“Comparing Hominid Skulls” Lab

PREDICT THE SPECIES OF YOUR FOSSIL SKULL:

I PREDICT IT IS A EARLY HOMINID.

Name: Jennifer Cicchini

LAB

 SOLVING GENETICS PROBLEMS

Introduction:

One of the facts of life involves the different types of offspring that can be produced as a result of sexual reproduction. Offspring may have traits of one parent, both parents, or neither parent. This depends upon the genes the offspring receive from the parents.  In order to determine the probable appearance of an offspring, we must first know something about the parents. What genes do they possess? We must know what possible genes can be found in the gametes of each parent and the possible ways these genes may combine during fertilization.

In today's lab, your goal will be to learn how to solve genetics problems. First, here are some terms for easy reference:

Phenotype - The physical appearance of an organism.

Genotype - The genes of an organism.

Allele - One of two or more forms of the same gene.

Locus - A specific location on a chromosome where a gene is found.

Heterozygote - An organism having two different alleles for a trait.

Homozygote - An organism having two of the same alleles for a trait.

Homologous Chromosomes - Chromosomes having genes for the same kinds of characteristics that pair during meiosis.

Genetics Problem Solving:

In this activity, the class will be divided into groups. Each group will be assigned a  problem to solve and write on the whiteboard. Everyone should follow along with their explanation and write the results down for themselves on a separate sheet of paper. It may help to solve the problems using the following guidelines:

a.       Assign letters (alleles) to the various characteristics.

b.      Determine the phenotype and genotype of each parent and indicate a mating.

c.       Determine all the possible kinds of gametes each parent can produce.

d.      Determine all the possible allele combinations that can result when these gametes combine to form the offspring, using a Punnett square.

e.      Answer the question asked in the question.

Genetics Problems:

1.       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 , The genetic gametes would be R.r.

b.      RrYy The genetic gametes would be RY, ry, Ry,rY

c.       rrYy The genetic gametes would be rY, ry.

d.      RrYY The genetic gametes would be RY, rY

2.       For each of the following, state whether the genotype of a diploid or haploid cell is represented.

a.       D , A haploid cell is represented.

b.      GG, A diploid cell is represented.

c.       P, A haploid cell is represented.

d.      Ee, A diploid cell is represented.

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 are these colors inherited?  No calculations needed! Name the type of inheritance this represents.

Both guinea pigs have two different alleles. It is a polygenic inheritance.

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 only way for the offspring to be black is if each parent has the recessive allele (b), the parents are heterozygous. It is a incomplete dominance inheritance

.

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 phenotype of the offspring would be 3:1, ¾ would be yellow and ¼ would be green. This type of inheritance is dominant.

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 is also a dominate inheritance and the offspring phenotype would be 3:1, 3 white and 1 yellow.

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 is a codominace  inheritance and the phenotype ratio would be 25% red, 25 white and 50% pink.

8.       In humans, the condition for normal blood clotting dominates the condition for non-clotting or hemophilia. Both alleles are linked to the X chromosome. A male hemophiliac marries a woman who is a carrier for this condition. In this respect, a carrier is a woman who has an allele for normal blood clotting and an allele for hemophilia. What are the chances that if they have a male child he will be normal for blood clotting? What is this type of inheritance called?

The inheritance type is called sex linked , there is a 50% chance that the male child will be normal for blood clotting.

9.       A person with an allele for type A blood and type O blood marries someone with an allele for type B blood and type O blood. List the types of offspring they could have and the probability for each blood type in the offspring. (A allele = I^A, B allele = I^B, O allele = i) What is the expected phenotype ratio of the offspring? What is this type of inheritance called?

The expected phenotype ratio of the offspring would be 1/4 AB, 1/4 would be AO or type A, 1/4 BO or B, ¼ would be OO or O blood type. This type of inheritance is codominance.

10.   Skin color in humans becomes darker by the number of dominant alleles; AABBCC have the darkest skin and aabbcc have the lightest skin. Place these genotypes in sequence according to the color of skin expected for each. Place the darkest skin first. What is this type of inheritance called?

Genotypes: AaBbCc, AAbbcc, aabbCc, AaBBCc, AaBBCC.

The darkest to lightest genotypes in sequence are: AaBBCC, AaBBCc, AaBbCc, AAbbcc, aabbCc

This type of inheritance is polygenic.

Part 1. Flip-a-coin

1.       Thoroughly read Flip-a-coin Lab.

2.       Ideally, the lab is performed with two pretend parents. If you work alone, that is fine; you will represent both parents! All sorts of things are possible today with reproductive technology :-).

3.       It is most important to understand that each "parent" will flip a coin, and the heads or tails on that flip of the coin represents one allele from the egg or sperm. See video here for help understanding how to complete the data sheet, below. Apologies for any background “hum.”

4.       Record your data in the Flip-a-coin Genetics Data Sheet. When it opens, it will might be "Read-only." Save with a new name to another location and then type in your data. Then you will be able to save the changes you make.

5.       When all the coin tossing is finished you created a child!

6.       Sketch your child at age 16, including the characteristics you have determined by flipping the coin.

8.       Attach both the data sheet and a photo or scan of the drawing of your son or daughter to your blog site.

Eyebrows (II)	N	N	NN	not connected
Eyebrow color	H	h	Hh	same as hair
Eyes distance apart	e	E	eE	average distance
Eyes size	E	e	Ee	medium
Eyes shape	A	A	AA	almond
Eyes slant	h	h	hh	upward slant
Eyelashes	l	L	lL	long
Eye color	abc	ABC	aAbBcC	Hazel
Mouth size	m	m	mm	narrow
Lips	l	l	ll	thin
Protruding lower lip	h	h	hh	absent
Dimples	d	D	dD	present
Nose size	n	N	nN	average
Nose shape	r	r	rr	pointed
Nostril shape	r	R	rR	rounded
Earlobe Attachment	F	F	FF	free
Freckles on checks	f	f	ff	absent
Hair color	abcd	AbCD	aAbbcCdD	dark blonde

THIS IS A PHOTO I DREW OF WHAT MY 16 YR OLD SON WOULD LOOK LIKE ACCORDING TO THE GENETICS I ENDED UP WITH  FROM THE FLIP A COIN LAB.

PART TWO: DNA TECHNOLOGIES:

The DNA technology I chose is Recombinant DNA Technology; It is a method in which the chosen DNA of one organism (Donor) is introduced to combine with the DNA of another organism termed as recipient organism. The result of it is that, the recipient organism acquires the genetic abilities of the donor. Altering the genome of an organism through introducing genes of interest is termed as gene manipulation or DNA recombinant technology. According to this mechanism has the capability to engineer new organisms, it is termed as genetic engineering. The reason I picked this is because I am Diabetic and found it interesting how they make insulin using recombinant DNA. When they first started making insulin they used animal DNA such as pigs but now it made using human DNA.

 As the illustration below shows the process of recombinant DNA, The DNA of the donor organism and the gene of interest is isolated and cut into fragments through using restriction endoclease. The fragments of the vector DNA and the fragments of the donor DNA are combined into DNA ligase which. Then it becomes the Recombinant DNA (rDNA). The introduction of the rDNA into the E Coli causes a multiplication of rDNA after this it creates multiplication of identical copies of rDNA.

"Recombinant DNA Technology." Recombinant DNA Technology. Web. 5 Apr. 2015. <http://www.tutorsglobe.com/homework-help/botany/recombinant-dna-technology-71762.aspx>.

DNA LAB (online)

Part 1. Your Personal Chromosome

1.       My chromosome was number 6.The  eight genes on my chromosome are:

1. Hemochromatosis is a disorder of iron metabolism wherein the body retains excess iron.  Excess iron is deposited in a variety of organs leading to their failure, and resulting in serious illnesses including cirrhosis, hepatomas, diabetes, cardiomyopathy, arthritis, and hypogonadotropic hypogonadism.

2. Opioid Receptor Gene encodes the opioid receptor which is the primary site of action for the most commonly used opioids including morphine, heroine and methadone. This receptor is a membrane of the g-protein coupled protein receptor family.

3. Estrogen receptor is a ligand –activated transcription factor composed of several domains important for hormone binding, DNA binding. It is also responsible for alternate splicing results in several ESRI, Mrna transcripts.

4. Connective tissue growth factor: The protein encoded by this gene is a mitogen that is secreted by vascular endothelial cells. The protein plays a role in cell adhesion in many cell types and is related to platelet derived growth factor.

5. Tumor Necrosis Factor is a multifunctional pro inflammatory cytokine secreted mostly by monocytes and macrophages. The primary role of TNF is the regulation of immune cells.

6. Advanced glycosylation end product is an important receptor for the amyloid beta peptide and it increases in Alzheimer’s disease. This protein is increased in neurons close to deposits of amyloid beta peptides and to neurofibrillary tangles.

7. Vascular endothelial growth factor is produced by many cell types including tumor cells, macrophages, Platelets, keratinocytes and renal mesangial cells.  They also play a role in normal physiological functions such as bone formation, wound healing and development.

8. Interlukin is a pro inflammatory cytokine that is secreted primarily by activated T cells and that has been implicated in several inflammatory diseases. It also stimulates a variety of cells to produce inflammatory mediators.

Genes cite source: "DNA Interactive: Discovering the DNA Structure and beyond." DNA Interactive: Discovering the DNA Structure and beyond. Web. 4 Apr. 2015. <http://www.dnai.org/>.

Pick one favorite of these eight genes.

1.       I found all the genes that I researched were really interesting. They all play an interesting role in our chromosomes. I did find that the one I seemed to relate to is the estrogen receptor and the different conditions that involve the gene like Breast cancer and artherosclerosis.