GENETICS AND INHERITANCE LAB

Part I Allele(s) from Mother Allele(s) from Father Genotype Phenotype Sex of child:    Tails   Boy Face shape R r Rr Round Chin Shape (I) V v Vv Very Prominent Chin Shape (II) r R rR Round  Cleft chin A A AA Absent Skin color Abcd abCd AabbcCdd Very Light Brown Hair type C c Cc Wavy Widow’s Peak w w ww Absent Eyebrows (I) B B BB Bushy Eyebrows (II) N n Nn Not Connected Eyebrow color H h Hh Red Eyes distance apart E E EE Close Together Eyes size e E eE Medium Eyes shape a A aA Almond (wide) Eyes slant h h hh Upward slant Eyelashes l l ll Short Eye color aAa AAA aAAAaA Light Brown Mouth size m m mm Narrow  Lips L L LL Large Protruding lower lip H H HH Protruding lower Lip Dimples d D dD Present Nose size n n nn Small Nose shape r R rR Rounded Nostril shape R R RR Rounded Earlobe Attachment f F fF Free Freckles on checks F F FF Present Hair color H h Hh Red
Part II      Genetic Problem I: For each of the diploid genotypes presented below, determine the genetic make up for all of the possible gametes that would result through the process of meiosis. Remember, each egg or sperm must have one of each letter. That letter can be upper or lower case.   A) Rr Answer = Rr r R Rr Rr Rr Rr B) RrYy Answer = RY, Ry, rY, ry Y y R RY Ry r rY ry C) rrYy Answer = rY, ry, rY, ry Y y r rY ry r rY ry D) RrYY Answer = RY,RY,rY,ry Y Y R RY RY r rY rY 2   For each of the following, state whether the genotype of a diploid or haploid cell is represented. A) Haploid B) Diploid C) Haploid D) Diploid
3     Yellow guinea pigs crossed with white ones always produce cream colored offspring. Two cream colored  guinea pigs when crossed produced yellow, cream and white offspring in the ratio of l yellow: 2 cream: l  white. Explain how these colors are inherited. No calculations needed! Name the type of inheritance this  represents.   The type of inheritance this represents is incomplete dominance. The colors are inherited because each guinea pig inherits 2 genes fro each trait from their parent.
4      In sheep, white is due to a dominant gene (B), black to its recessive allele (b). A white ewe mated to a  white ram produces a black lamb. What are the genotypes of the parents? You might need to construct  Punnet squares experimenting with different crosses to come up with this answer. Name the type of   inheritance this represents.  The genotypes of the parents are Bb,BB,bb. The type of inheritance this represents is Dominant.  B b B BB Bb b bB bb
5     In peas, yellow color (G) is dominant to green color (g). A heterozygous yellow is crossed with a green.  What is the expected phenotype ratio of the offspring? Name the type of inheritance this represents.  The type of inheritance this represents is Dominant. The expected phenotype ratio of the offspring is 3:1 (75% will be green and 25% will be yellow) G g G GG Gg g gG gg
6     White color (Y) is dominant to yellow color (y) in squash. A heterozygous white fruit plant is crossed with   a yellow fruit plant. What is the expected phenotype ratio of the offspring?  What is this type of inheritance called? This type of inheritance is dominant-recessive.  The expect phenotype ratio of the offspring is 3:1. Y y Y YY Yy y yY yy
7      In certain flowers, a cross between homozygous red and a homozygous white will always result in a pink     flower. A cross is made between two pink flowers. What is the predicted phenotype ratio of the colors red, pink and white appearing in the offspring?  What is this type of inheritance called?          This type of inheritance is called incomplete dominance.  R R Y R Y YR YR Y YY YR Y YR YR Y YY YR

DNA Technology

Process of Extracting DNA from a Human:

The follwoing Materials and Equipment are needed in the process of extracting DNA.  

• Sissors
• Buccal Swab
• Eppendorf Tube (Sample Tube)
• Warm Water Bath
• Micropippettors
• MicroCentrifuge

1. Lysis Solution
2. Concentrated Salt Solution
3. Isopropyl Alcohol

Step 1:  Take the Buccal Swab and swab across the inside of the cheek

Step 2:  After swabbing your cheek, use scissors to cut off the excess of the stem leaving only the tip to fall off into the Eppendorf Tube.

Step 3:  Using a Micropoppettor transfer Lysis Solution into the Eppendorf Tube.  (Lysis comes from the Greek word (to separate)).   Lysis contains two ingedents; degerents and Proteinase K.  Degerents disrupt the cell and nuclear envelope causing the cell to burst.  The proteinas K cause the Histone to free DNA.

Step 4:  The Eppendorf Tube is then placed into a Warm Water Bath which will allow DNA to be freed from cells. 

Step 5:  The Buccal Swab is removed from the Warm Water Bath and Concentrated Salt Solution is added to the Eppendorf Tube.  This causes the proteins & debris to clump together separating from the DNA.

Step 6:  The tube is then placed into a MicroCentrifuge with a second Eppendorf tube that is filled with water (to help with balance) and spun at a high rate of speed.  This will allow the heavy clumps of proteins & debris to sink to the bottom while strands of DNA are disrupted through the liquid.

Step 7:  Once the spinning process is completed the top liquid is then removed and placed into a clean Eppendorf Tube.

Step 8:  In the clean tube that you have just put the DNA liquid in, you will add Isopropyl Alcohol.

Step 9:   Since DNA is not soluble, by adding Isoproply Alcohol, you'll be able to see DNA with the naked eye. This only possible due to the fact the DNA will be in a clump.

Step 10:  Once again, place tube in MicroCentrifuge which will allow DNA to sink to the bottom of the tube.  When done, remove liquid from tube allowing DNA to dry.  After this the DNA can be frozen or a different solution may be added.

What specific type of tissue is that inside the mouth? 

Epithelial Tissue - Stratified Squamous

Gel Electrophoresis

Gel Electrophoresis is a process of adding electrical currents into a strand of DNA allowing it to be moved.  In manipulating the DNA, it is was determined that short strands moved quicker than the long strands.  Over a period of time short stands of DNA were found to move farther away from their starting point than long strands.  The "Gel" is a filter made of Jell-o or sponge like material with many holes at the top.  The DNA that had the same length would move at the same rate of speed and end up grouping together.  DNA is almost impossible to see by the naked eye therefore staining must happen.  By using this method DNA would sort themselves. 

There are at least 10 steps if not more in the Electrophoresis process.

1.   Make a gel substance with the same consistancy as jell-o.  This can be done by taking a table spoon of powder Agarose, flask, gel mold gel comb, buffer liquid (Salt water solution) & microwave. 
2.  With a micropippettor add a loading buffer to the DNA sample.  The loading buffer contains dye.  
3. Transfer the DNA sample into the well of the gel using a micropippettor.
4. Next take a sample from the DNA size standard and put into the well of the gel.  Again using a new micropippettor.
5. The Eletrophoresis box should have a red (Positive current) and a black (negative current) cable.  Remember on your power source make sure black goes to black and red goes to red.  
6. Pour your liquid buffer (Salt water solution) into eletrophoresis box.
7. Gengle set your gel mold down into the liquid buffer in the eletrophoresis box.  Make sure your wells are closet to the negative charge cable.
8. Turn on eletrophoresis box.  Notice, repelled by the negative charge, the DNA moves through the gel towards the positive current at the other end.  
9. Remove gel from mold and place in stain solution.  After 1/2 hour to an hour place on UV light box.
10. In order to see DNA with the naked eye one must use a stain in the DNA loading buffer.  The stain solution is Ethidium Bromide.  Ethidium Bromide binds to the DNA allowing visibility under a fluorescent light.  (Whenever using a light box remember to wear eye protection.)
11. Turn on Light Box

Hughes, Howard. "Gel Electrophoresis." Gel Electrophoresis. Web. 15 Nov. 2015.

 

 

 

 

 

Lab 10 Cell Reproduction and DNA

EIGHT GENES OF MY CHROMOSOME

The number for my Chromosome was 18 

• MIR8078 Also known as hsa-mir-8078

1. MiRNA's are trnascribed by RNA polymerase II as part of capped and polyadenylated primary transpcripts (pri-miRNAs) that can be either protein-coding or non-coding.
2. The mature miRNA is incorporated into a RNA-induced silencing complex (RISC), which recognizes target mRNAs throught imperfect base pairing with the miRNAs and most commonly results in translational inhibition or destablilization of the target mRNA. [provided by RefSeq, Sep 2009]  

•  COLEC12 Also known as CLP1; NSR2; SRCL; SCARA4

1. This gene encodes a member of the C-lectin family, proteins that possess collagen-like sequences and carbohydrate recognition domains.
2. A protein that can bind to carbohydrate antigens on microorganisms, facilitating their recognition and removal. [provided by RefSeq, Sep 2009]   

• YES1 Also known as Yes; c-yes; HsT441; P61-YES

1.  This gene is the cellular homolog of the Yamaguchi sarcoma virus oncogene.
2. The encoded protein has tyrosin kinase activity and belongs to the src family of proteins. 
   [provided by RefSeq, Sep 2009]

• TYMS Also known as TS; TMS; HST422

1. Thymidylate synthase catalyzes the methylation of deoxyuridylate to deoythymidylate using 5, 10 methylene-THF as a cofactor. 
2. the enzyme has been of interest as a target for cancer chemotherapeutic agents.  [provided by RefSeq, Sep 2009]  

•  USP14 Also known as TGT

1.  the gene encodes a member of the ubiquitin-specific processing (UBP) family of porteases that is a deubiquitinating enzyme (DUB) with Cys domains.
2. Mice with a mutation that results in reduced expression of the ortholog of this protein are retarded for growth, develop severe tremors by 2 to 3 weeks of age followed by hindlimb paralysis and death by 6 to 10 weeks of age. [provided by RefSeq, Sep 2009]

•  CETN1 Also known as CEN1; CETN

1.  the protein encoded by this gene plays important roles in the determination of centrosome position and segregation, and in the process of microtubule severing.
2. This protein is localized to the centrosome of interphase cells, and redistributes to the region of the spindle poles during mitosis, reflecting the dynamic behavior of the centrosome during the cell cycle.  [provided by RefSeq, Sep 2009] 

• LPIN 2 

1.  Mouse studies suggest that this gene functions during normal adipose tissue developement and may play a role in human triglyceride metabolism.
2. This gene represents a candidate gene for human lipodystrophy, characterized by loss of body fat, fatty liver, hypertriglyceridemia, and insulin resistance.  [provided by RefSeq, Sep 2009] 

•  MIR6718 Also known as hsa-mir6718

1.  MicroRNA's (miRNAs) are short non-coding RNAs that are involved in post-transcritional regulation of gene expression in multicellular organisms by affecting both the stability and translation of mRNAs.
2. The primary transcript is cleaved by the Drosha ribonuclease III enzyme to produce a stem-loop precursor miRNA (pre-miRNA), which is further cleaved by a cytoplasmic dicer ribonuclease to generate the mature miRNA.  [provided by RefSeq, Sep 2009]   

Part II 

My opinion of Smart Sparrow as a learning tool:

I enjoyed using Smart Sparrow for the second part of this lab and getting a better perspective on DNA and how a Double Helix is formed.  I found that having instant correction and direction was very helpful and allowed me to make the corrections.  I would not mind using this functions for futher labs.