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AP Biology Unit 8: Gene Expression & Regulation Notes Pt. 1

AP Biology Unit 8: Gene Expression & Regulation Notes Pt. 1

1- Cyclin in the PDGF could be considered as what type of cyclins (G1, G1/S,...)? 

- Cyclin in the PDGF could be considered as a G1/S cyclin. Since they are responsible for stimulating the division of human fibroblast cells in culture, the peak is in between the G1 and S phase.

2- Explain how the complex of MPF gets inactivated 

- The MPF complex is inactivated when the cyclin is broken down. APC/C is an enzyme that attaches an ubiquitin which tags the cyclin as something to be destroyed. The tag is attached to securin on the chromatids. The securin normally binds and inactivates separase but when the securin is sent for recycling, the separase is activated. The securin with a ubiquitin is sent to the proteasome or recycle bin and is broken down. The separase breaks down the cohesin that is holding the sister chromatids together, leading to the separation.

3- What is the role of p53 gene and how it performs this role?

- The p53 gene is a protein that is released in response to DNA being damaged. It suppresses tumors and works to prevent damaged DNA from performing mitosis. The cell is stopped at the G1 checkpoint by the production of CKI which blocks the activity. This activity being blocked allows for the DNA to be repaired. The p53 then activates the DNA repairing enzymes. In the case that the DNA is too damaged beyond repair, the p53 will trigger apoptosis to ensure that the damaged DNA is not passed onto daughter cells.

Type down some differences between the meiosis and mitosis, that you talked about in your group:

Mitosis creates two genetically identical daughter cells that are identical to the parent cell and are diploid. They make somatic cells or any cells that are genetically identical to the original parent cell. They only go through mitosis once. 

Meiosis creates four haploid cells that have genetic variation because there are steps like crossing over and the independent assortment that happens in meiosis 1. They are haploid cells that are only produced by sex organs for sexual reproduction. The haploid cells that are produced are gametes. They go through two steps, meiosis one and meiosis two. Meiosis one causes genetic variation but meiosis two is just like mitosis, with the lack of genetic material being duplicated in interphase. 


1- Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf?

Gametes Male AaTt - AT,At,aT,at

Gametes Female AaTt - AT,At,aT,at


AT

At

aT

at

AT

AATT

AATt

aATT

AaTt

At

AATt

AAtt

aATt

Aatt

aT

AaTT

AatT

aaTT

aatT

at

AaTt

Aatt

aaTt

aatt

The possible gametes for each parent are AT,At,aT,at. Since terminal flower and dwarf are both recessive traits, the genotype would have to be aatt. Using the punnett square, we are able to find that 1 out of 16 offspring would be predicted to have terminal flowers and be dwarf. This means that 25 out of 400 offspring would be predicted to have terminal flowers and be dwarf.  

2- If P1 is PpYyRr X Ppyyrr. What fraction of offspring from this cross are predicted to exhibit the recessive phenotypes for at least two of the three characters?

Gametes Male PpYyRr: PYR,PYr,PyR,Pyr,pYR,pYr,pyR,pyr

Gametes Female Ppyyrr: Pyr,pyr


PYR

PYr

PyR

Pyr

pYR

pYr

pyR

pyr

Pyr(1/2)

PPYyRr

PPYyrr

PPyyRr

PPyyrr

PpYyRr

PpYyrr

PpyyRr

Ppyyrr

pyr(1/2)

PpYyRr

PpYyrr

PpyyRr

Ppyyrr

ppYyRr

ppYyrr

ppyyRr

ppyyrr


P

p



Y

y



R

r

P

PP

Pp


y

Yy

yy


r

Rr

rr

p

Pp

pp


y

Yy

yy


r

Rr

rr

The offspring that are predicted to exhibit the recessive phenotypes would need two recessive alleles. Since two of the three characters have to be shown in this question, the possible combinations are PPyyrr, Ppyyrr, ppYyrr, ppyyRr, ppyyrr, Ppyyrr. The first punnett square shows that there is 6/16 chance which is 0.375 or ⅜.

Another way to find the probability of pp, when Pp and Pp are crossed, it is a ¼ chance. When Yy and yy are crossed, there is a 50% chance that the recessive trait is expressed. Then when Rr and rr are crossed, there is also a  50% chance that the recessive trait is expressed. The chance all 3 are recessive is 0.0625. Since we are only looking for two out of the three recessive traits to be exhibited, you only have to multiply 2 so 0.5 times 0.5 which is 0.25 and 0.5 times 0.25 which is 0.125. When added together, the total is 0.375 or ⅜. 

3- Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii). What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?


P

p



Y

y



I

i

p

Pp

pp


Y

YY

Yy


i

Ii

ii

p

Pp

pp


y

Yy

yy


i

Ii

ii

The possibility of recessive pp is 50%. The possibility of recessive yy is 25%. The possibility of recessive ii is 50%. The probability of having recessive pp and yy is 0.125.  The probability of having recessive pp and ii is 0.25. The sum of these probabilities are 0.375. 

4- complete the cross for the snapdragon flower that your teacher has started. Write down the genotypic and phenotypic ratios for the F2.

F1

r

r


F2

R

r

R

Rr

Rr


R

RR

Rr

R

Rr

Rr


r

Rr

rr

Since the offspring of the first breeding creates all heterozygous pink plants, the pink snapdragon is self pollinated, and creates the F2 which has RR as 25%, rr as 25% and Rr as 50%. The genotype ratio is RR:rr:Rr=1:1:2. The phenotypic ratio of f2 is red:white:pink=1:1:2. Two offspring are pink, one is red and one is white since it shows incomplete dominance.

Meselson and Stahl cracked the puzzle

Matt Meselson and Franklin Stahl originally met in the summer of 1954, the year after Watson and Crick published their paper on the structure of DNA. Although the two researchers had different research interests, they became intrigued by the question of DNA replication and decided to team up and take a crack at determining the replication mechanism. Now watch the following video and answer the questions:

1- Explain about 3 models proposed for the structure of DNA based on Watson and Crick' s announcement of DNA as genetic material. Use some ideas from the video with a timestamp (one time stamp is enough to show the authenticity of your work).

  1. Semiconservative: unfold in middle, make copies on either side 

  2. Conservative: Wraps around histone protein and makes exact duplicate

  3. Dispersive: Cut into 10-12 nucleotides and each segment is copied 

2- Based on Meselson and Stahl’s experiment which model is accepted for the replication of double helix DNA? 

  1. Based on Meselson and Stahl’s experiment, the semiconservative model is accepted for the replication of double helix DNA. Since the experiment growing e. coli with heavy nitrogen showed that the light nitrogen DNA increased as each generation is duplicated, it proves that DNA is replicated by first unfolding in the middle, and then making copies on either side.

3- Suppose that Meselson and Stahl have first grown the cells in N14 containing medium and then moved them into N15 containing medium before taking the samples, what would have been the result? (talk about bands density and location in the tube. You may draw in own notebook using two colors to help yourself solving this problem)

  1. With the conditions stated above, the results would have been identical to the original experiment, only with the data sets swapping the results. The band density in the N15 is increased as there is more duplication, so the amount of hybrid nitrogen will decrease, as the amount of N15 or heavy Nitrogen increases. The tube will show increased concentrations of the heavier N15 after each duplication since there is more N15 being duplicated semiconservatively. 

Chi Squared Practice Problems


1. Chi Squared Problems: you will be given the table of critical values and formulas, like on your test. Ex:

a. In a heterozygous, heterozygous dihybrid cross, the following data was obtained:

dominant for both traits: 570, dominant for trait 1 and recessive for trait 2: 185

dominant for trait 2 and recessive for trait : 190, recessive for both traits: 55

Perform a chi-square analysis to see if the data above agrees with the predicted outcome of this cross.

In this experiment, the null hypothesis is supported    because 1.092 is less    than 7.82. 

b. In 1901, Bateson reported    the first post-Mendelian study    of a cross involving two characters. White leghorn chickens, having white feathers and large "single" combs, were crossed to Indian Game Fowl, having dark feathers and small "pea" combs. The F1 were white with pea combs, so it is assumed that these are the dominant traits. The F2 distribution was: 111 white pea, 37 white single, 34 dark pea, and 8 dark single. (You must show your Chi squared table and all your calculations like part a)

Group

Observed

Expected

O-E

(O-E)^2

(O-E)^2/E

Dom both traits

(white pea)

111

107

(9/16 X 190)

4

(111-107)

16

(4)^2

0.1495

(16/107)

Dom trait 1, rec trait 2

(white single)

37

36

(3/16 X 190)

1

(37-36)

1

(1)^2

0.0277

(1/36)

Dom trait 2, rec trait 1

(dark pea)

34

36

(3/16 X 190)

-2

(34-36)

4

(-2)^2

0.1111

(4/36)

Rec both traits

(dark single)

8

12

(1/16 X 190)

-4

(8-12)

16

(-4)^2

1.3333

(16/12)

4 groups, DOF is 3




Chi squared: 

1.6216

Since the degree of freedom is 3, and the p value is 0.05, the critical value is 7.82. The calculated chi squared value in this question is 1.6216, which is less than the critical value. This means that the null hypothesis is supported since it follows the 9:3:3:1 ratio for a dihybrid cross. 

c. A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of basic codominance. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome? (you must show your Chi square table and all of your calculations)

Group

Observed

Expected

O-E

(O-E)^2

(O-E)^2/E

Trait 1 

(stripes)

50

44

(1/4 X 176)

6

(50-44)

36

(6)^2

0.818

(36/44)

Traits 2

(spots)

41

44

(1/4 X 176)

-3

(41-44)

9

(-3)^2

0.205

(9/44)

Traits 1 and 2

(stripes and spots)

85

88

(2/4 X 176)

-3

(85-88)

9

(-3)^2

0.102

(9/88)

3 groups, DOF is 2




Chi squared: 

1.125

Since the degree of freedom is 2, and the p value is 0.05, the critical value is 5.99. The calculated chi squared value in this question is 1.125, which is less than the critical value. This means that the null hypothesis is supported since it follows the 1:1:2 ratio for a codominance cross. According to the Chi-square test, she did get the predicted outcome since the difference is statistically insignificant. 


1- A white-eyed female Drosophila is mated with a red-eyed(wild type) male. What phenotypes and genotypes do you predict for the offspring?

The maleoff spring will all have white eyes. Both female offspring will have red eyes since they are hybrid. 50% of the offspring is female with XwXw+ and 50% of the offspring is male XwY. 100% of the female offspring is heterozygous XwXw+ and 100% of the male offspring is XwY. 50% red phenotype, 50% white phenotype. 


Xw

Xw

Xw+

XwXw+

XwXw+

Y

XwY

XwY

2- Neither Tim nor Rhoda has Duchene muscular dystrophy, but their firstborn son does have it. What is the probability that a second child of this couple will have the disease? What is the probability if the second child is a boy that has the disease? A girl?


XD

Xd

Xd

XDXd

XdXd

Y

XDY

XdY

Probability that a second child of this couple will have the disease is 50%. 

Probability that a second child is a boy that has the disease is 50%.

Probability that a second child is a girl that has the disease is 0%.


Summarize what you have learned. How do you recognize a pedigree related to mitochondrial inheritance?

I learned that mitochondrial DNA only comes from the egg, which is the zygote produced by the mother. This causes all offspring to have maternal mtDNA. This inheritance is called maternal inheritance and is a type of extranuclear inheritance. The pedigree that is displayed in the video is an example of mitochondrial inheritance. The affected father does not have any impact on his children as long as the mother is unaffected. Since the mother is unaffected, all of the offspring do not display the trait that makes the individual affected. When the father is unaffected and the mother is affected, all of the offspring are affected since all of the children get mtDNA from the mother. The pedigree looks similar to the autosomal dominant pedigree. The key point that sets them apart is that despite the father in 1,1 showing symptoms, that none of the children shows signs of the affected trait. The most important part of the video was the importance of examining the pedigree while looking at the sample fully as well as examining any consistencies before making any conclusions.

1- Based on the information provided by the teacher for question 1, what is the Phenotype of the parents for a, b and c? What is the probability of getting a colorblind son for c? (son is colorblind)

Phenotype of mother a: not colorblind

Phenotype of father a: colorblind

Phenotype of mother b: not colorblind (carrier)

Phenotype of father b: not colorblind

Phenotype of mother c: not colorblind (carrier)

Phenotype of father c: colorblind

Probability of getting a colorblind son for c is 50% from all possible offspring. 

2- If parental flies had been true-breeding for gray body with vestigial wings and black body with normal wings, which phenotypic class(es) would be largest among the testcross offspring?

  • gray body with vestigial wings: b+b+vgvg, gamete: b+vg

  • black body with normal wings: bbvg+vg+, gamete: bvg+

  • The genotype of f1 is bb+vgvg+. 

  • This has many gametes that are bvg,b+vg,bvg+ and b+vg+. 

  • The test cross would be black body with vestigial wings: bbvgvg, gamete: bvg

  • The genotypes possible for f2 is bbvgvg, bbvgvg+, bb+vgvg, and bb+vgvg+. 

  • The phenotypic classes that will be seen largest are black with normal wings, and grey with vestigial wings. This is because they are the parental types and it is because they are the original genes that were linked together in the P or parental generation. 

f1

b+vg


f2

bvg

bvg+

b+vg

b+vg+

bvg+

bb+vgvg+


bvg

bbvgvg

bbvgvg+

bb+vgvg

bb+vgvg+

Worksheet on heredity


1- Gene A, B, and C are located on the same chromosome. Testcross shows that  the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes? Explain

The higher the recombination frequency, the farther the genes are located on the same chromosome. Since A and C have a lower frequency of 12% when compared to the A and B with a higher frequency of 28%. The order can be either ACB or CAB. The gene C can be either in between A and C or before A. To know the linear order of the genes, the recombination frequency for genes B and C has to be determined. 

2- About 5% of individuals with Down syndrome have a chromosomal translocation in which a third copy of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent’s gonad, how could it lead to Down syndrome in a child?

During meiosis, the third copy of chromosome 21 being attached to chromosome 14 will be treated as one chromosome. This means that translocation of this chromosome would occur. There will be a gamete with only one chromosome 21, but there will be another chromosome with a normal chromosome 21 along with the extra chromosome 21 that was attached to chromosome 14. In the case that the gamete with two chromosome 21s is fertilized or fertilizes a gamete with a normal number of chromosomes which is 1, the offspring would have three chromosome 21s, which is called trisomy 21 or in other words, down syndrome. 

3- The ABO blood type locus has been mapped on chromosome 9. A father who has type AB blood and a mother who has type O blood have a child with trisomy 9 and type A blood. Using this information, can you tell in which parent the nondisjunction occured? Explain your answer.

Yes, it is possible to tell which parent cell that the nondisjunction occurred. The father who has blood type AB has A and B antigens. The mother who has blood type O has no antigens. When this is crossed, there is a 50% chance that the offspring has type A with the genotype AO and a 50% chance that the offspring has type B with the genotype BO. The A blood type means that the mothers chromosome 9 did not split properly. If nondisjunction took place in chromosome 9 of the father, the child would have blood type ABO. Since this is not the case, the mothers O blood type with no antigens had nondisjunction, causing the child to have AOO as their genotype. 

4- The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Explain how the activation of this gene could contribute to the development of cancer.

The activation of the gene that codes for intracellular tyrosine kinase can lead to the development of cancer because they are able to send many signals with only one dimerization of a tyrosine. Since there are many kinases, there is also a lot of cell division, since receptor tyrosine kinases play a large role in the rapid division of cells. This extremely rapid speed of cell division may be harmful since there can be a cell that is not properly functioning being replicated over and over. This rapid growth of cells that are faulty can contribute to the development of cancer if the cells do not go through apoptosis. 

5- Gene dosage, the number of active copies of a gene, is important to proper development. Identify and describe two processes that establish the proper dosage of certain genes.

One process that establishes the proper dosage of certain genes is when one x chromosome is inactivated in females. This is because females have 2 x chromosomes, with one being passed down from each parent. Having two active x chromosomes is most likely too much to enable the proper development. The other process that establishes the proper dosage of certain genes is triple x syndrome or trisomy x. Although the individual is a female with three x chromosomes, there are no developmental issues seen. This is because two of the x chromosomes are inactivated, to fit the proper dosage of the sex chromosomes.

6- Reciprocal crosses between two primrose varieties, A and B, produced the following results: A female x B male gives offspring with all green (non variegated)  leaves;   B female x A male  gives offspring with spotted (variegated) leaves. Explain these results.

    The reason the plants seem to be different colors is because the number of chlorophyll in the chloroplast. In the cross A female x B male, the green leaves are caused by the abundance of the chlorophyll. In the cross B female x A male, the variegated leaves are caused by the mix of cells with an abundance of the chlorophyll as well as a lack of chlorophyll, and also a mix of both. The wild type is green leaves caused by the abundance of the chlorophyll, while the mutant type is white leaves caused by the lack of the chlorophyll. Since chlorophyll is located in the chloroplast and chloroplast is maternally inherited, the offspring's phenotype is identical to the mother’s phenotype. This makes A a primrose with all green leaves, and B a primrose with spotted or variegated leaves. 

7- Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not?

    Mitochondrial genes are inherited maternally. They are critical to the metabolism of the cell but there are numerous mitochondria inherited from the mother, with the gamete which is an egg. Although there may be mutations in the mitochondria, each cell has many mitochondria. Since there most likely enough mitochondria that function normally, the gene mutations are generally not lethal SInce enough cellular respiration is taking place in the cell. 


DNA structure worksheet

1- How did Griffith experiment rule out the possibility that the R cells could have simply used the capsules of the dead S cells to become pathogenic?

  1. His experiment rules out this possibility by checking the offspring. If the cell reproduces with a capsule and gives capsule offspring, the R cell did not use the capsule of the dead S cell. It is DNA fragmentation that gave them the ability to produce a capsule on their own, allowing the offspring to also exhibit this trait. 

2- Griffith did not expect transformation to occur in his experiment. What results was he expecting? Explain your answer

  1. He was expecting that the mouse would not die since when performing the experiments separately, the heated S strains did not kill the mouse, since they were denatured. Additionally, the R strains did not kill the mouse since they are not virulent. When combined, he expected that the mouse would not die. However, this was not the case, since the DNA fragments from the heat killed S strains were transformed to the R strains, making the latter a S strain and causing it to have a capsule and be virulent, finally resulting in it killing the mouse. 

3- For Hershey and Chase (bacteriophage experiment) how would the results have differed if proteins carried the genetic information?

  1. If the protein carried the genetic information, the radioactivity would have transferred from the protein to the e coli cell, causing the radioactivity to only be seen in the pellet. The results would have been similar to the results of the DNA being changed to radioactive DNA. 

4- A fly has the following percentages of nucleotides in its DNA: 27.3% A, 27.6% T, 22.5% G, and 22.5% C. How do these numbers demonstrate the Chargaff’s rule about base ratios?

  1. Chargaff’s rule states that the A and T bases are equal, as well as the C and G bases in any organism, although the base composition of DNA varies between species. These numbers demonstrate the Chargaff’s rule about base ratios since the ratio between C and G bases are equal. The ratio between A and T bases are roughly equal, this slight imbalance may have been because of a mutation, or a mistake in the genetic coding. 

5- Given a polynucleotide sequence such as GAATTC, can you tell which is the 5’ end? If not, what further information do you need to identify the ends?

  1. No, you can not tell which is the 5’ end just from the polynucleotide sequence. You need the 5’ and 3’ given in the problem. The opposing polynucleotide sequence would be CTTAAG and the 5’ and 3’ ends would be antiparallel. 

1- Suppose instead of N15, Meselson and Stahl had started the cell growing with N14 and then moved the cells into N15-containing medium before taking the samples, how could results have changed?

  1. The results would have been identical to the original experiment, only with the data sets swapping the results. The band density in the N15 is increased as there is more duplication, so the amount of hybrid nitrogen will decrease, as the amount of N15 or heavy Nitrogen increases. The tube will show increased concentrations of the heavier N15 after each duplication since there is more N15 being duplicated semiconservatively. 

 

Original

Sample 1

Sample 2

Sample 3

100% N14

100% hybrid 

50% hybrid

25% hybrid



50% N15

75% N15

 

2- Write a summary.

  1. DNA replication is the process where DNA copies itself. In eukaryotes, this is done in the S phase in interphase, while in prokaryotes, this is done through binary fission. There were 3 theories which were the semiconservative, the conservative and the dispersive. Through the Messelson-Stahl experiment, the N 15 would become hybrid and eventually, there would be a greater sample of N14 since the semiconservative model was proven to be correct. DNA is antiparallel, meaning that there is a strand that runs from the 5’ to 3’ while the opposite strand runs from 3’ to 5’. Bases can only be added to the 3’ end, so one strand is a leading strand while the other is a lagging strand. 


Class worksheet on DNA replication and gene expression 

1- From 2 strands of the original DNA in the sketchbook, which one’s copying needs Okazaki fragments. You must be able to explain your answer.

- A would need Okazaki Fragments. The reason for this is the orientation of the 5’ and 3’ ends. Bases can only be added to the 3’ end and strand A would have a complementary strand that is from 3’ to 5’ based on the orientation of the DNA as well as the location of the origin of replication. Since the right side of A runs from 5’ to 3’, it would need okazaki fragments. It is like backstitching.

2- Suppose that DNA Pol I (ONE) is not functional in a given cell. How could that affect the synthesis of a leading strand?

- If DNA pol 1 is not functional, then the RNA primer is not digested and it is not replaced with DNA. This could affect the synthesis of the lagging strand because the strand that is created will have DNA as well as RNA primers. This may impact the synthesis of the leading strand since it would have only DNA while the lagging strand would be composed of DNA as well as RNA primers. In order to have one continuous strand, the DNA polymerase 1 is necessary also for the synthesis of the leading strand because it would have to be combined with the other sections of the strand to form a proper double strand.

3- what are 2 properties (one functional and one structural) distinguish euchromatin from heterochromatin?

- One functional property that distinguishes euchromatin from heterochromatin is that in heterochromatin, the genes could not be expressed while in euchromatin, the genes inside the DNA could be expressed. One structural property that distinguishes euchromatin from heterochromatin is that in heterochromatin the gene is more compact while in euchromatin, the genes are narrow but not compact. 

4- what did you learn from srb and Horowitz experiment?

- From the srb and Horowitz experiment, I learned that the test involved genes and using proteins to see whether or not they are necessary for the growth of the neurospora crassa. It showed that the fungus grew in the wild type in all 4 test tubes. Then it was mutated once to show that the minimal medium did not grow the fungus, this was because enzyme one was disabled. Then the second mutation showed that the minimal medium as well as the mm plus ornithine did not grow the fungi. This was because the second enzyme was disabled. Then the third mutation showed that all of the test tubes except mm plus arginine were not growing. This was discovered to be the cause of the 3rd enzyme. 

1- If you compare the sequence of the mRNA to that of nontemplate DNA strand, what similarity or difference would you notice?

The sequence of the mRNA and the nontemplate DNA strand are identical with the exception of the mRNA uses uracil while the nontemplate DNA strand uses thymine. The template strand is the original that is used to make the mRNA so it is complementary to the nontemplate strand and the mRNA strand. The difference is that RNA and DNA use different bases so the only difference is the presence of uracil or thymine. The direction of the nontemplate DNA and the mRNA are in the same direction of 5’ and 3’. They are coding strands. 

2- suppose that you have an mRNA with a poly-G sequence of 30 nucleotides long. What and how many amino acids would you expect to see after translation? You can look up the codon table for mRNA)

I would expect to see 10 amino acids after translation. GGG translates into glycine so there would be 10 glycines in a row after translation. The strand would look like 

Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly

3- For each of the following DNA strands, transcribe and translate (which one will make the longest amino acid chain?)

The longest amino acid chain is the 1st one because there are 4 codons and the last is a stop. 

Template 3’- ACACGACCTTAC-5’

mRNA     5’- UGU  GCU  GGA  AUG-3’

Nontemp 5’- TGTGCTGGAATG-3’

Amino acid: Cys  Ala  Gly  Met


Template 3’-AAATACAAAGGG-5’

mRNA     5’- UUU  AUG  UUU  GGG-3’

Nontemp 5’- TTTATGTTTGGG-3’

Amino acid: Phe  Met  Phe  Gly 


Template 3’-AACATTCCAAAAGGGGACTAA-5’

mRNA     5’- UUG  UAA  GGU  UUU  CCC  CUG  AUU-3’

Nontemp 5’- TTGTAAGGTTTTCCCCTGATT-3’

Amino acid: Leu  END


Transcription worksheet 

1- Promoter is located at the upstream end of a transcription unit.

Upstream promoter and 5’ cap

Downstream 3’ poly a tail

2- What does a promoter do?

A promoter is necessary for the eukaryotic cell when mRNA synthesis is going to take place. It is the region of the DNA strand on the 3’ end where RNA polymerase binds to in order to begin transcription. This promoter also contains the TATA box, where all of the nucleotides are thymine and adenine. 

3- How does RNA polymerase start transcribing a gene at the correct place on the DNA of prokaryotes versus eukaryotes?

 RNA polymerase starts transcribing a gene at the correct place on the DNA of prokaryotes using the origin of replication. Since prokaryotes only have a single ring of DNA, it is simple to replicate the DNA. RNA polymerase starts transcribing a gene at the correct place on the DNA of eukaryotes differently. On the template DNA, there is a promoter on the 3’ end of the strand. RNA polymerase is able to attach to the DNA here, and begins transcription. 

4- What happens if a harmful ray causes some change in the TATA box. How would that affect transcription of the gene?

If there is a harmful ray that causes some change in the TATA box, the transcription of the gene will most likely take longer. Since there are many bubbles of DNA replication, it will not completely stop the RNA synthesis.The change in the TATA box can cause it to disappear, which means that the promoter will be missing a piece that is necessary for the RNA polymerase to attach to the DNA. This may cause this specific location to not be functioning as an origin of replication. 

5- What are some important functions that 5’ cap and poly-A tail share/do? 

5’ cap function;

  • Indicates the start location of RNA synthesis. 

  • Allows for RNA polymerase to attach and begin transcription. 

  • Allows the beginning of the mRNA strand to leave the nucleus.

  • Contains the TATA box.

Poly-A tail function;

  • Indicates the end location of RNA synthesis. 

  • Allows for the DNA strand to be cut and end the transcription.

  • Allows the end of the mRNA strand to leave the nucleus. 

  • Contains many A nucleotides. 

5’ cap and poly-A tail share; 

  • Facilitate the export of mRNA.

  • Protects the mRNA from hydrolytic enzymes

  • Let the ribosomes attach to the 5’ end. 


DO NOW: Why should the gene expression be under regulation? (Why a gene cannot be expressed all the time)

I think that gene expression should be under regulation because there are genes that do not need to be expressed all of the time. One example is the gene expression of one of the x chromosomes in a female. There is only one x chromosome that is expressed, the other is suppressed. The reaction could be harmful to the cell, it can also be wasteful since there the process of making the gene would not be necessary.

1- What is the function of a promoter and operator?

  • Operator function: a segment of the DNA which acts like a switch to regulate the concentration of the enzyme. It is usually positioned within the promoter of the DNA. 

  • Promoter function: a segment of the DNA where RNA polymerase is able to attach on to, in order to produce more mRNA through transcription. 

2- The following are the steps that happen when we have a high level of tryptophan. Put them in order

B- Tryptophan does not need to be produced by the operon

C- Tryptophan will bind to the repressor protein, changing its conformation

A- The trp repressor-tryptophan complex can now bind to the operator of the trp operon

D- RNA polymerase is blocked from transcribing the genes needed to synthesize tryptophan

NOW LET’S WATCH A VIDEO

3- What happens if lactose is present and glucose is scarce? Put the following list in order. Start with the effects of lactose levels then proceed to the effects of glucose levels.

C- The enzymes needed for lactose metabolism must be transcribed when lactose is present

E- Lactose binds to the LacI repressor, changing LacI’s shape and making it fall off the operator

A- Now that LacI has been removed for the operator, RNA polymerase can proceed with transcription

G- The three enzymes involved in the metabolism of lactose are transcribed and expressed at moderate levels

D- cAMP levels increase because glucose is scarce (ATP is not being produced through cell respiration)

B- cAMP binds to CAP regulatory protein, causing it to bind to the promoter of the lac operon

F- CAP binding causes RNA Polymerase to bind to the promoter (higher affinity) and transcribe the gene at a higher level than before

4- What happens to the lac operon if both lactose and glucose are scarce?

If glucose is scarce, then the active receptor attaches to the operator and prevents the mRNA from being produced. This prevents the proteins for lactose breakdown from being produced. CAP is activated by binding with cAMP when glucose is scarce. The CAP which is activated can attach to the promoter of the lac operon which increases the affinity of the RNA polymerase. This can increase the transcription rate but there is no transcription taking place, so the level of glucose and lactose will not change. 

5- If a mutation happens to the lac operator so that the active repressor cannot bind. How would this affect the cell’s production of beta-galactosidase?

There will be increased amounts of beta-galactosidase since there is no way to prevent the RNA production. This means the cell will continue to make beta-galactosidase even when lactose runs out. This will be a waste of the cell's resources since there is no inhibition. 

1- What are the effects of histone acetylation and DNA methylation?

The effects of histone acetylation are that acetyl groups are attached to the positively charged lysines in histone tails. This causes the chromatin structure to loosen which allows for transcription to take place on the DNA strand. The effects of DNA methylation are that the methyl group is added to a base in DNA that is associated to reduce transcription. This condenses the chromatin and can cause long term inactivation of the genes. 


2- What are the general and specific transcription factors? (their roles and functions)

The general and specific transcription factors are necessary to initiate transcription in eukaryotic cells. General transcription factors are necessary for transcription of all genes necessary to code for proteins. Their function is to assemble the transcription initiation complex at the promoter at all genes.  Specific transcription factors are necessary for the stimulation or repression of transcription, within the DNA. They are most likely distal control elements, which are enhancers. They bind to control elements that are associated to a specific gene. Once it is bound, it increases or decreases the transcription rate. Activators are used to increase transcription while repressors are used to decrease the transcription. 


3 If you are comparing the nucleotide sequences of the distal control elements in the enhancers of three genes that are expressed only in muscle, what would you expect to find? Why?

When comparing the nucleotide sequence of the distal control element in the enhancers of the three genes that are expressed only in muscle, I would expect the genes to be similar to the enhancers. This similarity allows the same specific transcription factor to bind to the enhancer of all of the genes. This will stimulate the gene expression of this trait. 


1- Compare and contrast miRNA and siRNA

miRNA has a hairpin like structure and is able to break down mRNA and prevent translation. siRNA can inhibit gene expression by using RNAi. They have different RNA precursors. 

2- If the mRNA being degraded coded for a protein that promotes cell division in a multicellular organism, what would happen if a mutation disabled the gene encoding the miRNA that triggers this degradation?

In the case of the mutation that disables the gene encoding the miRNA that triggers this degradation, there will be excessive cell division in the multicellular organism. Since the degrading protein is disabled, there will be increased protein that promotes the cell division. This may cause faulty cells to duplicate, which may result in cancer. 

3- How can the inactivation of the X chromosome be related to the non-coding RNA?

The inactivation of the X chromosome can be related to the non-coding RNA because X-chromosome inactivation relies on the histones. When the histones are tightly wrapped and compressed, the x chromosome is heterochromatin which prevents translation and transcription. 

1- As you know mitosis gives rise to two daughter cells which are genetically identical to the parent cell. So how come mitotic divisions products are not composed of identical cells?

Mitotic divisions are not identical because differentiation creates specific cells that are necessary for specific needs. This means that there are proteins that are activated in some locations, and others that don’t. This heavily relies on transcription factors. 

2- signal molecules  released by an embryonic cell can induce changes in a neighboring cell without entering the cell. Explain about the process involved in this kind of signal transduction.

Single molecules released can induce change through cell to cell recognition. Using ligands, the cells can interact as long as a receptor is present and can accept the signal molecule. THis can cause changes to the cell that receives the ligand, to be able to differentiate the cell using determination. 

3- Why do we call maternal effect genes as polarity genes?

Maternal effect genes are called polarity genes because they are able to determine which side of the embryo becomes the anterior and posterior ends. This is done through differential gene expression to control what traits are expressed, and where they are expressed. The mother provides the mRNA necessary for this process through the egg cell. 

4- In the figure 18.17b of the powerpoint you can see that the lower cell is synthesizing signaling molecules, whereas the upper cell is expressing receptors for these molecules. In terms of gene regulation, explain  how these cells came to synthesize different molecules.

In terms of gene regulation, these cells came to synthesize different molecules because they have different roles. Determination is able to commit a cell to its final state, so it can activate or inactivate mRNA sequences to produce transcription factors. Before induction, the maternal determination can cause the cell to have special proteins to cause the various molecules. 


AP Biology Unit 8: Gene Expression & Regulation Notes Pt. 1

1- Cyclin in the PDGF could be considered as what type of cyclins (G1, G1/S,...)? 

- Cyclin in the PDGF could be considered as a G1/S cyclin. Since they are responsible for stimulating the division of human fibroblast cells in culture, the peak is in between the G1 and S phase.

2- Explain how the complex of MPF gets inactivated 

- The MPF complex is inactivated when the cyclin is broken down. APC/C is an enzyme that attaches an ubiquitin which tags the cyclin as something to be destroyed. The tag is attached to securin on the chromatids. The securin normally binds and inactivates separase but when the securin is sent for recycling, the separase is activated. The securin with a ubiquitin is sent to the proteasome or recycle bin and is broken down. The separase breaks down the cohesin that is holding the sister chromatids together, leading to the separation.

3- What is the role of p53 gene and how it performs this role?

- The p53 gene is a protein that is released in response to DNA being damaged. It suppresses tumors and works to prevent damaged DNA from performing mitosis. The cell is stopped at the G1 checkpoint by the production of CKI which blocks the activity. This activity being blocked allows for the DNA to be repaired. The p53 then activates the DNA repairing enzymes. In the case that the DNA is too damaged beyond repair, the p53 will trigger apoptosis to ensure that the damaged DNA is not passed onto daughter cells.

Type down some differences between the meiosis and mitosis, that you talked about in your group:

Mitosis creates two genetically identical daughter cells that are identical to the parent cell and are diploid. They make somatic cells or any cells that are genetically identical to the original parent cell. They only go through mitosis once. 

Meiosis creates four haploid cells that have genetic variation because there are steps like crossing over and the independent assortment that happens in meiosis 1. They are haploid cells that are only produced by sex organs for sexual reproduction. The haploid cells that are produced are gametes. They go through two steps, meiosis one and meiosis two. Meiosis one causes genetic variation but meiosis two is just like mitosis, with the lack of genetic material being duplicated in interphase. 


1- Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf?

Gametes Male AaTt - AT,At,aT,at

Gametes Female AaTt - AT,At,aT,at


AT

At

aT

at

AT

AATT

AATt

aATT

AaTt

At

AATt

AAtt

aATt

Aatt

aT

AaTT

AatT

aaTT

aatT

at

AaTt

Aatt

aaTt

aatt

The possible gametes for each parent are AT,At,aT,at. Since terminal flower and dwarf are both recessive traits, the genotype would have to be aatt. Using the punnett square, we are able to find that 1 out of 16 offspring would be predicted to have terminal flowers and be dwarf. This means that 25 out of 400 offspring would be predicted to have terminal flowers and be dwarf.  

2- If P1 is PpYyRr X Ppyyrr. What fraction of offspring from this cross are predicted to exhibit the recessive phenotypes for at least two of the three characters?

Gametes Male PpYyRr: PYR,PYr,PyR,Pyr,pYR,pYr,pyR,pyr

Gametes Female Ppyyrr: Pyr,pyr


PYR

PYr

PyR

Pyr

pYR

pYr

pyR

pyr

Pyr(1/2)

PPYyRr

PPYyrr

PPyyRr

PPyyrr

PpYyRr

PpYyrr

PpyyRr

Ppyyrr

pyr(1/2)

PpYyRr

PpYyrr

PpyyRr

Ppyyrr

ppYyRr

ppYyrr

ppyyRr

ppyyrr


P

p



Y

y



R

r

P

PP

Pp


y

Yy

yy


r

Rr

rr

p

Pp

pp


y

Yy

yy


r

Rr

rr

The offspring that are predicted to exhibit the recessive phenotypes would need two recessive alleles. Since two of the three characters have to be shown in this question, the possible combinations are PPyyrr, Ppyyrr, ppYyrr, ppyyRr, ppyyrr, Ppyyrr. The first punnett square shows that there is 6/16 chance which is 0.375 or ⅜.

Another way to find the probability of pp, when Pp and Pp are crossed, it is a ¼ chance. When Yy and yy are crossed, there is a 50% chance that the recessive trait is expressed. Then when Rr and rr are crossed, there is also a  50% chance that the recessive trait is expressed. The chance all 3 are recessive is 0.0625. Since we are only looking for two out of the three recessive traits to be exhibited, you only have to multiply 2 so 0.5 times 0.5 which is 0.25 and 0.5 times 0.25 which is 0.125. When added together, the total is 0.375 or ⅜. 

3- Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii). What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?


P

p



Y

y



I

i

p

Pp

pp


Y

YY

Yy


i

Ii

ii

p

Pp

pp


y

Yy

yy


i

Ii

ii

The possibility of recessive pp is 50%. The possibility of recessive yy is 25%. The possibility of recessive ii is 50%. The probability of having recessive pp and yy is 0.125.  The probability of having recessive pp and ii is 0.25. The sum of these probabilities are 0.375. 

4- complete the cross for the snapdragon flower that your teacher has started. Write down the genotypic and phenotypic ratios for the F2.

F1

r

r


F2

R

r

R

Rr

Rr


R

RR

Rr

R

Rr

Rr


r

Rr

rr

Since the offspring of the first breeding creates all heterozygous pink plants, the pink snapdragon is self pollinated, and creates the F2 which has RR as 25%, rr as 25% and Rr as 50%. The genotype ratio is RR:rr:Rr=1:1:2. The phenotypic ratio of f2 is red:white:pink=1:1:2. Two offspring are pink, one is red and one is white since it shows incomplete dominance.

Meselson and Stahl cracked the puzzle

Matt Meselson and Franklin Stahl originally met in the summer of 1954, the year after Watson and Crick published their paper on the structure of DNA. Although the two researchers had different research interests, they became intrigued by the question of DNA replication and decided to team up and take a crack at determining the replication mechanism. Now watch the following video and answer the questions:

1- Explain about 3 models proposed for the structure of DNA based on Watson and Crick' s announcement of DNA as genetic material. Use some ideas from the video with a timestamp (one time stamp is enough to show the authenticity of your work).

  1. Semiconservative: unfold in middle, make copies on either side 

  2. Conservative: Wraps around histone protein and makes exact duplicate

  3. Dispersive: Cut into 10-12 nucleotides and each segment is copied 

2- Based on Meselson and Stahl’s experiment which model is accepted for the replication of double helix DNA? 

  1. Based on Meselson and Stahl’s experiment, the semiconservative model is accepted for the replication of double helix DNA. Since the experiment growing e. coli with heavy nitrogen showed that the light nitrogen DNA increased as each generation is duplicated, it proves that DNA is replicated by first unfolding in the middle, and then making copies on either side.

3- Suppose that Meselson and Stahl have first grown the cells in N14 containing medium and then moved them into N15 containing medium before taking the samples, what would have been the result? (talk about bands density and location in the tube. You may draw in own notebook using two colors to help yourself solving this problem)

  1. With the conditions stated above, the results would have been identical to the original experiment, only with the data sets swapping the results. The band density in the N15 is increased as there is more duplication, so the amount of hybrid nitrogen will decrease, as the amount of N15 or heavy Nitrogen increases. The tube will show increased concentrations of the heavier N15 after each duplication since there is more N15 being duplicated semiconservatively. 

Chi Squared Practice Problems


1. Chi Squared Problems: you will be given the table of critical values and formulas, like on your test. Ex:

a. In a heterozygous, heterozygous dihybrid cross, the following data was obtained:

dominant for both traits: 570, dominant for trait 1 and recessive for trait 2: 185

dominant for trait 2 and recessive for trait : 190, recessive for both traits: 55

Perform a chi-square analysis to see if the data above agrees with the predicted outcome of this cross.

In this experiment, the null hypothesis is supported    because 1.092 is less    than 7.82. 

b. In 1901, Bateson reported    the first post-Mendelian study    of a cross involving two characters. White leghorn chickens, having white feathers and large "single" combs, were crossed to Indian Game Fowl, having dark feathers and small "pea" combs. The F1 were white with pea combs, so it is assumed that these are the dominant traits. The F2 distribution was: 111 white pea, 37 white single, 34 dark pea, and 8 dark single. (You must show your Chi squared table and all your calculations like part a)

Group

Observed

Expected

O-E

(O-E)^2

(O-E)^2/E

Dom both traits

(white pea)

111

107

(9/16 X 190)

4

(111-107)

16

(4)^2

0.1495

(16/107)

Dom trait 1, rec trait 2

(white single)

37

36

(3/16 X 190)

1

(37-36)

1

(1)^2

0.0277

(1/36)

Dom trait 2, rec trait 1

(dark pea)

34

36

(3/16 X 190)

-2

(34-36)

4

(-2)^2

0.1111

(4/36)

Rec both traits

(dark single)

8

12

(1/16 X 190)

-4

(8-12)

16

(-4)^2

1.3333

(16/12)

4 groups, DOF is 3




Chi squared: 

1.6216

Since the degree of freedom is 3, and the p value is 0.05, the critical value is 7.82. The calculated chi squared value in this question is 1.6216, which is less than the critical value. This means that the null hypothesis is supported since it follows the 9:3:3:1 ratio for a dihybrid cross. 

c. A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of basic codominance. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome? (you must show your Chi square table and all of your calculations)

Group

Observed

Expected

O-E

(O-E)^2

(O-E)^2/E

Trait 1 

(stripes)

50

44

(1/4 X 176)

6

(50-44)

36

(6)^2

0.818

(36/44)

Traits 2

(spots)

41

44

(1/4 X 176)

-3

(41-44)

9

(-3)^2

0.205

(9/44)

Traits 1 and 2

(stripes and spots)

85

88

(2/4 X 176)

-3

(85-88)

9

(-3)^2

0.102

(9/88)

3 groups, DOF is 2




Chi squared: 

1.125

Since the degree of freedom is 2, and the p value is 0.05, the critical value is 5.99. The calculated chi squared value in this question is 1.125, which is less than the critical value. This means that the null hypothesis is supported since it follows the 1:1:2 ratio for a codominance cross. According to the Chi-square test, she did get the predicted outcome since the difference is statistically insignificant. 


1- A white-eyed female Drosophila is mated with a red-eyed(wild type) male. What phenotypes and genotypes do you predict for the offspring?

The maleoff spring will all have white eyes. Both female offspring will have red eyes since they are hybrid. 50% of the offspring is female with XwXw+ and 50% of the offspring is male XwY. 100% of the female offspring is heterozygous XwXw+ and 100% of the male offspring is XwY. 50% red phenotype, 50% white phenotype. 


Xw

Xw

Xw+

XwXw+

XwXw+

Y

XwY

XwY

2- Neither Tim nor Rhoda has Duchene muscular dystrophy, but their firstborn son does have it. What is the probability that a second child of this couple will have the disease? What is the probability if the second child is a boy that has the disease? A girl?


XD

Xd

Xd

XDXd

XdXd

Y

XDY

XdY

Probability that a second child of this couple will have the disease is 50%. 

Probability that a second child is a boy that has the disease is 50%.

Probability that a second child is a girl that has the disease is 0%.


Summarize what you have learned. How do you recognize a pedigree related to mitochondrial inheritance?

I learned that mitochondrial DNA only comes from the egg, which is the zygote produced by the mother. This causes all offspring to have maternal mtDNA. This inheritance is called maternal inheritance and is a type of extranuclear inheritance. The pedigree that is displayed in the video is an example of mitochondrial inheritance. The affected father does not have any impact on his children as long as the mother is unaffected. Since the mother is unaffected, all of the offspring do not display the trait that makes the individual affected. When the father is unaffected and the mother is affected, all of the offspring are affected since all of the children get mtDNA from the mother. The pedigree looks similar to the autosomal dominant pedigree. The key point that sets them apart is that despite the father in 1,1 showing symptoms, that none of the children shows signs of the affected trait. The most important part of the video was the importance of examining the pedigree while looking at the sample fully as well as examining any consistencies before making any conclusions.

1- Based on the information provided by the teacher for question 1, what is the Phenotype of the parents for a, b and c? What is the probability of getting a colorblind son for c? (son is colorblind)

Phenotype of mother a: not colorblind

Phenotype of father a: colorblind

Phenotype of mother b: not colorblind (carrier)

Phenotype of father b: not colorblind

Phenotype of mother c: not colorblind (carrier)

Phenotype of father c: colorblind

Probability of getting a colorblind son for c is 50% from all possible offspring. 

2- If parental flies had been true-breeding for gray body with vestigial wings and black body with normal wings, which phenotypic class(es) would be largest among the testcross offspring?

  • gray body with vestigial wings: b+b+vgvg, gamete: b+vg

  • black body with normal wings: bbvg+vg+, gamete: bvg+

  • The genotype of f1 is bb+vgvg+. 

  • This has many gametes that are bvg,b+vg,bvg+ and b+vg+. 

  • The test cross would be black body with vestigial wings: bbvgvg, gamete: bvg

  • The genotypes possible for f2 is bbvgvg, bbvgvg+, bb+vgvg, and bb+vgvg+. 

  • The phenotypic classes that will be seen largest are black with normal wings, and grey with vestigial wings. This is because they are the parental types and it is because they are the original genes that were linked together in the P or parental generation. 

f1

b+vg


f2

bvg

bvg+

b+vg

b+vg+

bvg+

bb+vgvg+


bvg

bbvgvg

bbvgvg+

bb+vgvg

bb+vgvg+

Worksheet on heredity


1- Gene A, B, and C are located on the same chromosome. Testcross shows that  the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes? Explain

The higher the recombination frequency, the farther the genes are located on the same chromosome. Since A and C have a lower frequency of 12% when compared to the A and B with a higher frequency of 28%. The order can be either ACB or CAB. The gene C can be either in between A and C or before A. To know the linear order of the genes, the recombination frequency for genes B and C has to be determined. 

2- About 5% of individuals with Down syndrome have a chromosomal translocation in which a third copy of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent’s gonad, how could it lead to Down syndrome in a child?

During meiosis, the third copy of chromosome 21 being attached to chromosome 14 will be treated as one chromosome. This means that translocation of this chromosome would occur. There will be a gamete with only one chromosome 21, but there will be another chromosome with a normal chromosome 21 along with the extra chromosome 21 that was attached to chromosome 14. In the case that the gamete with two chromosome 21s is fertilized or fertilizes a gamete with a normal number of chromosomes which is 1, the offspring would have three chromosome 21s, which is called trisomy 21 or in other words, down syndrome. 

3- The ABO blood type locus has been mapped on chromosome 9. A father who has type AB blood and a mother who has type O blood have a child with trisomy 9 and type A blood. Using this information, can you tell in which parent the nondisjunction occured? Explain your answer.

Yes, it is possible to tell which parent cell that the nondisjunction occurred. The father who has blood type AB has A and B antigens. The mother who has blood type O has no antigens. When this is crossed, there is a 50% chance that the offspring has type A with the genotype AO and a 50% chance that the offspring has type B with the genotype BO. The A blood type means that the mothers chromosome 9 did not split properly. If nondisjunction took place in chromosome 9 of the father, the child would have blood type ABO. Since this is not the case, the mothers O blood type with no antigens had nondisjunction, causing the child to have AOO as their genotype. 

4- The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Explain how the activation of this gene could contribute to the development of cancer.

The activation of the gene that codes for intracellular tyrosine kinase can lead to the development of cancer because they are able to send many signals with only one dimerization of a tyrosine. Since there are many kinases, there is also a lot of cell division, since receptor tyrosine kinases play a large role in the rapid division of cells. This extremely rapid speed of cell division may be harmful since there can be a cell that is not properly functioning being replicated over and over. This rapid growth of cells that are faulty can contribute to the development of cancer if the cells do not go through apoptosis. 

5- Gene dosage, the number of active copies of a gene, is important to proper development. Identify and describe two processes that establish the proper dosage of certain genes.

One process that establishes the proper dosage of certain genes is when one x chromosome is inactivated in females. This is because females have 2 x chromosomes, with one being passed down from each parent. Having two active x chromosomes is most likely too much to enable the proper development. The other process that establishes the proper dosage of certain genes is triple x syndrome or trisomy x. Although the individual is a female with three x chromosomes, there are no developmental issues seen. This is because two of the x chromosomes are inactivated, to fit the proper dosage of the sex chromosomes.

6- Reciprocal crosses between two primrose varieties, A and B, produced the following results: A female x B male gives offspring with all green (non variegated)  leaves;   B female x A male  gives offspring with spotted (variegated) leaves. Explain these results.

    The reason the plants seem to be different colors is because the number of chlorophyll in the chloroplast. In the cross A female x B male, the green leaves are caused by the abundance of the chlorophyll. In the cross B female x A male, the variegated leaves are caused by the mix of cells with an abundance of the chlorophyll as well as a lack of chlorophyll, and also a mix of both. The wild type is green leaves caused by the abundance of the chlorophyll, while the mutant type is white leaves caused by the lack of the chlorophyll. Since chlorophyll is located in the chloroplast and chloroplast is maternally inherited, the offspring's phenotype is identical to the mother’s phenotype. This makes A a primrose with all green leaves, and B a primrose with spotted or variegated leaves. 

7- Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not?

    Mitochondrial genes are inherited maternally. They are critical to the metabolism of the cell but there are numerous mitochondria inherited from the mother, with the gamete which is an egg. Although there may be mutations in the mitochondria, each cell has many mitochondria. Since there most likely enough mitochondria that function normally, the gene mutations are generally not lethal SInce enough cellular respiration is taking place in the cell. 


DNA structure worksheet

1- How did Griffith experiment rule out the possibility that the R cells could have simply used the capsules of the dead S cells to become pathogenic?

  1. His experiment rules out this possibility by checking the offspring. If the cell reproduces with a capsule and gives capsule offspring, the R cell did not use the capsule of the dead S cell. It is DNA fragmentation that gave them the ability to produce a capsule on their own, allowing the offspring to also exhibit this trait. 

2- Griffith did not expect transformation to occur in his experiment. What results was he expecting? Explain your answer

  1. He was expecting that the mouse would not die since when performing the experiments separately, the heated S strains did not kill the mouse, since they were denatured. Additionally, the R strains did not kill the mouse since they are not virulent. When combined, he expected that the mouse would not die. However, this was not the case, since the DNA fragments from the heat killed S strains were transformed to the R strains, making the latter a S strain and causing it to have a capsule and be virulent, finally resulting in it killing the mouse. 

3- For Hershey and Chase (bacteriophage experiment) how would the results have differed if proteins carried the genetic information?

  1. If the protein carried the genetic information, the radioactivity would have transferred from the protein to the e coli cell, causing the radioactivity to only be seen in the pellet. The results would have been similar to the results of the DNA being changed to radioactive DNA. 

4- A fly has the following percentages of nucleotides in its DNA: 27.3% A, 27.6% T, 22.5% G, and 22.5% C. How do these numbers demonstrate the Chargaff’s rule about base ratios?

  1. Chargaff’s rule states that the A and T bases are equal, as well as the C and G bases in any organism, although the base composition of DNA varies between species. These numbers demonstrate the Chargaff’s rule about base ratios since the ratio between C and G bases are equal. The ratio between A and T bases are roughly equal, this slight imbalance may have been because of a mutation, or a mistake in the genetic coding. 

5- Given a polynucleotide sequence such as GAATTC, can you tell which is the 5’ end? If not, what further information do you need to identify the ends?

  1. No, you can not tell which is the 5’ end just from the polynucleotide sequence. You need the 5’ and 3’ given in the problem. The opposing polynucleotide sequence would be CTTAAG and the 5’ and 3’ ends would be antiparallel. 

1- Suppose instead of N15, Meselson and Stahl had started the cell growing with N14 and then moved the cells into N15-containing medium before taking the samples, how could results have changed?

  1. The results would have been identical to the original experiment, only with the data sets swapping the results. The band density in the N15 is increased as there is more duplication, so the amount of hybrid nitrogen will decrease, as the amount of N15 or heavy Nitrogen increases. The tube will show increased concentrations of the heavier N15 after each duplication since there is more N15 being duplicated semiconservatively. 

 

Original

Sample 1

Sample 2

Sample 3

100% N14

100% hybrid 

50% hybrid

25% hybrid



50% N15

75% N15

 

2- Write a summary.

  1. DNA replication is the process where DNA copies itself. In eukaryotes, this is done in the S phase in interphase, while in prokaryotes, this is done through binary fission. There were 3 theories which were the semiconservative, the conservative and the dispersive. Through the Messelson-Stahl experiment, the N 15 would become hybrid and eventually, there would be a greater sample of N14 since the semiconservative model was proven to be correct. DNA is antiparallel, meaning that there is a strand that runs from the 5’ to 3’ while the opposite strand runs from 3’ to 5’. Bases can only be added to the 3’ end, so one strand is a leading strand while the other is a lagging strand. 


Class worksheet on DNA replication and gene expression 

1- From 2 strands of the original DNA in the sketchbook, which one’s copying needs Okazaki fragments. You must be able to explain your answer.

- A would need Okazaki Fragments. The reason for this is the orientation of the 5’ and 3’ ends. Bases can only be added to the 3’ end and strand A would have a complementary strand that is from 3’ to 5’ based on the orientation of the DNA as well as the location of the origin of replication. Since the right side of A runs from 5’ to 3’, it would need okazaki fragments. It is like backstitching.

2- Suppose that DNA Pol I (ONE) is not functional in a given cell. How could that affect the synthesis of a leading strand?

- If DNA pol 1 is not functional, then the RNA primer is not digested and it is not replaced with DNA. This could affect the synthesis of the lagging strand because the strand that is created will have DNA as well as RNA primers. This may impact the synthesis of the leading strand since it would have only DNA while the lagging strand would be composed of DNA as well as RNA primers. In order to have one continuous strand, the DNA polymerase 1 is necessary also for the synthesis of the leading strand because it would have to be combined with the other sections of the strand to form a proper double strand.

3- what are 2 properties (one functional and one structural) distinguish euchromatin from heterochromatin?

- One functional property that distinguishes euchromatin from heterochromatin is that in heterochromatin, the genes could not be expressed while in euchromatin, the genes inside the DNA could be expressed. One structural property that distinguishes euchromatin from heterochromatin is that in heterochromatin the gene is more compact while in euchromatin, the genes are narrow but not compact. 

4- what did you learn from srb and Horowitz experiment?

- From the srb and Horowitz experiment, I learned that the test involved genes and using proteins to see whether or not they are necessary for the growth of the neurospora crassa. It showed that the fungus grew in the wild type in all 4 test tubes. Then it was mutated once to show that the minimal medium did not grow the fungus, this was because enzyme one was disabled. Then the second mutation showed that the minimal medium as well as the mm plus ornithine did not grow the fungi. This was because the second enzyme was disabled. Then the third mutation showed that all of the test tubes except mm plus arginine were not growing. This was discovered to be the cause of the 3rd enzyme. 

1- If you compare the sequence of the mRNA to that of nontemplate DNA strand, what similarity or difference would you notice?

The sequence of the mRNA and the nontemplate DNA strand are identical with the exception of the mRNA uses uracil while the nontemplate DNA strand uses thymine. The template strand is the original that is used to make the mRNA so it is complementary to the nontemplate strand and the mRNA strand. The difference is that RNA and DNA use different bases so the only difference is the presence of uracil or thymine. The direction of the nontemplate DNA and the mRNA are in the same direction of 5’ and 3’. They are coding strands. 

2- suppose that you have an mRNA with a poly-G sequence of 30 nucleotides long. What and how many amino acids would you expect to see after translation? You can look up the codon table for mRNA)

I would expect to see 10 amino acids after translation. GGG translates into glycine so there would be 10 glycines in a row after translation. The strand would look like 

Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly  Gly

3- For each of the following DNA strands, transcribe and translate (which one will make the longest amino acid chain?)

The longest amino acid chain is the 1st one because there are 4 codons and the last is a stop. 

Template 3’- ACACGACCTTAC-5’

mRNA     5’- UGU  GCU  GGA  AUG-3’

Nontemp 5’- TGTGCTGGAATG-3’

Amino acid: Cys  Ala  Gly  Met


Template 3’-AAATACAAAGGG-5’

mRNA     5’- UUU  AUG  UUU  GGG-3’

Nontemp 5’- TTTATGTTTGGG-3’

Amino acid: Phe  Met  Phe  Gly 


Template 3’-AACATTCCAAAAGGGGACTAA-5’

mRNA     5’- UUG  UAA  GGU  UUU  CCC  CUG  AUU-3’

Nontemp 5’- TTGTAAGGTTTTCCCCTGATT-3’

Amino acid: Leu  END


Transcription worksheet 

1- Promoter is located at the upstream end of a transcription unit.

Upstream promoter and 5’ cap

Downstream 3’ poly a tail

2- What does a promoter do?

A promoter is necessary for the eukaryotic cell when mRNA synthesis is going to take place. It is the region of the DNA strand on the 3’ end where RNA polymerase binds to in order to begin transcription. This promoter also contains the TATA box, where all of the nucleotides are thymine and adenine. 

3- How does RNA polymerase start transcribing a gene at the correct place on the DNA of prokaryotes versus eukaryotes?

 RNA polymerase starts transcribing a gene at the correct place on the DNA of prokaryotes using the origin of replication. Since prokaryotes only have a single ring of DNA, it is simple to replicate the DNA. RNA polymerase starts transcribing a gene at the correct place on the DNA of eukaryotes differently. On the template DNA, there is a promoter on the 3’ end of the strand. RNA polymerase is able to attach to the DNA here, and begins transcription. 

4- What happens if a harmful ray causes some change in the TATA box. How would that affect transcription of the gene?

If there is a harmful ray that causes some change in the TATA box, the transcription of the gene will most likely take longer. Since there are many bubbles of DNA replication, it will not completely stop the RNA synthesis.The change in the TATA box can cause it to disappear, which means that the promoter will be missing a piece that is necessary for the RNA polymerase to attach to the DNA. This may cause this specific location to not be functioning as an origin of replication. 

5- What are some important functions that 5’ cap and poly-A tail share/do? 

5’ cap function;

  • Indicates the start location of RNA synthesis. 

  • Allows for RNA polymerase to attach and begin transcription. 

  • Allows the beginning of the mRNA strand to leave the nucleus.

  • Contains the TATA box.

Poly-A tail function;

  • Indicates the end location of RNA synthesis. 

  • Allows for the DNA strand to be cut and end the transcription.

  • Allows the end of the mRNA strand to leave the nucleus. 

  • Contains many A nucleotides. 

5’ cap and poly-A tail share; 

  • Facilitate the export of mRNA.

  • Protects the mRNA from hydrolytic enzymes

  • Let the ribosomes attach to the 5’ end. 


DO NOW: Why should the gene expression be under regulation? (Why a gene cannot be expressed all the time)

I think that gene expression should be under regulation because there are genes that do not need to be expressed all of the time. One example is the gene expression of one of the x chromosomes in a female. There is only one x chromosome that is expressed, the other is suppressed. The reaction could be harmful to the cell, it can also be wasteful since there the process of making the gene would not be necessary.

1- What is the function of a promoter and operator?

  • Operator function: a segment of the DNA which acts like a switch to regulate the concentration of the enzyme. It is usually positioned within the promoter of the DNA. 

  • Promoter function: a segment of the DNA where RNA polymerase is able to attach on to, in order to produce more mRNA through transcription. 

2- The following are the steps that happen when we have a high level of tryptophan. Put them in order

B- Tryptophan does not need to be produced by the operon

C- Tryptophan will bind to the repressor protein, changing its conformation

A- The trp repressor-tryptophan complex can now bind to the operator of the trp operon

D- RNA polymerase is blocked from transcribing the genes needed to synthesize tryptophan

NOW LET’S WATCH A VIDEO

3- What happens if lactose is present and glucose is scarce? Put the following list in order. Start with the effects of lactose levels then proceed to the effects of glucose levels.

C- The enzymes needed for lactose metabolism must be transcribed when lactose is present

E- Lactose binds to the LacI repressor, changing LacI’s shape and making it fall off the operator

A- Now that LacI has been removed for the operator, RNA polymerase can proceed with transcription

G- The three enzymes involved in the metabolism of lactose are transcribed and expressed at moderate levels

D- cAMP levels increase because glucose is scarce (ATP is not being produced through cell respiration)

B- cAMP binds to CAP regulatory protein, causing it to bind to the promoter of the lac operon

F- CAP binding causes RNA Polymerase to bind to the promoter (higher affinity) and transcribe the gene at a higher level than before

4- What happens to the lac operon if both lactose and glucose are scarce?

If glucose is scarce, then the active receptor attaches to the operator and prevents the mRNA from being produced. This prevents the proteins for lactose breakdown from being produced. CAP is activated by binding with cAMP when glucose is scarce. The CAP which is activated can attach to the promoter of the lac operon which increases the affinity of the RNA polymerase. This can increase the transcription rate but there is no transcription taking place, so the level of glucose and lactose will not change. 

5- If a mutation happens to the lac operator so that the active repressor cannot bind. How would this affect the cell’s production of beta-galactosidase?

There will be increased amounts of beta-galactosidase since there is no way to prevent the RNA production. This means the cell will continue to make beta-galactosidase even when lactose runs out. This will be a waste of the cell's resources since there is no inhibition. 

1- What are the effects of histone acetylation and DNA methylation?

The effects of histone acetylation are that acetyl groups are attached to the positively charged lysines in histone tails. This causes the chromatin structure to loosen which allows for transcription to take place on the DNA strand. The effects of DNA methylation are that the methyl group is added to a base in DNA that is associated to reduce transcription. This condenses the chromatin and can cause long term inactivation of the genes. 


2- What are the general and specific transcription factors? (their roles and functions)

The general and specific transcription factors are necessary to initiate transcription in eukaryotic cells. General transcription factors are necessary for transcription of all genes necessary to code for proteins. Their function is to assemble the transcription initiation complex at the promoter at all genes.  Specific transcription factors are necessary for the stimulation or repression of transcription, within the DNA. They are most likely distal control elements, which are enhancers. They bind to control elements that are associated to a specific gene. Once it is bound, it increases or decreases the transcription rate. Activators are used to increase transcription while repressors are used to decrease the transcription. 


3 If you are comparing the nucleotide sequences of the distal control elements in the enhancers of three genes that are expressed only in muscle, what would you expect to find? Why?

When comparing the nucleotide sequence of the distal control element in the enhancers of the three genes that are expressed only in muscle, I would expect the genes to be similar to the enhancers. This similarity allows the same specific transcription factor to bind to the enhancer of all of the genes. This will stimulate the gene expression of this trait. 


1- Compare and contrast miRNA and siRNA

miRNA has a hairpin like structure and is able to break down mRNA and prevent translation. siRNA can inhibit gene expression by using RNAi. They have different RNA precursors. 

2- If the mRNA being degraded coded for a protein that promotes cell division in a multicellular organism, what would happen if a mutation disabled the gene encoding the miRNA that triggers this degradation?

In the case of the mutation that disables the gene encoding the miRNA that triggers this degradation, there will be excessive cell division in the multicellular organism. Since the degrading protein is disabled, there will be increased protein that promotes the cell division. This may cause faulty cells to duplicate, which may result in cancer. 

3- How can the inactivation of the X chromosome be related to the non-coding RNA?

The inactivation of the X chromosome can be related to the non-coding RNA because X-chromosome inactivation relies on the histones. When the histones are tightly wrapped and compressed, the x chromosome is heterochromatin which prevents translation and transcription. 

1- As you know mitosis gives rise to two daughter cells which are genetically identical to the parent cell. So how come mitotic divisions products are not composed of identical cells?

Mitotic divisions are not identical because differentiation creates specific cells that are necessary for specific needs. This means that there are proteins that are activated in some locations, and others that don’t. This heavily relies on transcription factors. 

2- signal molecules  released by an embryonic cell can induce changes in a neighboring cell without entering the cell. Explain about the process involved in this kind of signal transduction.

Single molecules released can induce change through cell to cell recognition. Using ligands, the cells can interact as long as a receptor is present and can accept the signal molecule. THis can cause changes to the cell that receives the ligand, to be able to differentiate the cell using determination. 

3- Why do we call maternal effect genes as polarity genes?

Maternal effect genes are called polarity genes because they are able to determine which side of the embryo becomes the anterior and posterior ends. This is done through differential gene expression to control what traits are expressed, and where they are expressed. The mother provides the mRNA necessary for this process through the egg cell. 

4- In the figure 18.17b of the powerpoint you can see that the lower cell is synthesizing signaling molecules, whereas the upper cell is expressing receptors for these molecules. In terms of gene regulation, explain  how these cells came to synthesize different molecules.

In terms of gene regulation, these cells came to synthesize different molecules because they have different roles. Determination is able to commit a cell to its final state, so it can activate or inactivate mRNA sequences to produce transcription factors. Before induction, the maternal determination can cause the cell to have special proteins to cause the various molecules.