A cell that contains organelles called chloroplasts
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1.Treat 25 mg of tissue with Proteinase K by adding 20 ml of stock PK solution (20 mg/ml) and 180 ul water. Function: proteinase destroys protein.
2.Let sit at 55 degrees for at least 1 hour. Function: This temperature is optimal for enzyme reaction; if too high will degrade DNAase.
3.Add 1/10 volume of Kac (potassium acetate) (5 M; pH 5.6) (in this example, 20 ul but remember the 1/10 proportion as a general rule). . Mix well and place on ice for 10 minutes. Function: precipitates membrane, C, H, O, lipids, proteins, and monomers.
4.Add 200 ml of phenol saturated with 10mM Tris. ( IN HOOD) Function: removes nonpolar proteins, lipid residue.
5.Cap tube: mix well by inverting tube.
6.Spin 5 minutes in microfuge.
Function: separates aqueous (top) and phenol (bottom) layers 7.Remove upper aqueous layer (CONTAINS DNA) to a new tube.
8.Add 200 ul of chloroform to aqueous layer (IN HOOD). Function: removes excess phenol from DNA.
9.Cap tub; mix well by inverting. Function: sets an emulsion; biphasic solution.
10.Remove upper aqueous layer (CONTAINS DNA) to a new tube.
11.Add 500 ul of EtOH and cap tube (again, 500 in this example; as a proportion use twice the total amount). Function: precipitates DNA; DNA is insoluble in ethanol.
12.Cap tube; mix well by inverting tube several times.
13.Spin for 5 minutes in microfuge. 13000 rpm. (10 min. at 6000.) Function: collects DNA in bottom of the tube.
2.Let sit at 55 degrees for at least 1 hour. Function: This temperature is optimal for enzyme reaction; if too high will degrade DNAase.
3.Add 1/10 volume of Kac (potassium acetate) (5 M; pH 5.6) (in this example, 20 ul but remember the 1/10 proportion as a general rule). . Mix well and place on ice for 10 minutes. Function: precipitates membrane, C, H, O, lipids, proteins, and monomers.
4.Add 200 ml of phenol saturated with 10mM Tris. ( IN HOOD) Function: removes nonpolar proteins, lipid residue.
5.Cap tube: mix well by inverting tube.
6.Spin 5 minutes in microfuge.
Function: separates aqueous (top) and phenol (bottom) layers 7.Remove upper aqueous layer (CONTAINS DNA) to a new tube.
8.Add 200 ul of chloroform to aqueous layer (IN HOOD). Function: removes excess phenol from DNA.
9.Cap tub; mix well by inverting. Function: sets an emulsion; biphasic solution.
10.Remove upper aqueous layer (CONTAINS DNA) to a new tube.
11.Add 500 ul of EtOH and cap tube (again, 500 in this example; as a proportion use twice the total amount). Function: precipitates DNA; DNA is insoluble in ethanol.
12.Cap tube; mix well by inverting tube several times.
13.Spin for 5 minutes in microfuge. 13000 rpm. (10 min. at 6000.) Function: collects DNA in bottom of the tube.
As this family has an Autosomal Recessive Disorder, we must know that this disorder will only be expressed, in the family members who are homozygous recessive (aa), in this way, to know the probability of IV-4 and IV-5 to present the disorder, we must evaluate the probabilities of combinations of their parents' alleles in a diagram called a punnett square.
In the pedigree above, we can see that the dark colored balls and squares represent the members of the family who have the disorder and therefore have the "aa" genotype.
Thus, when observing the crossings throughout the family, we know that the father of IV-4, has the genotype "Aa", however we do not know which genotype of the mother of IV-4, but as she does not have the disorder, she may have the genotype "Aa", or "AA", for that reason, we must make the punnett square for both situations and see what the probabilities of IV-4 will present the disorder.
As you can see in the figure attached below, if IV-4's mother has the "Aa" genotype, IV-4 will have a 25% chance of presenting the disorder. If IV-4's mother has the "AA" genotype, IV-4 has a 0% chance of inheriting the disorder.
In the case of IV-5, we can see that his mother has the disorder and therefore has the genotype "aa", we know that his father does not have the disorder, but we do not know if the father's genotype is "AA" or "Aa "and therefore we must evaluate both situations.
As you can see in the figure below called "IV-5", if IV-5's father has the "AA" genotype, IV-5 will have a 0% chance of inheriting the disorder. However, if IV-5's father has the "aa" genotype, IV-5 will have a 50% chance of inheriting the disorder.
