statement needs clarification but is likely misleading.