PCR lab

SICKLE CELL DISEASE

Sickle cell disease is a genetic disorder caused by a mutation in the HBB gene coding for b-globin , one of the two polypeptides that combine to form the functional hemoglobin protein. Hemoglobin is the predominant protein found in red blood cells. It binds the oxygen that enters your lungs and carries it to all cells for use in cellular respiration. After delivering oxygen to the cells, hemoglobin transports carbon dioxide waste back to the lungs, where it is exhaled. Like all proteins, the precise folding and three-dimensional shape of hemoglobin is critical to its function. The functional protein is a helical structure consisting of two interwoven a-globin subunits and two b-globin subunits. The subunits are held together by precise hydrophobic (water-fearing) and hydrophilic (water-loving) interactions between the amino acids of the four polypeptide chains.

The allele responsible for sickle cell disease contains a single nucleotide-pair substitution resulting in a missense mutation . The wild-type allele ( HbA ) codes for a hydrophilic glutamic acid residue (Glu or E) in the seventh position of the b-globin polypeptide. The mutant allele ( HbS ) results in substitution of this hydrophilic glutamic acid with a hydrophobic valine (Val or V). In the absence of oxygen, the hydrophobic valine ends up on the outer surface of the hemoglobin protein, exposed to the aqueous environment of the cytosol. Thus, under low oxygen conditions, sickle cell hemoglobin molecules aggregate, or clump together, within red blood cells due to hydrophobic interactions between exposed valine residues of different hemoglobin molecules.

Red blood cells are very small cells that normally have a round shape so they can flow easily through the tiniest blood vessels. Aggregation of sickle cell hemoglobin molecules within red blood cells results in an abnormal sickle-shape of the cells. Individuals with sickle cell disease typically have reduced red blood cell counts (anemia) due to the reduced lifespan of the abnormally shaped cells. The unusual shape also makes it difficult for the cells to pass through small capillaries, resulting in blockage of blood flow and oxygen deprivation to major organs, leading to sickle cell crisis. Sickle cell crisis is very painful. It can cause permanent organ damage and even death.

Sickle cell disease is considered an autosomal recessive disorder. The HBB gene is located on human chromosome 11 (not sex-linked) and individuals must possess two copies of the HbS allele to exhibit the complete disease phenotype. However, HbA/HbS heterozygotes may show some symptoms after prolonged exposure to very low oxygen conditions. Because half of the hemoglobin in a heterozygote’s red blood cells contain the abnormal valine, some clumping of hemoglobin does occur, but not nearly to the extent seen in HbS/HbS homozygotes. Thus, the HbA allele shows incomplete dominance over the HbS allele since the heterozygote has an intermediate phenotype compared to the two homozygotes. Heterozygotes are said to have sickle cell trait to differentiate from the more severe sickle cell disease observed in HbS/HbS homozygotes.

Curiously, carrying at least one HbS allele imparts significant resistance to malaria, a mosquito-borne disease caused by a parasite that infects red blood cells. Thus, heterozygotes have a unique selective advantage over either homozygote in regions of the world where malaria is prevalent, notably sub-Saharan Africa, South America, and to a lesser extent, the Middle East and India. This finding explains why a seemingly harmful allele has been maintained for thousands of years in populations that live in these regions of the world. The allele has more recently spread throughout the world by migration. Even in the United States, the risk of inheriting sickle cell disease is quite high (https://www.cdc.gov/ncbddd/sicklecell/data.html). An estimated 1 in 13 African Americans has sickle cell trait and 1 in 365 African Americans has sickle cell disease. Approximately 1 in 16,300 Hispanic Americans has sickle cell disease. Individuals can be easily genotyped from blood samples using the biotechnology techniques of PCR, restriction enzyme digestion, and DNA gel electrophoresis. In this lab, we will use these techniques to genotype and diagnose individuals from a fictitious family suspected to be carrying the sickle cell allele.

HHMI BioInteractive Videos on Sickle Cell Disease:

https://www.biointeractive.org/classroom-resources/sickle-cell-disease

https://www.biointeractive.org/classroom-resources/making-fittest-natural-selection-humans

Review Video on Transcription and Translation:

https://www.youtube.com/watch?v=gG7uCskUOrA

PCR – POLYMERASE CHAIN REACTION

Deoxyribonucleic acid (DNA) molecules are the macromolecules/biomolecules that carry the genetic information of living organisms. In eukaryotes, the genetic information is divided into multiple chromosomes. Each chromosome is one extremely long, double-stranded polymer of nucleotides containing many genes. An average human chromosome is approximately 130 million base pairs long and contains about 850 protein-coding genes. The covalent bond joining adjacent nucleotides in DNA is called a phosphodiester bond. Phosphodiester bonds are formed by the enzyme DNA polymerase . The phosphodiester bonds between nucleotides in DNA molecules are very stable unless they are physically stretched or exposed to enzymes called nucleases . Complementary base pairs between the two strands of DNA are held together by comparatively weak hydrogen bonds which can be disrupted by heat or cellular enzymes called helicases .

Polymerase chain reaction (PCR) revolutionized the field of molecular biology, earning its inventors a Nobel Prize in 1993. PCR is used to make many copies of a specific DNA sequence in just a few hours. Recall that the human genome contains billions of base pairs and over 20,000 unique genes. Using PCR, a single gene can be easily amplified from a complex mixture of cellular DNA for genetic analysis. There are many other applications of PCR that require minor modifications of the basic procedure; tiny amounts of DNA from a crime scene can be used to identify a criminal, minute amounts of viral or bacterial genetic material can be detected in human samples, and DNA from ancient fossils can even be amplified for sequencing.

While PCR makes use of many of the same principles as cellular DNA replication, there are several important differences. PCR requires a pair of short, single-stranded DNA primers (typically 15 – 22 nucleotides long) to define the boundaries of the region being copied. The primers ensure that only the target sequence is amplified. Another critical difference is the use of a thermostable enzyme called Taq DNA polymerase to build the new DNA molecules. Taq DNA polymerase was isolated from a species of bacterium, Thermus aquaticus, that lives in the hot springs of Yellowstone National Park. This enzyme can withstand temperatures of 95 ○C (near boiling) without denaturing.

DNA replication by PCR occurs in three steps. First, high temperature (94 – 95 ○C) is used to denature the template DNA, breaking the hydrogen bonds between base pairs in the DNA sample to create a single-stranded DNA template. Next, the temperature is lowered to 50 – 60 ○C to allow annealing (complementary base pairing) of the primers to opposing strands of the template DNA. Finally, the temperature is increased to 72 ○C for optimal extension of the primers in the 5’ 3’ direction by Taq DNA polymerase. The three steps produce two copies of each target DNA molecule, but after repeating the three steps for 30 cycles, up to a billion copies of each target sequence can be produced.

Review Video on Replication:

https://www.youtube.com/watch?v=TNKWgcFPHqw&feature=emb_logo

HHMI BioInteractive Video on PCR:

https://www.biointeractive.org/classroom-resources/polymerase-chain-reaction-pcr

RESTRICTION ENZYME ANALYSIS

Nucleases are enzymes that are capable of breaking (hydrolyzing) phosphodiester bonds in DNA molecules. Nucleases can be classified into two major groups: exonucleases and endonucleases. If the enzyme digests nucleotides from the ends of the DNA molecule, it is called an exonuclease. If the enzyme digests DNA by breaking phosphodiester bonds in the interior of a DNA molecule, it is called an endonuclease . Most endonucleases are not specific. This means that they can break phosphodiester bonds at any sequence in a DNA molecule.

A special class of endonucleases has been isolated from a variety of bacteria, including several cyanobacteria (i.e., blue-green algae). These special enzymes, termed restriction enzymes or restriction endonucleases (RE) , digest DNA by breaking bonds only within a specific short sequence of nucleotides. These are called recognition sequences , and usually range in size from 4 to 8 base pairs but can be as long as 23 base pairs. Different restriction endonucleases most often recognize different nucleotide sequences; however, a few of them can recognize the same nucleotide sequence. These restriction enzymes are called isoschizomers .

Restriction endonucleases confer an adaptive advantage on bacteria by digesting foreign DNA usually from an invading bacteriophage (i.e., bacterial virus). The resulting DNA fragments can then be further degraded and destroyed by exonucleases. In molecular biology laboratories, these enzymes are used to cut DNA in a precise and predictable manner. They are extensively used in recombinant DNA technologies and can sometimes be used to determine sequence variations (alleles) between individuals. Sequence variations that can be detected by restriction enzyme analysis are referred to as restriction fragment length polymorphisms (RFLPs).

Restriction endonucleases derive their names from the organisms from which they are isolated. For example, Eco RI was the first (# I ) RE isolated from the bacterium E scherichia co li, strain R Y13. Dde I was the first (# I ) RE isolated from the sulfur-reducing anaerobic bacterium D esulfovibrio de sulfuricans.

EcoRI recognizes the following nucleotide sequence:

2

5’ – G A A T T C – 3’ EcoRI 5’ – G A A T T C – 3’

3’ – C T T A A G – 5’ 3’ – C T T T A A G – 5’

Each strand of DNA is cut at the phosphodiester bond between the G and A bases (indicated by the arrow signs). Notice that the sequence GAATTC is the same on both strands when each strand is read 5′→ 3′. Such symmetrical sequences are called palindromes . This enzyme cuts the double strands asymmetrically, leaving protruding ends. These protruding bases are referred to as sticky ends because they will easily form new hydrogen bonds with other complementary sticky ends. In the research lab, compatible sticky ends enable researchers to “paste” together DNA sequences from different sources. This is known as genetic engineering .

DdeI recognizes a different palindromic nucleotide sequence. “N” can be any nucleotide base:

5’ – C T N A G – 3’ DdeI 5’ – C T N A G – 3’

3’ – G A N T C – 5’ 3’ – G A N T C – 5’

DNA AGAROSE GEL ELECTROPHORESIS

Once a segment of DNA has been amplified by PCR and/or digested by restriction endonucleases, the fragments must be visualized and analyzed. Fragments can be separated and identified by size using agarose gel electrophoresis . Agar is a large polysaccharide extracted from marine algae. Agarose is a purified form of agar. Gels are made by pouring melted agarose into casting trays and allowing it to solidify at room temperature. Each tray has a comb with teeth. When the agarose cools completely, it forms a semi-solid gel (similar in appearance and texture to Jello™). The comb is removed, leaving spaces called wells . The gels are immersed in an ionic buffer. The buffer has a pH above 8.0. DNA at this pH is negatively charged because the phosphates in the DNA backbone have lost hydrogen ions. DNA samples mixed with certain dyes are loaded into the wells of the gel. The dye molecules serve as the indicator of the movement of invisible DNA molecule through the gel as an electric current is applied through the gel. The negatively charged DNA will migrate from the anode (the negative electrode, with black color-coding) to the cathode (the positive electrode, with red color-coding) along with the current.

Separation of the DNA fragments occurs as they migrate through the network of agarose molecules. Smaller fragments slip through the network faster than larger molecules. The tightness of the network (i.e., agarose concentration) also affects the rate of migration. Gels with higher agarose concentrations favor the separation of smaller fragments, while lower agarose concentrations favor the separation of larger fragments. Therefore, the rate of migration is a function of fragment size, as well as the concentration of agarose.

WARNING

Gel box

Power supply

Negative (black) Positive (red)

Wells Gel

Gel tray

After separation, the DNA fragments can be visualized inside the gel by a different DNA-binding dye. If incubated with this special stain, the DNA fragments trapped inside an agarose gel bind to the dye and appear as bands. The pattern of DNA bands is characteristic for a specific DNA sample and the restriction enzymes used to cleave it. A banding pattern can be referred to as a DNA fingerprint .

By electrophoresing a series of DNA fragments of known size (a DNA size standard or ladder ) along with the DNA samples of interest, the sizes of unknown fragments can be estimated. In this exercise, a commercially obtained DNA marker will be used. This particular marker DNA is composed of a series of fragments increasing in size by approximately 100 base pairs (bp). Fragment sizes range from approximately 100 bp to 1500 bp. By comparing the distance traveled by unknown fragments with that of known DNA ladder fragments, the sizes of the unknown fragments can be estimated.

Interactive Overview of the Lab:

https://edpuzzle.com/media/5ef24a931ce74b3f43e57ec3 I. Sequence Analysis of the HBB Gene

PROCEDURE (work in groups)

In the box below you will find the double-stranded DNA sequence corresponding to the mRNA produced from the HBB wild-type allele (HbA). Note that the mRNA sequence is not shown. The one-letter code for the amino acid sequence of the b-globin polypeptide is shown in red, below the DNA sequence.

1  5’-CATTTGCTTCTGACACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCATCTGACTCCTGAGGAGAAGTCT

   3’-GTAAACGAAGACTGTGTTGACACAAGTGATCGTTGGAGTTTGTCTGTGGTACCACGTAGACTGAGGACTCCTCTTCAGA

                                                        M  V  H  L  T  P  E  E  K  S 

80  GCCGTTACTGCCCTGTGGGGCAAGGTGAACGTGGATGAAGTTGGTGGTGAGGCCCTGGGCAGGCTGCTGGTGGTCTACCCT
    CGGCAATGACGGGACACCCCGTTCCACTTGCACCTACTTCAACCACCACTCCGGGACCCGTCCGACGACCACCAGATGGGA
     A  V  T  A  L  W  G  K  V  N  V  D  E  V  G  G  E  A  L  G  R  L  L  V  V  Y  P  

161 TGGACCCAGAGGTTCTTTGAGTCCTTTGGGGATCTGTCCACTCCTGATGCTGTTATGGGCAACCCCAAGGTGAAGGCTCAT
    ACCTGGGTCTCCAAGAAACTCAGGAAACCCCTAGACAGGTGAGGACTACGACAATACCCGTTGGGGTTCCACTTCCGAGTA
     W  T  Q  R  F  F  E  S  F  G  D  L  S  T  P  D  A  V  M  G  N  P  K  V  K  A  H 

242 GGCAAGAAAGTGCTCGGTGCCTTTAGTGATGGCCTGGCTCACCTGGACAACCTCAAGGGCACCTTTGCCACACTGAGTGAG
    CCGTTCTTTCACGAGCCACGGAAATCACTACCGGACCGAGTGGACCTGTTGGAGTTCCCGTGGAAACGGTGTGACTCACTC
     G  K  K  V  L  G  A  F  S  D  G  L  A  H  L  D  N  L  K  G  T  F  A  T  L  S  E  

323 CTGCACTGTGACAAGCTGCACGTGGATCCTGAGAACTTCAGGCTCCTGGGCAACGTGCTGGTCTGTGTGCTGGCCCATCAC
    GACGTGACACTGTTCGACGTGCACCTAGGACTCTTGAAGTCCGAGGACCCGTTGCACGACCAGACACACGACCGGGTAGTG
     L  H  C  D  K  L  H  V  D  P  E  N  F  R  L  L  G  N  V  L  V  C  V  L  A  H  H  

404 TTTGGCAAAGAATTCACCCCACCAGTGCAGGCTGCCTATCAGAAAGTGGTGGCTGGTGTGGCTAATGCCCTGGCCCACAAG
    AAACCGTTTCTTAAGTGGGGTGGTCACGTCCGACGGATAGTCTTTCACCACCGACCACACCGATTACGGGACCGGGTGTTC
     F  G  K  E  F  T  P  P  V  Q  A  A  Y  Q  K  V  V  A  G  V  A  N  A  L  A  H  K  

485 TATCACTAAGCTCGCTTTCTTGCTGTCCAATTTCTATTAAAGGTTCCTTTGTTCCCTAAGTCCAACTACTAAACTGGGGGA
   ATAGTGATTCGAGCGAAAGAACGACAGGTTAAAGATAATTTCCAAGGAAACAAGGGATTCAGGTTGATGATTTGACCCCCT
     Y  H

566 TATTATGAAGGGCCTTGAGCATCTGGATTCTGCCTAATAAAAAACATTTATTTTCATTGCAA-3’
    ATAATACTTCCCGGAACTCGTAGACCTAAGACGGATTATTTTTTGTAAATAAAAGTAACGTT-5’

1. Identify and label the following on the previous page:

A. location of the start codon
B. location of the stop codon
C. the 5’ untranslated region
D. the 3’ untranslated region
E. the seventh amino acid (Glu/E) that is altered in sickle cell disease
F. the forward PCR primer 
5’ -CATTTGCTTCTGACACAACTG - 3’
G. the reverse PCR primer (remember this will be on the opposite strand of DNA)
5’ - CCTTGAGGTTGTCCAGGTG - 3’
H. the recognition site for the DdeI restriction enzyme. Remember, N can be any base. 


DdeI recognition site:

			5’ –  C T N A G  –  3’
			3’ –  G A N T C  –  5’
2. The DNA sequence of the HbS (sickle cell) allele coding for the first ten amino acids of the mutant b-globin protein is shown below, corresponding to the shaded region of the DNA sequence on the previous page. Can you find the DdeI recognition site in the HbS DNA sequence? Write the mRNA sequence and one-letter amino acid code beneath the DNA sequence. Circle the amino acid substitution. Note: the bottom strand is the template strand. 


HbS DNA sequence:         	5’ – ATGGTGCATCTGACTCCTGTGGAGAAGTCT – 3’
                	3’ - TACCACGTAGACTGAGGACACCTCTTCAGA – 5’

HbS mRNA sequence:



HbS amino acid sequence: 



3. Using the position of the PCR primers on the gene map on the previous page, determine the size of the expected PCR product in base pairs. Hint: place a mark after every 10 base pairs once you begin counting. Will the size of the PCR product be different for the HbA and HbS alleles?


4. On the maps below, mark the approximate position of the DdeI restriction site and indicate the length in base pairs (bp) of all fragments produced by the PCR followed by DdeI restriction digestion for the HbA and HbS alleles. The arrows represent the PCR primers.

                       HbA (normal allele)			            HbS (sickle cell allele)








5.  Using your PCR/DdeI map of the HbA and HbS alleles from 4, complete the following table for what you would expect to see after gel electrophoresis for each phenotype:



	


Phenotype

	


Genotype

	


Predicted PCR/Restriction Fragment Lengths (bp)



	

Normal hemoglobin
	


	




	

Sickle cell trait
	


	




	

Sickle cell disease
	


	











II. PCR Amplification



 PROCEDURE (work individually)

1. PCR is so sensitive that it can be used to amplify DNA from a single starting molecule. Under ideal conditions, each cycle of PCR will double the number of DNA copies in the reaction tube. Use the following table to predict the number of copies that can be produced after 10 cycles of PCR. The first few have been completed for you. Then, determine a mathematical expression (a formula) to help you calculate how many molecules will be produced after 30 cycles of PCR under ideal conditions.	


			


	


PCR Cycle

	


DNA Copies



	

START
	

1


	

1
	

2


	

2
	

4


	

3
	

8


	

4
	




	

5
	




	

6
	




	

7
	




	

8
	




	

9
	




	

10
	






	Mathematical expression (formula): 	

	DNA copies = ________________,
		where n = ______________.

	DNA copies after 30 cycles = 
						________________________________

2. Using Excel, create a graph showing the relationship between the number of PCR cycles and the number of DNA copies produced for 10 cycles of PCR only. Use a smooth curve. Remember TAILS!
		Independent variable – 
		Dependent variable –
3. Calculate the appropriate annealing temperature for the PCR primers you will use to genotype the family members for sickle cell. An easy rule of thumb is to add 4 ○C for every G or C in the primer, and 2 ○C for every A or T, then subtract 5 ○C.

Forward primer: 5’ - CATTTGCTTCTGACACAACTG - 3’

Annealing temperature =  (4 x ________ ) + (2 x ________ ) – 5 = ________ ○C
                                                # G & C              # A & T

		Reverse primer: 5’ – CCTTGAGGTTGTCCAGGTG – 3’

Annealing temperature =  (4 x ________ ) + (2 x ________ ) – 5 = ________ ○C
                                                # G & C              # A & T
4.  Use the following interactive to explore the technique and analysis of PCR reactions. 

PCR Interactive:

https://www.labxchange.org/library/items/lb:LabXchange:bc4846e2:lx_simulation:1



	


Tube

	


DNA Sample

	


Template DNA

	


Forward primer

	


Reverse primer

	


Nucleotide mix

	


Taq DNA pol

	


Final volume



	

1
	

Sam
	

14 ml
	

2.5  ml
	

2.5  ml
	

5  ml
	

1  ml
	

25  ml


	

2
	

Gloria
	

14 ml
	

2.5  ml
	

2.5  ml
	

5  ml
	

1  ml
	

25  ml


	

3
	

Nakeesha
	

14 ml
	

2.5  ml
	

2.5  ml
	

5  ml
	

1  ml
	

25  ml


	

4
	

Marcus
	

14 ml
	

2.5  ml
	

2.5  ml
	

5  ml
	

1  ml
	

25  ml




5. You have been provided with DNA samples for four members of the Ryan family. Sam is the father and Gloria is the mother. While neither Gloria nor Sam have sickle cell disease, both have a family history of sickle cell and are concerned that their children might inherit the disease. Nakeesha is their 4-year-old daughter, and Marcus is a baby boy. The PCR reactions for determining the genotypes of the family members are shown below:





PCR program:
 		30 cycles of (30 sec @ 95 ○C, 30 sec @ ____ ○C, 1 min @ 72 ○C)
		* the annealing temperature comes from your calculations in 3. 
6.  At the end of the PCR reaction, how many copies of each template molecule of DNA should be in each tube?

7.  What is the expected size (in base pairs) of the PCR products? Remember, these have not yet been cut by the restriction enzyme.




III. Restriction Digestion of PCR Products



 PROCEDURE (work individually)

1.  Use the following interactive to explore the function of restriction endonucleases. Note, however, that you will be cutting your PCR products from step II, not plasmid DNA.

Restriction Digest Interactive:

https://www.labxchange.org/library/items/lb:LabXchange:783397ff:lx_simulation:1


2.  Observe the setup for DdeI restriction digest of the PCR products in the table below. How much water needs to be added to each tube to produce the correct final volume? 


	


Tube

	


PCR Sample

	


H2O

	


PCR Product

	


Buffer

	


DdeI Enzyme

	


Final Volume



	

1
	

Sam
	

___ ml
	

2.5  ml
	

1  ml
	

1  ml
	

10  ml


	

2
	

Gloria
	

___ ml
	

2.5  ml
	

1  ml
	

1  ml
	

10  ml


	

3
	

Nakeesha
	

___ ml
	

2.5  ml
	

1  ml
	

1  ml
	

10  ml


	

4
	

Marcus
	

___ ml
	

2.5  ml
	

1  ml
	

1  ml
	

10  ml









Incubate the reactions at 37 ○C for 1 hour. 


IV. Gel Electrophoresis and Analysis of Results


· PROCEDURE (work in groups)
Lane 1: DNA Size Standard
Lane 2: Sam
Lane 3: Gloria
Lane 4: Nakeesha
Lane 5: Marcus
1,000
900
800
700
600
500
400
300
200
100


1.  You will first need to add 2 ml of loading dye to each sample. The loading dye contains glycerol to allow the sample to sink to the bottom of the well, as well as a blue dye to track the progress of electrophoresis. Complete the interactive below to experience loading your PCR/DdeI digest samples, running the gel, and visualizing the DNA bands on the gel:




Gel Electrophoresis Interactive:

https://www.labxchange.org/library/
items/lb:LabXchange:ef33215a:lx_
simulation:1
 
On the right are the gel results for your experiment. Complete the analysis questions based on these results.
2.  Using a ruler, measure the distance traveled in millimeters (mm) for each band in Lane 1: DNA Size Standard. Measure from the bottom edge of the well to the center of each band. Record your results in the table on the following page. 


	


Length of DNA (base pairs)

	


Migration Distance (mm)



	

1,000
	




	

900
	




	

800
	




	

700
	




	

600
	




	

500
	




	

400
	




	

300
	




	

200
	




	

100
	






      Next, graph your results below. Note that the y-axis is in log (base 10) scale. This will help produce a straight line. Remember TAILS! Draw a line of best fit. Is length of DNA directly or inversely proportional to migration distance? 
__________________________________
Finally, measure the distance migrated for each band in lane 2 and label these points on the graph. What is the size (in base pairs) of each band? How do these compare to the values predicted in part I, step 4?
__________________________________


10
20
30
40
50
60
80
70
90
100










3.  Determine the genotypes (use HbA and HbS for the alleles) and diagnosis for each family member (unaffected, sickle cell trait, or sickle cell disease):


	


	


Genotype

	


Diagnosis (Phenotype)



	


Sam

	


	




	


Gloria

	


	




	


Nakeesha

	


	




	


Marcus

	


	







4.  Use a Punnett square to determine the probability of Sam and Gloria having an unaffected child, a child with sickle cell trait, or a child with sickle cell disease. You will need the genotypes of Sam and Gloria determined in the previous step.

	
		% unaffected :	_______________
			
		% sickle cell trait: 	_______________

		% sickle cell disease:	_______________

 5.  Complete the pedigree below using the provided key:

  
Male
Female
Unaffected
Sickle cell trait
Sickle cell disease








6.  Sam’s brother died of sickle cell disease but neither of his parents had the disease. Gloria’s sister tested unaffected, and both of her parents had sickle cell trait. Add these additional family members to the pedigree above. 


V. COVID-19 Testing Using Q-RT-PCR (Extension)


If you are interested, watch this video on how quantitative-reverse transcriptase-PCR (Q-RT-PCR) is being used to test for SARS-CoV-2, the virus that causes COVID-19. Q-RT-PCR is a faster, more precise version of PCR that can be used to identify and quantitate the amount of a specific RNA sequence in a sample. The video also includes a review of the virus structure and the basics of PCR.


Extension Video on COVID-19 Testing:


https://www.minipcr.com/news/covid-19-qpcr-testing/


Questions: 
1. Explain how you determined the sickle cell genotype for individuals of the Ryan family. What specific feature of the sickle cell mutation allowed you to use this strategy to distinguish between the HbA and HbS alleles?


2. PCR is a synthetic version of the DNA replication process that occurs in E. coli (and our own) cells. Both forms of replication rely on many of the same principles, but there are also differences. Use the table below to compare bacterial DNA replication and PCR:



	


	



E. coli Replication

	


PCR



	


DNA template

	


	




	

Number of DNA template copies produced
	


	




	

Separates hydrogen bonds between DNA strands
	


	




	

Provides 3’ -OH for extension by DNA polymerase 
	


	




	

DNA polymerase used for extension
	


	




	

Optimal temperature for extension
	


	






3.  Describe one cycle of PCR including the 3 steps involved. Be sure to include and underline the 4 main ingredients of a PCR reaction.




4.  Why do you think G and C nucleotides contribute more to the annealing temperature than A and T nucleotides? Consider the structure of DNA and the chemical nature of base pairing. 


5.  Why did you incubate the DdeI restriction enzyme reaction at 37 ○C? What is the purpose of the buffer in a restriction enzyme reaction?


6.  Which electrode (positive or negative) is placed at the bottom of the gel, farthest from the wells where the DNA samples are loaded? Why?


7.  If DNA is colorless when dissolved in water, how were you able to see the DNA after it ran through the gel?


8.  What is the purpose of the DNA size standard (also referred to as a marker or ladder)? 




References:


This lab was adapted from:

 miniPCRTM Sickle Cell Genetics Lab: Diagnosing Baby Marie – Instructor’s Guide © 2018 by Amplyus LLC. Retrieved from https://www.minipcr.com/wp-content/uploads/Sickle-Cell-Lab-1.0_instructors-guide_vF.pdf