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1. Locate the whitefish blastula cells that are labelled by the numbered arrows. From the cellular structures you see in each of these cells, identify the stage of the cell cycle that each of these white fish blastula cells is undergoing.

2. Locate the white fish blastula cell that is outlined by a box and pointed to by the arrow numbered 4. In the box provided make a cell drawing of this cell. This cell was measured using the ocular graticule and found to be 15 divisions across its longest side. You will need to identify the stage of mitosis that this white fish blastula cell is undergoing and include this information in the title of your cell drawing.

3. Include a scale bar in your cell drawing of the cell pointed to by the arrow numbered 4. Show your full working for the scale bar or the section will not be graded.

Stages of Cell Division

(This question covers content from Practical 5 and 6)

The image on the following page is of a whitefish blastula undergoing various stages of cell division. The cells were stained with iron haematoxylin and viewed using the 40x objective and 10x ocular lens. Particular cells in the image of the whitefish blastula have been identified with numbered arrows; when answering this question make sure you correctly locate the specified cell.

NOTE: For a satisfactory grade to be awarded you must justify your answer with a concise explanation linking the structures you can see in the image to the stage of mitosis.

Number 2

Metaphase

*Condensed chromosomes

*The chromosomes are aligned at the Metaphase plate i.e. imaginary plane

*Each sister chromatid has a kinetochores which are attached to a microtubule coming from opposite spindle poles.

Number 3

Early Prophase

*Condensation of chromatin fibres starts and formation of discrete chromosomes take place.

*At this stage, chromosomes are visible and have two identical sister chromatids.

*Microtubules formation takes place

*Nucleolus and nuclear envelope disappears

HINT: Be sure to follow the guidelines for cell drawings covered in Practical 6.

TIP: View the image on your computer screen to answer this question, and when you draw cell 4, as some structures may be lost when you print the

Document.

Include a scale barin your cell drawing of the cell pointed to by the arrow numbered 4. Show your full working for the scale bar or the section will not be graded. 

             Total magnification is 400 X as there is 40X Objective lens and 10X Ocular lens.

             For 400X, 1 division of a graticule is 2.5 µm.

             There are 15 divisions.

             Conversion factor at 400 X = 2.5 µm

             So, 15*2.5 µM = 37.5 µm 

             Scale bar Formula:

             X (mm) / 1 mm = 500 µm / 37.5 µm

             X (mm) * 0.55 mm = 500 µm / 37.5 µm

             X (mm) = 7.33 µm i.e. 7 µm 

  1. Osmosis in Red Blood Cells.

(This question covers content from Practical 6)

A researcher performed an experiment to investigate the movement of water across the phospholipid bilayer of human red blood cells from healthy donors. The researcher was given three beakers containing 20 mL of a glucose solution; they were labelled solution A, solution B and solution C. Each solution contained varying concentrations of glucose. After measuring the baseline diameter of red blood cells, 50 µL of fresh whole blood was added to each labelled beaker of glucose solution. After 5, 15 and 25 minutes red blood cell diameter measurements were repeated. At each time point, the diameter of 40 red blood cells were measured using the ocular graticule of a light microscope (Olympus WF) at 1000x magnification. The experiment was repeated for solution B and solution C.

Movement of Water across Phospholipid Bilayer

The results of the experiment are presented in Table 1.  

Table 1: Diameter (mean) of red blood cells (n = 40) from healthy donors after immersion in either glucose solution A, B or C. 

Time

Mean diameter of red

Mean diameter of

Mean diameter of red

(min)

blood cells in glucose

red blood cells in

blood cells in glucose

solution A

glucose solution B

solution C

(µm)

(µm)

(µm)

0

7.0

7.0

7.0

5

7.2

7.6

6.9

15

7.0

8.1

6.2

25

7.1

8.8

5.8

Using the graph paper provided, draw a graph of the data presented in Table 1. Include a figure legend with your graph. See notes at the end of this question for HINTSand 

NOTE: Be sure to follow the guidelines for drawing graphs and writing figure legends, as are outlined in Practical 2 of your LFS103 Practical Manual. Also check the marking criterion on the rubric attached to this assessment Task 1 (5) as this will be used to assess the quality of your response to this question. 

TIP: Review your Question 1 feedback on Task 1(3).

It helps in detection of some disorders or to check abnormality of the cells.

HINT: You will need to compare intracellular and extracellular concentrations to establish tonicity, then link this to the concentration gradient of water and ultimately the net movement of water molecules. Include in your description the correct terminology that describes the shape of the RBC after immersion in such osmotic solutions)

Solution A

Glucose solution A is isotonic to the red blood cells as not much difference is seen in change of the diameter of the red blood cells with respect to time. Concentration gradient of water is same in and out of the cell. So, the flow of water in and out of the cell is happening at the same rate.

Solution B

Glucose solution B is hypotonic to the red blood cells as the mean diameter of red blood cells got increased with change in the time. Water concentration is higher outside the cell that means the glucose concentration is lower than solution C. So, water will move inside the cell due to which the size of the cell increased.

Solution C

Glucose solution C is hypertonic to the red blood cells as the mean diameter of these cells got decreased with change in time. As in this case, the concentration of the glucose is higher outside the cell, so due to osmotic forces, the water comes out of these cells undergo crenation.

Tissues: Tissue 1.

(This question covers content from Practical 7) 

The histology image below is a tissue sample viewed using a compound light microscope at a total magnification of 400X. Consider this image and answer

Tissue Classification


Questions 3 a), b), c) and d).

Classify this tissue subtype: 

Pseudo stratified ciliated columnar epithelium 

  1. Name a location in the human body where this tissue type would be present: 

Trachea 

  1. Name the cell indicated by the arrow and define its primary function: 

Goblet cells – They secrete mucin 

  1. Give a functionof this tissue and define how the tissue structure enables this function: 

Function:

It is the type of epithelium cells and consisting of single layer of cells. It seems that there are many layers of the cell but it is a simple epithelium layer. The epithelial cells are tightly packed. These cells are connected to adjacent cells by intercellular connections. It is the type of epithelium cells and consisting of single layer of cells. These are present in the trachea and consist of cilia and goblet cells. The inspired air is cleansed through ciliated cells and goblet cells secrete mucus which prevent invasion of foreign agents in the trachea. All the cells are attached to underline basement membrane. They are mostly found in the lining of the trachea. (Eroschenko, 2018) 

Identifying Tissues: Tissue 2.

The histology image below is a tissue sample viewed using a compound light microscope at a total magnification of 1000X. Consider this image and answer Questions 3 e), f), g) and h). 

  1. Classify this tissue subtype: 

Loose connective tissue 

  1. What are the 3 general components of this tissue type? TIP: They are found in every subtype of this tissue. 

Protein fibres, ground substance and specialized cells. 

  1. Name the structure indicated by the arrow:

Collagen fibres 

  1. Define the main role for this cell/structure identified by the arrow (i.e. your answer to g) and relate to this tissue’s function. 

Collagen fibres are one of the main elements in loose connective tissue. They are spread in different directions and form a loose mesh-like tissue. These fibres are acidophilic in nature. The main function of the collagen is to provide resistance or protect the internal organs and provides support to internal organs with a fluid matrix. (Eroschenko, 2008) 

Identifying Tissues: Tissue 3 

The histology image shown below is of a trachea cross section stained with haematoxylin and eosin (H&E) and viewed under a microscope at a total magnification of 100X. You can clearly see various layers of different tissue types in the section.

  1. In the box provided and using scientific drawing instructions taught in Practical 7, draw a zone diagramof this trachea section. Label the tissue types pointed to by the arrows on your drawing. You DO NOT need to provide a scale bar for this drawing.

HINT: In your labels classify the subtypes of tissues and include details on the tissue subtypes where relevant (e.g. epithelial layers include cell shape and any specialised structures on the apical surfaces) – as covered in Week 7 and Practical 7.

NOTE: Remember to use the terminology for naming tissues presented in the course material. DO NOT use terms like mucosa, submucosa, adventitia, lamina propria as we did not discuss these in LFS103.

  1. Pseudo stratified columnar epithelium
  2. Tracheal cartilage
  3. Smooth muscles
  4. Adipose tissue
  1. pH and the buffer capacity of solutions.

(This question covers content from Practical 9)

An experiment was conducted to measure the buffering capacity of three unknown solutions A, B and C. Pooled freshly donated blood samples

(300 mL) from healthy volunteers (n = 40) were prepared for testing by adding anticoagulant, removing any naturally occurring buffering systems, and adding 10 mL of either solution A, B or C. Following measurement of the baseline pH of each blood sample with the added solution A, B or C; ten 1 mL aliquots of

Buffering Capacity Experiment

0.01M HCl (hydrochloric acid) were added to the sample. After the addition of each 1 mL HCl aliquot the pH was measured using a pH meter.

The results of the experiment are presented in the graph below.

  1. What is the pH of 0.01M HCl?

(Show your full working or the section will not be graded.)

As HCL is a strong acid, it will completely dissociate in the solution.

pH = - log [H3O+]

pH = - log (0.01 M)

pH = - (- 2) = 2

  1. Using the graph provided (Figure 1) determine the buffering capacity of each solution, i.e. how many mL of 0.01M HCl can be added beforethe pH of the solution changes dramatically (above 1 pH change);

Solution A – When we add 0.01M HCL up to 4 ml then there is a change in pH. It means the buffer capacity of solution A is lesser than the solution C which means when we add more than 4 ml 0.01 M HCL in the solution, then there will be a change in pH.

Solution B - When we add 0.01M HCL up to 7 ml then there is a change in pH. It means the solution B has a higher buffering capacity and it can resist change for a longer time.

Solution C - When we add slight 0.01M HCL then will be a drastic change in pH. It means the buffering capacity of solution C is very low. A slight addition of 0.01 M HCL can leads to change in the pH of the solution.

  1. You are told that the test solutions (A, B or C) are either pure water, diluted carbonic acid/bicarbonate buffer or concentrated carbonic acid/bicarbonate buffer.

Using the results from this experiment (Figure 1) predict the name of the unknown test solution that was added to each of the blood samples.

(Justify your answer by relating your prediction to the shape of the graph and the pH of the blood after aliquots of HCl were added or this section will not be graded.)

In your answer, you will need to refer to the products and/or reactants of the carbonic acid/bicarbonate buffer reaction, as outlined below:

CO2 + H2O ↔ H2CO3 ↔ H+ + HCO3‾


Blood sample with solution A

Solution A is concentrated carbonic acid/bicarbonate buffer as compared to solution B. So, less H+ ions are available to interact with conjugate bases. Therefore, after adding 4 ml of 0.01 M HCL, the pH starts changing. So, the solution has lower buffering capacity as compared to solution B.

Blood sample with solution B

Solution B is diluted carbonic acid/bicarbonate buffer as compared to solution A. If the solution is diluted then there must be more water in it. The conjugate base present in the water can interact and consumed H+ ions in the solution. It creates a gradual change in pH instead of drastic change in curve. So, this solution has high buffering capacity. In simple words, higher the H+ ions, higher will be its buffering capacity.

Experiment Methodology

Blood sample with solution C

         Solution C underwent drastic change in the pH upon adding even a slight 0.01M  HCL in the solution. It has a very poor buffering capacity.

  1. d) Using the reaction given for the carbonic acid/bicarbonate blood buffering system describe why the pH of solution B changed to 2.7 after 8mL of HCl were added.

HINT: Think about what has happened to the H+ ions as they have been added to the solution.

The conjugate base present in the water can interact and consumed H+ ions in the solution. It creates a gradual change in pH instead of drastic change in curve. So, this solution has high buffering capacity. In simple words, higher the H+ ions, higher will be its buffering capacity. After adding 8ml of HCL, there is a shortage of conjugate base, so H+ ions concentration is increased which results in change in pH.  

  1. Predict what the pH of the blood sample with solution C would be if after adding 10mL of HCl we added 10 mL of 0.01M NaOH (sodium hydroxide).

(Justify your answer or this section will not be graded)

The pH of the blood sample will remain the same as there is equal concentration of acid present as compared to concentration of the base. So, neutralization reaction takes place, hence resist change in the pH.

  1. Human Genetics and Patterns of Inheritance.

(This question covers content from Practical 10)

Incomplete Dominance

There are two versions for hair texture – “curly” and “straight”. However, many people have hair texture that appears to be a combination of the two forms and this phenotype is referred to as being “wavy”.

If a woman with straight hair has children with a man who has curly hair, then what is the chance that their children will have wavy hair?

NOTE: For full marks you should show all your workings and complete the Punnet square to help justify your answer.

  1. What is the genotype for the mother?

Let straight be the dominant trait and expressed as S

Let curly be a recessive trait and expressed as s

So, the genotype of the mother is SS

  1. What is the genotype for the father?

The genotype of the father is cc

Show your working for this question by completing the Punnet square below:

SS (Straight)

Ss(wavy)

Ss (wavy)

ss (curly)

 Phenotype: 1:2:1 

  1. Using the outcomes from your Punnet square, write the possible genotypes and their corresponding phenotypes for the offspring.

SS : The offspring has straight hairs

Ss : The offspring has wavy  hairs

Ss:  The offspring has curly hairs

There is 50% chance that the offspring has wavy hairs. The genotype will be Ss of the offspring.

The mother of a newborn baby has blood group B and her partner has blood group A. The baby when tested has blood group O. However, the alleged father voices his concerns about the blood group of the baby. In terms of blood grouping, he asks you if there is a chance that he is not the father of the newborn baby.

  1. What are all the possiblegenotypes of the mother and her partner (the alleged father)?

Experiment Results

(There may be more than one possible genotype in some cases.)

What is the genotype for the baby:

Genotype of the baby is ii

What is the genotype for the mother:

Genotype of the mother is IBi

Possible genotype(s) for the father:

Pssibilities :  IAIA  or  IAi

Complete the two Punnet squares below to establish if the partner has a chance of being the father or not.

If genotype of father is IAIA

He is not the father of the baby.

IBIA

IAi

IAIB

IAi

 If genotype of the father is IAi

He is the father of the baby.

IBIA

IAi

IBi

ii

The phenotypic ratio is 1:1:1:1. It has 25% of chance.

  1. Is there a chance he is not the father? Explain your response.

If the genotype of the father is IAIA then there will be possibility that he is not the father of the baby. As no combination is forming i.e. ii. The genotype of the baby is ii which cannot be formed by combination of the genotype of this person and the mother. ii is a recessive trait and denotes O blood group

Red-green colour blindness is an X-linked recessive disorder in humans. Your friend is the son of a normal vision father. His mother also has normal vision but his maternal grandfather was colour-blind.

  1. Determine all the possible genotypes of the mentioned family members and complete the Punnet square to determine the inheritance pattern of colour blindness in your friend’s family.

NOTE: For full marks you must show all possible genotypes of all the mentioned family members and complete the Punnet square to help justify your answer.

         Genotype for your friend’s father:

         XY – NORMAL

Genotype for your friend’s mother:

XX’ – Mother is a carrier of this trait.

Genotype of the maternal grandfather:

X’Y – colour blind

Show your working for this question by completing the Punnet square below:

XX

XX’

XY

X’Y

Using your completed Punnet square determine what is the chance that yourfriend is colour blind?

There is only 25% chance that the son is having colour blindness.

The pattern of inheritance for a trait can be described as dominant or recessive. The genes can be on sex chromosomes (X or Y chromosomes) and they are described as X-linked or Y-linked traits. If the genes are not located on the sex chromosomes they are located on the autosomal chromosomes.

A pedigree chart can be used to map out a particular trait of interest and link the gene(s) to family members.

  1. Study the pedigree and explain if the trait could be described as a Y-linked trait?

(Justify your answer or this section will not be graded, you can use family members in the pedigree to help explain your outcome)

If the first generation father has a Y linked trait then it must be forwarded to each of the male offspring in next generation. So, every male gets affected due to that. We can notice that the trait is going from 1st to 5th and then 12th.

  1. Study the pedigree and explain if the trait could be described as an X-linked recessive trait?

(Justify your answer or this section will not be graded, it may help if you use family members in the pedigree to help explain your outcome)

If the trait is X- linked recessive then it should be only expressed in males and females are the carrier of that trait.  In 2nd generation, 3rd gets affected and in 3rd generation, there will be 50% chance as the 2nd generation female is crossed with a normal male. So, 5th and 6th offspring will be affected.

  1. If the trait was described as autosomalrecessive what is the genotype for the individual at I 2 positions on the family pedigree?

(You can show your working in a punnet square to help justify your answer if needed)

I 2 (Indicated by *) Female will be carrier of this trait.

So, genotype of female will be XX* and of male will be X*Y

XX*

X*X*

XY

X*Y

So, there is 25% chance, the male is affected in 3rd generation and 25% chance that the female is also get affected in 3rd generation. And 25% chance each of male and female to be normal.

References: 

Eroschenko, V. (2008). DiFiore's Atlas of Histology with Functional Correlations (11 ed.). USA: Lippincott Williams and Wilkins.

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