Question

19. Two children are investigated for the expression of a gene (D) that encodes an important enzyme for muscle development. The results of the studies of the gene and its product follow.For child 2, the enzyme activity of each stage was very low and could be estimated only at approximately 0.1 unit at ages 1, 2, 3, and 4

a. For both children, draw graphs representing the developmental expression of the gene. (Fully label both axes.)

b.How can you explain the very low levels of active enzyme for child 2? (Protein degradation is only one possibility.)

c.How might you explain the change in the Southern blot for child 2 compared with that for child 1?

d.If only one mutant gene has been detected in family studies of the two children, define the individual children as either homozygous or heterozygous for gene D.

Probe: P gene D Autoradiograms Northern blots (samples at different Southern blot stages of development) 12 kb Who digest of genomic DNA 3 kb 3 kb Age: 1 yr 2 yr 3 yr yr Intensity scans of bands: (0.5) (1.0) (2.00 (2.00
child 1

child 2

Answer 1

a) For child1 For Child 2: Units of active enzyme Units of active enzyme 4 Time (Age) Time (Age) In case of child 1 units of enzyme gradually increases as their age increases, so that's why the graph is linear. But irn second case although the age increases, units of active enzyme remain in the same position. So in this case the graph is parallel to the X-axis. b) Cause of very low levels of active enzyme for child 2 The only cause of this low level of active enzyme for child 2 is protein degradation. It means the breakdown of proteins into smaller polypeptides or amino acids. Many rapidly degraded proteins function as regulatory molecules, such as transcription factors. The rapid turnover of these proteins is necessary to allow their levels to change quickly in response to external stimuli. Other proteins are rapidly degraded in response to specific signals, providing another mechanism for the regulation of intracellular enzyme activity. Besides this, faulty or damaged proteins are recognized and rapidly degraded within cells, thereby eliminating the consequences of mistakes made during protein synthesis. From the southern blotting image of child 2 proves that instead of whole DNA fragment, two small DNA molecule is produce from that DNA. As a result instead of being produce a whole functional protein, two small protein will produce whose functions are not same like that functional protein. So, thats why they does not play an important role like that protein produce in child 1. So, they will easily degraded c) Southern blot in child 2 Southern blot for child 1: Southem blot Southern blot 12 kb Xho digest of Tho digest of genomic DNA ic NA In the case of child 2, an additional restriction appear in that DNA, that cuts the fragment into two smaller fragments. But in first case, there is no additional restriction site appear in DNA, so the whole fragment remain in an intact condition.

d) In case of first child, the child is heterozygous for gene D. In heterozygous condition there is one dominant allele, that’s why the cell acts as a normal and a functional protein is produced.

But in case child 2, the child is homozygous for gene D. That means mutation occurs in both allele, so the protein which is produced in second case is truncated.