Question

bisc161 lab 7 - molecular genetics and mitosis

1. a group of scientists are investigating the arrest of a poacher caught with an elephant tusk. park rangers believed the tusk is from a recently discovered crime where three elephants from the same population were found dead with their tusks removed. the rangers gave the scientists dna samples from the tusks and the three dead elephants. the numbers in the table below represent the size of the 2 gene pieces cut by the restriction enzyme in base pairs.

draw the expected pattern of bands on the gel below that shows the dna fragments of the three elephants. a dna ladder for a control group has already been added to well 1. the dna from the seized elephant tusk is in well 2. if both dna fragments are the same length, you only need to draw one line, since they both move the same distance. (0.5 pt)

can the rangers arrest the suspected poacher? explain your answer. (0.5 pt)

Explanation:

To determine if the rangers can arrest the suspected poacher, we need to check if the DNA fragments from the seized tusk (in well 2) match the DNA fragments of any of the three dead elephants. DNA fingerprinting (using restriction fragment length polymorphism - RFLP in this case, where restriction enzymes cut DNA into fragments) is based on the principle that unique DNA fragment patterns (for specific genes) can identify an individual. If the fragment lengths (in base pairs) of the tusk DNA match those of one of the dead elephants for the genes FH19, FH67, and FH129, it means the tusk came from that elephant. Looking at the table:

• For Elephant 1: FH19 (189, 193), FH67 (120, 129), FH129 (132, 160)
• For Elephant 2: FH19 (193, 193), FH67 (113, 147), FH129 (160, 160)
• For Elephant 3: FH19 (183, 185), FH67 (103, 130), FH129 (152, 160)

Assume (since the gel - drawing part is about visual representation of fragment lengths and we focus on the matching logic) that if we compare the tusk DNA (well 2) fragment lengths (not shown in text but based on the table logic) - if, for example, the tusk DNA has fragment lengths that exactly match one of the elephants (e.g., if the tusk's FH19, FH67, and FH129 fragment lengths are the same as Elephant 2's 193 (for FH19 - since two 193s mean one band), 113 and 147 (FH67), 160 (FH129 - two 160s mean one band)). In a real - world DNA fingerprinting (like RFLP) scenario, matching fragment patterns across multiple genes provides strong evidence of identity.

Answer:

If the DNA fragment lengths (for genes FH19, FH67, and FH129) of the tusk in well 2 match the fragment lengths of one of the three dead elephants (e.g., if it matches Elephant 2: FH19 = 193 (one band as two same - length fragments), FH67 = 113 and 147, FH129 = 160 (one band)), then the rangers can arrest the suspected poacher. This is because matching DNA fragment patterns (across multiple genes) using techniques like RFLP (where restriction enzymes cut DNA into identifiable fragments) provides strong evidence that the tusk came from one of the dead elephants, linking the poacher to the crime scene (the dead elephants).