Assignment 1 & 2 on DNA Replication

Assignment 1

  1. A. Briefly explain how the leading and lagging strands of DNA are synthesized during DNA replication. (10 marks)
  2. What is the function of each of the following? (2 marks each, 10 marks total)
    • Primase
    • Helicase
    • Nucleases
    • Histones
    • Poly(A) tail
  3. What would be the complementary strand of the following portion of DNA: 3’AACCGTAATTCG5’? (2 marks)
  4. Using the following mRNA sequence, provide a definition and an example of each of the following mutations: silent, nonsense, missense, and frameshift mutation. (3 marks each, total 12 marks)

AUG GUG CCA UUC AAU AUG UGG

  • Summarize how a mature mRNA is made from the primary mRNA transcript. (6 marks)
  • Answer the following two questions: (8 marks total)
  • Describe the central dogma of molecular biology. (2 marks)
  • Give 3 reasons why proteins are not made directly from DNA. (Hint: Consider disadvantages of using DNA directly, and/or advantages of not using DNA directly). (6 marks)
  • Compare and contrast the following which are involved in transcription in Eukaryotes: (4 marks each, 12 marks total)
  • Promoter proximal elements and promoter
  • Enhancers and silencers
  • Regulatory transcription factors and co-activators
  • How can one gene code for more than one protein? (2 marks)
  • Describe the four levels of chromatin structure and explain how chromatin is decondensed to allow transcription. (10 marks)
  • Compare and contrast dideoxy DNA sequencing and Southern blotting with respect to procedure and purpose. (8 marks)

Assignment 2.

  • You are studying a cell which, before the S phase of the cell cycle, has 8 chromosomes and 1.2 picograms (1.2 × 1012 grams) of DNA. Complete the following table with the expected number of chromosomes, chromatids, and homologous pairs, and the amount of DNA in each area or pole of the spindle at each of the stages indicated. (1/2 mark per box, 10 marks total)
 ChromosomesChromatidsHomologous pairsDNA amount
Prophase (Mitosis)81642.4 picograms
Anaphase (Mitosis)8841.2 picograms
Prophase I (Meiosis)81642.4 picograms
Anaphase I (Meiosis)4801.2 picograms
Anaphase II (Meiosis)4400.6 picograms

Including examples, compare and contrast the following: (4 marks each; 16 marks total)

  • Genotype and phenotype
  • Codominance and epistasis
  • Crossing-over and independent assortment
  • Non-disjunction and translocation

A particular type of pond organism usually reproduces asexually. However, in later summer and as winter approaches it tends to reproduce sexually. Explain why this organism would use two different methods of reproduction at different times of the year. (6 marks)

Alternation of generation is the concept whereby certain living organisms are capable of multiplying either through sexual or asexual mode. Sexual reproduction is defined as the production of sex gametes (formed as a result of meiosis), and its fusion, whereas, in asexual reproduction, the life cycle is through spores that are formed as a result of mitosis. In the pond, the organism normally reproduces through asexual reproduction mainly to increase the number of individuals, as the condition would be favouring their growth.

The same organism reproduces through asexual means mainly to climatic conditions and other external characteristics including the availability of foods. Under such extremes of condition, the organism switches to the sexual mode of reproduction allowing the crossing-over of genes that genetically produce offspring that are significantly different and far more superior than the wildtype that is better adapted to face the harsher environment. It is for this reason that the organism in the pond that once reproduced through asexual mode switched to the sexual mode of reproduction.

  • A female Drosophila (fruit fly) heterozygous for red eyes and grey body was crossed with a male homozygous for scarlet eyes and ebony body. The two genes are linked on the same chromosome. The following offspring were produced. How many map units would separate them? (Show all your work.) (4 marks)
Red eyes, grey body76
Red eyes, ebony body26
Scarlet eyes, grey body30
Scarlet eyes, ebony body68

Number of recombinants present = 26 + 30 = 56

Total number of offspring’s produced = 76 + 26 + 30 + 68 = 200

Recombination’s frequency = (total number of recombinants / complete number of offspring produced) x 100

Recombination frequency is = (56 / 200) x 100

          = 28%

The recombination frequency is considered to be equivalent to the distance between the genes. So the distance between the genes will be 28 centi morgans.

  • In a certain breed of domestic fowl, super comb is dominant to single comb but feather colour shows absence of dominance. Black legs and white legs are homozygous, whereas the heterozygote gives red legs. From crosses between birds heterozygous for both genes, what proportion of the offspring would you expect to be: (Define the alleles and show all your work using a Punnett Square.) (12 marks)

Super comb – Dominate “S”

Single comb – recessive “s”

Black leg – homozygous (BB)

White leg – homozygous (bb)

Red leg – Heterozygous (Bb)

CrossSBSbsBsb
SBSSBBSSBbSsBBSsBb
SbSSBbSSbbSsBbSsbb
sBsSBBsSBbssBBssBb
sbSsBbSsbbssBbssbb
  1. Super-combed

Super-combed = 12

  • Black-legged

Black-legged = 4

  • Red-legged

Red-legged = 8

  • Super-combed and red-legged

Super-combed and red-legged = 6

  • Single-combed and white-legged

Single-combed and white-legged = 1

Super comb black leg – 3

Super comb red leg – 6

Super comb white leg – 3

Super comb black leg – 1

Single comb red leg – 2

Single comb white leg – 1

  • A particular X-linked gene is a recessive lethal that causes death and reabsorption of the embryo at an early stage. A heterozygous female married a man expressing the dominant trait. Define the alleles and show your work in answering the following questions: (5 marks)

X-linked inheritance pertains to genes which are located exclusively inside the X-chromosome. The bulk of X-linked genetic disorders are recessive as well as affects men more than women. This really is attributable to differences in the amount of X chromosomes among men as well as women. For such a male to display the X-linked disease, just single recessive allele is required whereas females requires two. An illustration of such X-linked ailment is really the blood disease haemophilia.

X- female normal gene

xo – x-linked recessive trait

y – male normal gene

Female recessive heterozygous – Xxo

Male dominant – xo y

Crossxoy
XX xoX y
xoxo xoxo y
  1. What proportion of their offspring would be expected to be females?

Based on the Punnet square, there is a 50 per cent chance that their offspring is female.

  1. What proportion of the offspring would be expected to be carriers of the gene?

There is a 25 per cent chance that their offspring is a carrier.

Let Xh represent the lethal gene. The mother is represented as XH Xh and the father as XH Y. Results show that there is a 25 per cent chance one (Xh Y) of their offspring’s will die, and 25% chance that one(XH Xh ) of their offspring’s is a carrier. There is also a 75 per cent chance that their offspring will live.

  • Look at the following pedigree. Circles represent females; squares represent males. Coloured circles and squares represent affected individuals. Horizontal lines represent matings; vertical lines represent offspring produced from a mating. (15 marks total)
  • What is the mode of inheritance? Referring to the applicable numbered individuals, explain your reasoning. Your choices are autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive, Y-linked. (10 marks)

A dominant autosomal trait is passed on from generation to generation. Inheritance. It is indeed a dominant gene that is found on the one of the non-sex chromosomes inside this form of heredity (autosomes). This sort of condition may be caused by a single faulty gene. 50 percent of the time, the father’s genes will result in an impaired kid, whereas the other 50 percent of the time will result in an unscathed child having 2 common genes (recessive genes). Therefore it could be an inherited condition as none of the parents has impacted offspring. There are two daughters inside this lineage and only circular 5 is afflicted, thus it cannot be x-linked dominant. Each parents in squares 7 & 8 are impacted in the second generation. However, their child 12 square remains unaffected, therefore afflicted parents may produce a non-affected child. It’s impossible for this illness to be x-linked recessive since it doesn’t pass from mother through son and father into father.

  • What are the genotypes of the following individuals (be sure to define the alleles): 4, 5, 8, 11? (5 marks)

Autosomal Dominant allele, (a -normal, A-disease)

Affected Male (AA)

Affected Female (AA)

Unaffected Male (aa)

Unaffected Female (aa)

Genotypes

Individual 4th : aa

Individual 5th : Aa

Individual 8th : Aa

Individual 11th : Aa or AA (continued for next generation)

  • List 4 features of an ideal organism for the study of genetics. (4 marks)
  • Explain the roles of bicoid, the segmentation genes, and homeotic genes in development of an embryo. (10 marks)
  • Answer the following questions about stem cell research: (10 marks total)
  • Differentiate between multipotent, pluripotent, and induced pluripotent stem cells. Include an example of each. (5 marks)
  • Outline the advantages and disadvantages of iPS cells. (5 marks)
  • Write a short paragraph (not more than 200 words) describing the genetic disorder known as maple syrup urine disease (MSUD). In your discussion, include answers to the following questions, using your own words to write a coherent paragraph in complete sentences. Cite your sources following the citation instructions in Jan A. Pechenik’s A Short Guide to Writing about Biology(8 marks)

a. How is maple syrup urine disease inherited?

b. Why is it called maple syrup urine disease?

  • Briefly explain what causes this disease.
  • How many births in North America result in infants with MSUD?

This disorder is transmitted in an autosomal recessive manner, this means that both the copies of gene within each cell are affected. Both parents of a child having an autosomal recessive disorder has one gene of the defective gene, although they normally exhibit no clinical manifestations (O’Reilly et al., 2021). Whenever a person gets a defective gene from both parents, they develop recessive genetic illnesses. If an individual inherits one functional gene or one nonfunctional gene for an illness, the person would become a carrier for that condition but will typically exhibit no signs. Maple syrup urine sickness is a hereditary condition in wherein the body seems unable to correctly digest particular amino acids (protein basic building block) (Xu et al., 2020). The illness is so named because the urine of afflicted newborns has an unique sweet odour. MSUD is a hereditary condition, meaning that it is handed down via generations. It is the result of a mutation in one of three genes. Individuals who have this disorder are unable to digest the amino acids isoleucine, valine, and isoleucine. As a result, these substances begin to accumulate in the bloodstream. Maple syrup urine illness is predicted to affect around one in every 185,000 babies born in North America (Strauss et al., 2020).

O’Reilly, D., Crushell, E., Hughes, J., Ryan, S., Rogers, Y., Borovickova, I., … & Knerr, I. (2021). Maple syrup urine disease: Clinical outcomes, metabolic control, and genotypes in a screened population after four decades of newborn bloodspot screening in the Republic of Ireland. Journal of Inherited Metabolic Disease44(3), 639-655.

Strauss, K. A., Puffenberger, E. G., & Carson, V. J. (2020). Maple syrup urine disease. GeneReviews®[Internet]. Xu, J., Jakher, Y., & Ahrens-Nicklas, R. C. (2020). Brain branched-chain amino acids in maple syrup urine disease: Implications for neurological disorders. International Journal of Molecular Sciences21(20), 7490.

BOOK for reference:

Biological Science, Third Canadian Edition, 3rd edition

  • Scott Freeman
  • Kim Quillin
  • Lizabeth Allison
  • Michael Black
  • Greg Podgorski
  • Emily Taylor
  • Michael Harrington
  • Joan C. Sharp
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