Review Sheet answers

1. Garden peas are a great visible example of heredity. The plants have many characteristics (pea color, pea shape, pod color and flower color) which are all determined by a single gene and assort independently.
   
2. Only dominant phenotypes can be heterozygous. In order to show a recessive phenotype the genotype MUST be homozygous because with even one dominant allele, the phenotype will be dominant.
   
3. No. you must possess two recessive alleles to show the allele phenotype.
   
4. A test cross is a cross done in order to determine the genotype of a parent. It is used when the parent displays the dominant trait, but it is unknown whether or not the genotype is homozygous or heterozygous. If the unknown parent is crossed with a recessive phenotype parent and all progeny display the dominant genotype, the unknown parent is homozygous dominant. If the cross results with half of the offspring with the recessive phenotype and half with the dominant, then the unknown parent is heterozygous.
   
5. The gene for the flower color is showing incomplete dominance. In this situation the two flowers are both homozygous (one for blue phenotype and the other for yellow phenotype). When the grower crosses both flowers all of the offspring are heterozygous and are thus green.
   
6. Mendel's law of independent assortment
   
7. Each offspring will receive one allele, this follows mendel's law of segregation.
   
8. When you cut one plant and replant it, you are not changing any genetic material. A seed may yield a plant that varies from the parent because of crossing over. During meiosis homologous chromosomes align and genetic material exchanges between chromosomes. Ultimately at the end of meiosis, all sister chromatids separate and you end up with four daughter cells which genetically different material.
   
9. No, genes assort independently. Normally, the expression of one gene does not determine the expression of another.
   
10. The F1 generation would yield all heterozygous offspring, and so the F2 generation would be determined by crossing two heterozygous plants. When you cross two heterozygous parents, the offspring will have the phenotypic ratio of 3(dominant) to 1(recessive). Therefore, 25% of the offspring will show the recessive trait which is 8/32.
    
11. This varies, but typically one.
    
12. Seed color – Yy, Seed shape – Rr
    

    
    

    	YR	Yr	yR	yr
    YR	YYRR	YYRr	YyRR	YyRr
    Yr	YYRr	YYrr	YyRr	Yyrr
    yR	YyRR	YyRr	yyRR	yyRr
    yr	YyRr	Yyrr	yyRr	yyrr

    Yellow, round: 9, Green, round: 3, Yellow, wrinkled: 3, Green, wrinkled: 1
    

    
     

    Two heterozygous parents crossed in a dihybrid cross is always 9:3:3:1
    

     

13. Gametes produced are:
    
    1. E,F
       
    2. E,e,F
       
    3. e, f
       
    4. E,e,F,f
       
14. Grandparents = ¼, great grandparents 1/8
    
15. No. Hand cells are autosomal cells (non sex cells). Changes in cells in your hand will only affect future hand cells. Offspring are only affected by daughter cells produced in meiosis, which takes place in sex cells.
    
16. 50% chance. Each time you conceive you have a 50/50 chance of having a boy or a girl. The sex of previous children do not affect the sex of future children.
    
17. A roans genotype is heterozygous, Rr. Two roans cannot produce all roan colored offspring, the offspring have a 25% chance of being red, 50% chance of being roan and 25% chance. Roan cows are displaying codominance, when the cow has one allele for red, and another for white, both are expressed resulting in roan color. Since only 50% of the offspring will be heterozygous, only 50% will be roan.
    
18. For some traits, there are more than two possible alleles. An example of this is blood types and fruit fly eye color.
    
19. Don't answer this question.
    
20. Some blood types express co-dominance, and other complete dominance. There are multiple alleles, and blood type is made up of two genes (one for ABO, the other Rh- or +).
    
21. Both parents are heterozygous. Mary has AO, John has BO, Joan has OO, James has AO, Pete has BO.
    
22. 50% chance of B, 50% chance of O.
    
23. No there was not. It would be impossible for Mr Doe and Mrs Doe to have a child with O blood type because Mrs Doe can only produce gametes with A or B which will always be dominant over O.
    
24. Yes, both parents could be heterozygous (BO) and would then be able to have children that are OO (25% chance).
    
25. Because females have 2 X chromosomes. If one of the X chromosomes has a mutation they have another correct allele on their other X chromosomes. Males only have one shot with certain traits because they only have 1 X chromosome. If there is a problem with that allele they will express the recessive phenotype.
    
26. See 25
    
27. If parents know their genotype, which was probably determined using a pedigree, they can calculate the probability of their child inheriting a specific trait. For some parents this could deter them from having children.
    
28. In science a large sample size is always important. It is impossible to say that something is statistically correct without repeating it many times, you always have to account for randomness and error.
    
29. G
    
30. Farmers could selectively breed for sheep that can only produce white offspring, which would be sheep with the genotype WW.
    
31. The best option would be to cross the two together, if there is only one phenotype visible than this is the dominant trait, if both phenotypes are present, let's say two hairy and two naked, than create an F2 generation to see what the phenotypic ratio would be.
    
32. Inbreeding is discouraged because it limits genetic diversity and recessive traits (which many diseases are) tend to surface more often.
    
33. If the trait never appears in males than most likely having it is so lethal that the males don't even make it to birth.
    

Females can be carriers if one X chromosome carries the allele for the recessive trait and the other carries the allele for the dominant trait. Since females have 2 X chromosomes the dominant trait will be expressed and the recessive will do unnoticed, but the female still has a 50/50 chance of passing the recessive trait on to the offspring. If the offspring is a male, they will only receive 1 X chromosome in total (because their father gave them their Y chromosome), and the recessive trait will be expressed.

Sex Linked Problems

1. Mate a colorblind male and a carrier female.
   a) What are the chances of having a normal boy?
   b) What are the chances that their sons will be colorblind?
   c) What are the chances of having a normal girl?
   d) What are the chances that their daughters will be carriers?
   
2. Red-green colorblindness is caused by a sex-linked recessive allele. A color blind man marries a woman with normal vision whose father was color blind. What is the probability that they will have a color blind daughter? What is the probability that their first son will be color blind?
   
3. A man with hemophilia (a recessive, X-linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait.
   a) What is the probability that a daughter of this mating will be hemophiliac?
   b) What are the chances of having a son with the disorder?
   c) What are the chances of their daughters being a carrier?
   
4. A recessive sex-linked gene (h) located on the X chromosome increases blood-clotting time. This causes the genetic disease, hemophilia.
   

   a) Explain how a hemophilic offspring can be born to two normal parents.
   

    b) Can a female ever develop hemophilia?
   

   
    

Practice Problems (More) and Solutions

Learning Genetics Can Be Fun

1. A pea plant with round seeds is crossed with a pea plant that has wrinkled seeds. For the cross, indicate each of the following:

a) the genotype of each of the parents if the round seed plant is heterozygous.

b) the gametes produced by the parents

c) the genotypes and phenotypes of the F₁ generation

d) the F₂ generation if two round plants from the F₁ generation were crossed

2. A plant of genotype TTYy was crossed with one of genotype ttYy. If the cross yielded 800 seeds, what number of what phenotypes probably resulted? What were their genotypes?

3. In summer squash, white fruit color is dominant. Yellow is recessive. A squash plant that is homozygous for white is crossed with a homozygous yellow one. Predict the appearance of

a) the F₁ generation (F₁ = all white)

b) the F₂ (F₂ = 3/4 white and 1/4 yellow)

c) the offspring of a cross between an F₁ individual and a homozygous white individual. (A crossbetween an F₁ individual and a homozygous white individual wouldproduce only white offspring.)

4. For Dalmatian dogs, the spotted condition is dominant to non-spotted.

a) Using a Punnett square, show the cross between two heterozygous parents.

b) A spotted female Dalmatian dog is mated to an unknown male. From the appearance of the pups, the owner concludes that the unknown male was a Dalmatian. The owner notes that the female had six pups, three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?

5. In horses, the trotter characteristic is dominant to the pacer characteristic. A male trotter mates with three different females, and each female produces a foal. The first female, a pacer, gives birth to a foal that is a pacer. The second female, also a pacer, gives birth to a foal that is a trotter. The third female, a trotter, gives birth to a foal that is a pacer. Determine the genotypes of the male, all three females, and the three foals sired.

6. Imagine for hair color that B gives brown hair and b gives blonde hair. Use a Punnett square to determine the following in a cross of two heterozygous parents.

a) What are the chances of the offspring being homozygous brown haired?

b) What are the chances of the offspring having blonde hair?

c) What are the chances of the offspring being heterozygous brown haired?

d) What is the genotypic ratio?

e) What is the phenotypic ratio?

f) Is there a heterozygous blonde haired offspring? Why?

g) If curly hair is dominant to straight hair, what letters will we use to show these genes?

h) A heterozygous curly haired male marries a straight haired female. What are their genotypes?

i) What would be the gametes for the male parent?

j) What would be the gametes for the female parent?

k) What are the chances of the offspring being homozygous curly haired?

l) What are the chances of the offspring having straight hair?

m) What are the chances of the offspring being heterozygous curly haired?

n) What is the genotypic ratio?

o) What is the phenotypic ratio?

p) Is there a heterozygous straight haired offspring? Why?

7. Blonde hair is due to a recessive gene, while brown hair is dominant. Agnes has brown hair, but

her mother was a blonde. Ralph, her husband, has brown hair also, and his father was blonde. Is it

possible for Ralph and Agnes to have blonde children? If so, what percentage?

8. Thalassemia is a serious human genetic disorder that causes severe anemia. The homozygous condition (T^mT^m) leads to severe anemia. People with thalassemia die before sexual maturity. The heterozygous condition (T^mTⁿ) causes a less serious form of anemia. The genotype TⁿTⁿ causes no symptoms of the disease. Indicate the possible genotypes and phenotypes of the offspring if a male with the genotype T^mTⁿ marries a female of the same genotype.

9. For guinea pigs, black fur is dominant to white fur color. Short hair is dominant to long hair. A guinea pig that is homozygous for white and homozygous for short hair is mated with a guinea pig that is homozygous for black and homozygous for long hair. Indicate the phenotype(s) of the F₁ generation. If two hybrids from the F₁, generation are mated, determine the phenotypic ratio of the F₂ generation.

10 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid. The mating with female B, which is black, solid, produces a single pup, which is white, spotted. The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted. Indicate the genotypes of the parents.

11. For human blood type, the alleles for types A and B are codominant, but both are dominant over the type O allele. The Rh factor is separate from the ABO blood group and is located on a separate chromosome. The Rh+ allele is dominant to Rh-. Indicate the possible phenotypes from the mating of a woman, type,O, Rh-, with a man, type A, Rh+.

12. In a disputed paternity case, a woman with blood type B has a child with type O, and she claimed that it had been fathered by a man with type A. What can be proved from these facts? (man could be father but this is not proof)

13. Multiple alleles control the intensity of pigment in mice. The gene D¹ designates full color, D² designates dilute color and D³ is deadly when homozygous. The order of dominance is D¹ > D² > D³. When a full color male is mated to a dilute color female, the offspring are produced in the following ratio: two full color to one dilute to one dead. Indicate the genotypes of the parents.

14. Multiple alleles control the coat color of rabbits. A gray color is produced by the dominant allele C. The C^ch allele produces a silver-gray color when present in the homozygous condition, C^chC^ch, called chinchilla. When C^ch is present with a recessive gene, a light silver-gray color is produced. The allele C^h is recessive to both the full color allele and the chinchilla allele. The C^h allele produces a white color with black extremities. This coloration pattern is called Himalayan. An allele C^a is recessive to all the other alleles. The C^a allele results in a lack of pigment, called albino. The dominance hierarchy is C > C^ch > C^h > C^a. The table below provides the possible genotypes and phenotypes for coat color in rabbits. Notice that four genotypes are possible for full color but only one for albino.

Phenotypes Genotypes

Full color CC, CC^ch, CC^h, CC^a

Chinchilla C^chC^ch

Light gray C^chC^h, C^chC^a

Himalaya C^hC^h, C^hC^a

Albino C^aC^a

a) Indicate the genotypes and phenotypes of the F₁ generation from the mating of a heterozygous Himalayan coat rabbit with an albino coat rabbit.

b) The mating of a full color rabbit with a light gray rabbit produces two full color offspring, one light gray offspring, and one albino offspring. Indicate the genotypes of the parents.

c) A chinchilla color rabbit is mated with a light gray rabbit. The breeder knows that the light grey rabbit had an albino mother. Indicate the genotypes and phenotypes of the F₁ generation from this mating.

d) A test cross is performed with light gray rabbit, and the following offspring are noted: five Himalayan color rabbits and five light gray rabbits. Indicate the genotype of the light-gray rabbit.

15. A geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. If oval radishes are crossed with oval radishes, the following phenotypes are noted: 100 long, 200 oval, and 100 round radishes. For the cross between a pure-breeding round and a pure-breeding long radish, use symbols to show the F₁ and F₂ generations.

16. Palomino horses are known to be caused by the interaction of two different genes. The allele C^r in the homozygous condition produces a chestnut, or reddish color, horse. The allele C^m produces a very pale cream color, called cremello, in the homozygous condition. The palomino color is caused by the interaction of both the chestnut and cremello alleles. Indicate the expected ratios in the F₁ generation from mating a palomino with a cremello.

17 In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. The recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light-brown, color is produced. The genes that produce black or tan coat color rely on the gene C, which produces pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) CCBB x Ccbb (b) ccBB x CcBb (c) CcBb x ccbb (d) CcBb x CcBb

18. For ABO blood groups, the A and B genes are codominant but both A and B are dominant over type O. Indicate the blood types possible from the mating of a male who is blood type O with a female of blood type AB. Could a female with blood type AB ever produce a child with blood type AB? Could she ever have a child with blood type O?

19. Baldness (H^B) is dominant in males but recessive in females. The normal gene (Hⁿ) is dominant in females, but recessive in males. Explain how a bald offspring can he produced from the mating of a normal female with a normal male. Could these parents ever produce a bald girl? Explain your answer.

20. Use the phenotype chart (pedigree) at the right to answer the following questions.

a) How many children do the parents A and B have?

b) Indicate the genotypes of the parents.

c) Give the genotypes of M and N.

21. For the family outlined, draw a pedigree

a) Generation I : A male with blood type O marries a female with blood type A.

b) Generation II : They have 4 children. The first born is a male with A type blood, the second born is a girl with O type blood, and the last two children were identical twin boys.

c) Generation III : The last boy married a woman with type B blood. They had two girls, one with AB blood type and one with B blood type.

SOLUTIONS:

http://kvhs.nbed.nb.ca/gallant/biology/genetics_learning_problems_121_solutions.html

Homework – READ THIS

Homework

1. Do your Genetics Review!!!!
   

   Next Friday we will be doing and in class review, but this will not be possible if you do not complete the review beforehand
   

2. Website
   
   1. Click on the link for practice problems. It will take you to a site where you can practice crosses!!!!
      
3. Textbook
   
   1. Page 159 Questions 2,3,5,9,12,14,16,17(a and b)
      

 

It is so important that you do your homework. Please you the tips sheet that I handed out in class, it will really really help you solve the problems.

I will be in the classroom every day at lunch Monday-Friday. Do not hesitate to come and ask questions or have me go over problems with you.

 

Mendel’s Laws

Mendel's Laws

Law of Segregation

• For each trait and organism has two hereditary factors now called genes
  
• Alternate forms of the gene are called alleles
  
• The two alleles separate during meiosis and only one is passed on to the gamete
  
• The two gametes unite during fertilization to produce an offspring with two genes
  
• If the two alleles differ, the phenotype will reflect the dominant allele
  
• Not all genes have dominant and recessive alleles
  

 

Law of Independent Assortment

The inheritance of one gene does not normally affect the inheritance of another gene

 

 

 

Genetic Disorders List

Disorder	Student
22q11.2 deletion syndrome	Arun
Angelman syndrome	Leah
Canavan Disease	Batula
Celiac Disease	Waseem
Charcot-Marie-Tooth Disease	Justin
Color Blindness	Stephanie
Cri du Chat	Abdul
Cystic Fibrosis	Kat
Down Syndrome	Alyssa
Duchenne muscular dystrophy	Lena
Hemophilia	Asiyaa
Klinefelter Syndome	Robin
Neurofibromatosis	Sadiq
Phenylketonuria	Kaamel
Prader-Willi syndrome	Zamzam
Sickle Cell disease	Habon
Spina Bifida	Sarah
Tay Sachs Disease	Ben
Turner Syndrome	Sal
Parkinsons Disease	Zeinab A
Leukemia	Reem
Obesity	Hira

Dwarfism .......................................................Katylyn

Alzheimers .....................................................Lara

Practice Crosses

Please use this website to practice any type of cross!! It gives you the questions and the solution!

http://www.ksu.edu/biology/pob/genetics/intro.htm

And you can follow this link to find all the hints and tricks to solving mendelian crosses!

http://kvhs.nbed.nb.ca/gallant/biology/genetics_problems_tips.html



Good Luck!!!!

Monohybrid Crosses

Monohybrid Crosses

Allele = two or more alternate forms of a gene. Alleles are located on the same location of homologous chromosomes

Dominant = the allele that determines expression

Recessive = these alleles are overruled by dominant alleles.

Genotype = which alleles the organism contains

Phenotype = observable traits of an organism

Homozygous= both alleles are the same

Heterozygous = the two alleles are different

You can determine the genotype and the phenotype of progeny by doing a monohybrid cross and using a Punnett square.

Using Mendel's work, we cross one round plant (Rr) with one wrinkled plant (rr). What are the potential genotypes and phenotypes of the F1 generation.

Solution:

	r	r
R	Rr	Rr
r	rr	rr

Genotypes : 2/4 are Rr, 2/4 are rr

Phenotypes: 2/4 are round, 2/4 are wrinkled.

Now let's cross two heterozygous plants. What does this mean?? Rr. What is the phenotype of these plants? Round (because it is dominant).

	R	r
R	RR	Rr
r	Rr	rr

Genotypes of the F1 generation? ¼ RR, 2/4 Rr, ¼ rr

Phenotypes of the F1 generation? ¾ round, ¼ wrinkled.

Mendel’s Laws

Mendel

Gregor Mendel was an Autrian monk who looked at heredity patterns in pea plants. He noticed that the pea plants had varying characteristics: round peas, wrinkled peas, green or yellow peas, different flower colors and different pod colors. He also noticed that when he crossed certain plants he ended up with a variety of results.

Please read pages 130-132 for a detailed outline on his experiments.

From his experiments Mendel devised a few laws of inheritance.

The first set of laws is the Law of Segregation

1. For each trait an organism has two hereditary factors called genes
   
2. Alternate forms of the gene are called alleles
   
3. The two alleles separate during meiosis and only one is passed on the gamete
   
4. The two gametes unite during fertilization to produce an offspring with two genes
   
5. If the two alleles differ, the phenotype will reflect the dominant allele
   
6. Not all genes have dominant and recessive alleles.
   

The second set of laws is the Law of Independent Assortment

1. The inheritance of one gene does not normally affect the inheritance of another gene
   

Some genes are linked to other genes. For instance, recessive genes found on the X chromosome will cause a recessive trait in a male but will only cause a recessive trait in a female if she is homozygous recessive.

Karyotype Chart

Karyotype Chart

Use karyotype charts to look at chromosomes. This chart shows all the chromosomes paired up with one and other (homologous chromosomes). It is especially important to see nondisjunction.

Nondisjunction – In meiosis during anaphase one, when homologous chromosomes are pulled to opposite poles of the cell, sometimes, both chromosomes move to one pole of the cell. The result of this is a daughter cell with an extra chromosome and a daughter cell with a missing chromosomes. When gametes fuse during fertilization you can end up with an offspring with an extra chromosome, or an offspring with a missing chromosomes.

Normally you have 2 of each chromosome. An extra chromosome (3 in total) is called trisomy. An example of this is Down Syndrome = trisomy on chromosome 21.

Kleinfelter syndrome is another example of trisomy. In this situation you have 2 X chromosomes and a Y chromosome.

Monosomy is when you are missing a chromosome. Turner syndrome is the result of only one X chromosome (not 2^nd X, or Y chromosome).

Answers to Game Day

Games Day – Here are the answers for studying purposes!!

INDICATE IF THIS OCCURS IN MITOSIS, MEIOSIS, OR BOTH

DAUGHTER CELL HAS 23 CHROMOSOMES - Meiosis

1 DIVISION OCCURS - Mitosis

CROSSING OVER - Meiosis

GENETICALLY DIFFERENT DAUGHTER CELLS - Meiosis

4 DAUGHTER CELLS - Meiosis

ANAPHASE PULLS SISTER CHROMATIDS APART - Both

2 DAUGHTER CELLS - Mitosis

BEGINS WITH A DIPLOID PARENT CELL - Both

PARENT CELL HAS 46 CHROMOSOMES - Both

HOMOLOUS CHROMOSOMES ALIGN - Meiosis

2 DIVISIONS OCCUR - Meiosis

DAUGHTER CELL HAS 46 CHROMOSOMES - Mitosis

DAUGHTER CELLS ARE SEX CELLS (GAMETES) - Meiosis

CHROMOSOMES ALIGN SINGLE FILE ALONG EQUATORIAL PLATE - Both

ANAPHASE PULLS HOMOLOGOUS CHROMOSOMES APART - Meiosis

CELLS SPLIT DURING CYTOKINESIS - Both

DAUGHTER CELLS ARE AUTOSOMAL CELLS - Mitosis

GENETICALLY IDENTICAL DAUGHTER CELLS - Mitosis

 

 

 

 

 

 

Draw out sequences of steps (WITH LABELS) of Mitosis

 

Draw out sequence of steps (WITH LABELS) of Meiosis

 

*****You must indicate how many chromosomes are in the parent and daughter cells.

 

You Can see the answers in the diagrams in the Mitosis and Meiosis section of this blog, or in the textbook!

 

Team Points

Team Marks

Points	Team 1	Team 2	Team 3	Team 4	Team 5	Team 6	Team 7
Diagrams	80	81	80	56	72	89	68
M,M,B	24	22	23	19	22	22	20
Trivia (Nov 20)	5	3	3	5	3	2	2
Trivia on cloning article	1	1	2	4	4	1	0

Team 1 = Sarah, Sal, Asiyaa and Zeinab

Team 2 = Arun, Waseem, Lena and Reem

Team 3 = Ahmad, Batula, Laura, Kat

Team 4 = Karly, Zamzam, Sadiq, Kaamel

Team 5 = Robin, Ben, Stephanie, Katlyn

Team 6 = Lea, Alyssa, Hira and Mohammaed

Team 7 = Lara, Habon, Justin, Mike and Abdul

Genetics Unit Assignment

Genetic Unit Assignment

 

For this unit you will get to choose which level of achievement you would like to meet.

 

Level 1 – Maximum mark on this assignment is 59%

• Choose a genetic disorder and describe its genetic component (trisomy, mutation, autosomal/sex-linked disease), the symptoms of the disorder and treatment for the disease.
  

Level 2 – Maximum mark on this assignment is 69%

• Choose a genetic disorder and describe its genetic component (trisomy, mutation, autosomal/sex-linked disease), the symptoms of the disorder and treatment for the disease.
  
• Discuss genetic technologies used in research, treatment and /or screening for this disease.
  

 

Level 3 – Maximum mark on this assignment is 79%

• Choose a genetic disorder and describe its genetic component (trisomy, mutation, autosomal/sex-linked disease), the symptoms of the disorder and treatment for the disease.
  
• Discuss genetic technologies used in research, treatment and /or screening for this disease.
  
• Discuss one ethical dilemma associated with the use of this genetic technology.
  

 

Level 4 – Maximum mark on this assignment is 100%

• Choose a genetic disorder and describe its genetic component (trisomy, mutation, autosomal/sex-linked disease), the symptoms of the disorder and treatment for the disease.
  
• Discuss genetic technologies used in research, treatment and /or screening for this disease.
  
• Discuss the merits of surrounding ethical dilemmas associated with the use of this genetic technology and defend one side of the argument.
  

 

You can also choose the format of your assignment

 

1. Written assignment – You may choose to write a traditional paper (for the level 1-2), or a letter to your MP (level 3-4) discussing your disease, associated genetic technologies and your defend one side of the ethical dilemma.
   
2. Oral presentation – You may choose to do an oral presentation in front of the class discussing your genetic disease, if you choose level 3 or 4 you should be prepared to take questions from the class and to lead a class discussion. (total time would be no more than 10 minutes with class discussion)
   
3. Creative Pamphlet – You may choose to make a pamphlet about the disease and technologies associated with it.
   

   
    

   
   

   	Level 1	Level 2	Level 3	Level 4
   Knowledge and Understanding	Genetic Disease is clearly defined, symptoms, treatments are included.	Genetic Disease is clearly defined, symptoms, treatments are included.	Genetic Disease is clearly defined, symptoms and treatments are included.	Genetic Disease is clearly defined, symptoms and treatments are included.
   Application	No genetic technologies are mentioned	Genetic technologies involved with this disease are included	Genetic technologies and one ethical dilemma associated with it are included.	Genetic technologies and defence of side of the ethical dilemma are included.
   Communication	Paper – 300 words length, some scientific terms are included             Presentation – well spoken, use of scientific terms                     Pamphlet – Scientific terms are used, at least two sided	Paper – 500 words in length, proper use of scientific terms, thoughts are organized           Presentation – at least 3 mins in length, well spoken, proper use of scientific terms, thoughts are well organized             Pamphlet – scientific terms are used, at least 3 sided, images are used to enhance information	Paper – Letter to your MP using proper business letter format, thoughts are very well organized, proper scientific language used     Presentation – at least 4 mins and with questions, very well spoken, proper use of scientific language, thoughts are very well organized       Pamphlet – Very well organized, good use of effective images, minimum 3 sided.	Paper – Letter to your MP using proper business letter format, thoughts are extremely organized, language is clear and proper scientific terms are used Presentation – at least 4 mins and lead a class discussion reviewing both sides of the ethical argument. Very well spoken, use of proper scientific language, thoughts are well organized Pamphlet – professional looking pamphlet with excellent use of images and scientific language

Meiosis

Meiosis (not be confused with mitosis)

What: Formation of GAMETES. In this situation the parent cell has 46 chromosomes (diploid) and will eventually make 4 daughter cells with 23 chromosomes (haploid).
Where : sex cells

Meiosis has two stages, Meiosis 1 and Meiosis 2

The first stage of Meiosis is DIFFERENT from Mitosis.

Prophase 1
Chromosomes shorten and thicken
centrioles move to opposite poles

Metaphase 1
Homologous chromosomes (one from each parent) line up
Crossing over occurs (exchange of genetic material between homologous chromosomes)

Anaphase 1
Homolous chromosomes are pulled to opposite poles of the cell

Telophase and Cytokinesis 1
Nuclear membrane forms and two daughter cells are formed (each has 23 chromosomes duplicated)




Meiosis 2 - This process looks almost identical to mitosis, the difference is the number of chromosomes. Since meiosis 1 ended with each cell containing 23 duplicate chromosomes, that is how meiosis 2 will begin.

Mitosis

Mitosis

******* here is a great link showing the mitosis and meiosis!!

http://www.pbs.org/wgbh/nova/baby/divi_flash.html

What: Duplication of cells. One parent cells divides and makes two genetically identical daughter cells. The parent cell, and the daughter cells all have 46 chromosomes and are diploid.

When: Cells divide all the time to replace dead or damaged cells and to facilitate growth (think about a root of a plant!)

Before mitosis DNA replicates.

Prophase
Chromosomes shorten and thicken
Centrioles move to opposite poles
Nuclear membrane fades

Metaphase
Chromosomes line up in single file on the equatorial plate
Spindle fibers come from centrioles and attach to centromere

Anaphase
Centromere divides
Sister chromatids are pulled apart to opposite poles

Telophase
Chromosomes are on opposite poles
Spindle fibers dissolve
Nuclear membrane forms

Cytokinesis
Cytoplasma divides
2 new cells form
In plants a cell plate forms between the two cells and turns into a cell wall, in animals cells furrow into two cells.

Cell Cycle

Cell Cycle

• The Cell goes through a cycle
• Most of the time the cell is in interphase which is where the cell does its work
• Specific interphase is dependent on cell type (ie Liver or Nerve, etc)
• During interphase the cell also prepares for cellular division by DNA duplication
• Cellular division occurs after DNA replication, afterward the cell cycle repeats
• A cell can only divide so many times, this is determined by the cellular clock
• The more specialized the cell, the less it divides. For example, nerve cells, muscle cells and secretory cells do not divide as much as skin cells and cells from the digestive tract.
• Cells that divide continuously include spermatocytes (one billion a day!!!!) and cancer cells

Cell Specialization

• After fertilization cells begin to rapidly divide and eventually differentiate into specialized cells.
• Every cell contains the same genetic material
• In specialized cells certain genes are activated, while others are inactivated.
• Gene expression determines cell type
• Once cells become specialized they can only produce more specialized cells (liver cells can only make other liver cells)
• Cells that are not specialized and that make all types of cells are called TOTIPOTENT
• an example of these cells are STEM CELLS

DNA Structure

The human cell has 2 sets of 23 chromosomes, in total 46 chromosomes.

Before cell division, DNA duplicates and you have two sister chromatids attached by a centromere.

DNA Structure

• Structure of DNA discovered by James Watson and Francis Crick
• Rosalind Franklin used X-ray diffraction to help determine structure of DNA
• DNA is composed of nitrogenous bases, deoxribose sugar and phosphate=nucleotide
• Sugar and phosphates form the backbone of the helix (spine of the ladder)
• Nitrogenous bases from one side of the ladder bond to nitrogenous bases on the other side
• They are bonded by hydrogen bonds which are weak bones between a positive hydrogen molecule and a negative charge on an electronegative nitrogen or oxygen molecule
• The four nucleotides are adenosine, guanine (purines) and cytosine and thymine (pyrimidines).
• Adenosine complimentary bonds to Thymine
  
• Guanine complimentary bonds to cytosine

Cancer

Normal, healthy, cells know it is time to divide because of cell to cell communication. When neighbouring cells are dying or damaged cells receive signals to generate new cells.

Have you ever played guitar? If you have, you will notice that your skin cells divide to create calluses in order to protect your fingertips!

Normal, healthy, cannot divide in isolation. They can only divide when other specialized cells are close by.

Cancer cells are cells that are able to divide in isolation and at an accelerated rate. They are not responding to cell to cell communication.
The lack of communication causes a second proble. Healthy cells adhere to each other in order to help in communication. Cancer cells do not always adhere to healthy cells, or to other cancer cells. This means that they can dislodge from a tumor and move around the body which means metastasis is occurring. This makes locating the source of the cancer very difficult.

Why is cancer so dangerous?
Cancer cells are rapidly dividing which means they are utilizing a ton of energy and resources. Cancer cells do not specialize, meaning that they do not contribute to the health of maintenance of the body. If you have cancer cells in your liver, they are taking a ton of energy and resources from the liver to fuel duplication, but do not contribute at all to the function of the liver.


Here is a link to an awesome website all cancer!

http://www.cancerquest.org/

Cloning

Cloning is a relatively new genetic process where scientists are able to create genetically identical offspring from a single cell! Scientists use cloning to make beautiful plants and "ideal" animals like super cows (produce A LOT of milk). Here is a link to an article about cloning!

http://www.sciencenewsforkids.org/articles/20040128/Feature1.asp

Your textbook also has some good information on cloning (pages 94-98).

This is a picture of Dolly, the first mammal ever cloned. Beside her is her first born lamb, Bonnie.

Welcome to your Classroom Blog!!

Hello Class!!

I am happy to tell you that I have set up this blog in order to help with course material! Even though I strongly recommend that you take notes in class, most (definitely NOT all) of your notes will also be posted here!

I will post basic notes, diagrams, study tips, useful website links and more!! Please check this blog regularly!! I will do my best to post notes before class. You should copy and paste the class notes into a word document and print them off, unless you prefer to study from the computer.

I will also post diagrams here, but will do my best to distribute hard copies of diagrams in class. If you miss a class, or lose the handout, this is where you will find the diagram.

Finally, I will be posting links to websites that will be useful study tools and great resources for your assignments. You should USE them! But do not limit yourself to these sites....there are lots of goodies out there!!

Cheers!

Miss Sassine