you would like to obtain additional data to make a firm diagnosis for two patien
ID: 176536 • Letter: Y
Question
you would like to obtain additional data to make a firm diagnosis for two patients so that you can come up with some possible treatments for your patients. for each patient answer questions about what additional data would be useful and why. be specific!!
1. patient pedigree constructed from her family history: what types of info will the pedigree for this patient give you?
2. DNA sequences of specific genes:
a - which specific genes will you sequence and why? (assume that as a control, you also sequence non-affected relatives of the patient)
b - if there are mutations in the coding region of the genes you sequence, what will that tell you?
c - if there are mutations in the non-coding regions, what will that tell you?
3. full genome sequence of patient: what potentially useful info will this give you?
4. human cancer database: what potentially useful info will this give you?
These are data that I have and conclusions that I have for each patient so far.
test
average
A
B
glucose
70-100 mg/dL
35 mg/dL
45 mg/dL
lactic acid
5-10 mmol/dL
20 mmol/dL
15 mmol/dL
blood pH
7.3-7.4
7.2
7.25
SpO2
95-100%
95%
98%
pyruvate
35-50 mg/dL
28 mg/dL
30 mg/dL
fructose-2-6-bisphosphate
10-20 mg/dL
15 mg/dL
16 mg/dL
-> both patient A and B are hypoglycemic because of low glucose. so anaerobic break down of glucose to generate energy in muscles provide higher lactate level with low blood pH. the increased lactate concentration will decrease pyruvate and fructose-2-6-bisphosphate.
enzyme
kinetic parameter
average
A
B
glucose transporter-1 (GLUT-1)
Vmax(f)
Km glucose
0.01-0.02
0.09-0.10
0.08
0.09
0.02
0.10
hexokinase (HK)
Vmax (f)
Km glucose
0.05-0.07
0.10-0.30
0.50
0.18
0.09
0.12
phosphofructose kinase (PFK)
Vmax (f)
Km F6P
Km ATP
0.02-0.05
0.50-0.75
0.45-0.65
0.27
0.40
0.55
0.05
0.55
0.55
triose phosphate isomerare (TPI)
Vmax (f)
Km DHAP
0.50-0.80
1.0-3.0
0.58
1.9
0.70
2.5
glyceraldehyde 3-phosphate dehydrogenase (GAPDH)
Vmax (f)
Km GAP
Km NAD+
0.90-1.20
0.20-0.50
0.09-0.20
1.00
0.29
0.10
1.10
0.25
0.09
pyruvate kinase (PK)
Vmax (f)
Km PEP
3.0-5.0
0.30-0.60
13.8
0.40
5.0
0.3
lactate dehydrogenase
Vmax (f)
Km Pyr
8.0-12.0
0.20-0.50
21.8
0.45
13.0
0.20
-> patient A: Km of GLUT-1 is comparable but consistently rate of reaction is fast. has low affinity for glucose as HK has higher Km, but it makes up with much high Vmax. glycolysis occuring at smooth rate (GAPDH). pyruvate gets reduced to lactate under anaerobic condition by LDH action (LDH Vmax is very high).
-> patient B: everything is mostly normal, but the reactions are much slower compared to patient A and normal average data. so the body may not be able to cope up with glucose demand if B moves a lot or exercise a lot.
enzyme
regulator (activator or inhibitor)
average
A
B
hexokinase (HK)
G6P
0.01-0.03
0.02
0.02
phosphofructose kinase (PFK)
ATP
F2, 6BP
1.50-3.00
0.01-0.03
1.75
0.02
8.00
0.001
pyruvate kinase (PK)
ATP
2.00-4.50
2.05
2.50
-> patient A: every criteria falls under the normal healthy average range.
-> patient B: high PFK, may be due to high rate of glycolysis in a cancer cells
enzyme
relative abundance
average
A
B
GLUT-1
mRNA
protein
500-1000
1.0-2.0
2000
5
550
1.5
hexokinase (HK)
mRNA
protein
200-500
0.5-2.0
1400
3.5
220
0.4
phosphofructose kinase (PFK)
mRNA
protein
300-750
2.0-5.0
6000
20.0
550
2.5
triose phosphate isomerase (TPI)
mRNA
protein
50-75
0.5-0.7
55
0.45
60
0.55
glyceraldehyde 3-phosphate dehy. (GAPDH)
mRNA
protein
550-1000
1.0-2.0
520
0.9
600
1.1
pyruvate kinase (PK)
mRNA
protein
250-550
2.0-5.0
1500
12
300
2.4
lactate dehydrogenase (LDH)
mRNA
protein
50-75
0.1-0.3
200
0.6
60
0.1
-> patient A: high lucose uptake because GLUT-1 is high. HK high because glucose is most important substrate of HK and patient has high glucose uptake. high PFK leads to faster metabolism which is not good. high LDH means blood flow deficiency, cerebrovascular accident such as stroke, heart attack or some type of cancer. high PK means problems with red blood cells.
-> patient B: low glucose uptake cmpared to patient A, but falls into normal range. HKf alls into normal because glucose uptake is normal. PFK is also nomral, so matabolism should be normal.
test
average
A
B
glucose
70-100 mg/dL
35 mg/dL
45 mg/dL
lactic acid
5-10 mmol/dL
20 mmol/dL
15 mmol/dL
blood pH
7.3-7.4
7.2
7.25
SpO2
95-100%
95%
98%
pyruvate
35-50 mg/dL
28 mg/dL
30 mg/dL
fructose-2-6-bisphosphate
10-20 mg/dL
15 mg/dL
16 mg/dL
Explanation / Answer
1.
The pedigree of the patients A and B predicts the transfer of genes responsible for the synthesis of hexokinase protein, phosphofructokinase enzyme, triose phosphate isomerase enzyme, glyceraldehyde-3-phosphate dehydrogenase enzyme, pyruvate kinase enzyme, and lactate dehydrogenase enzyme from one generation to the next. The presence of mutations in the concerned genes, inheritance of mutations across the generations can be observed from the pedigree. The hypoglycemic condition and lactic acid, pyruvate and fructose-2,6 bis-phosphate levels indicate that the glucose breakdown is occurring as anerobic. Oxygen specificity is also in the normal range. These levels can be monitored in the pedigree all along the inheritance.
2.
a)
The genes of hexokinase, phospho-fructo kinase, pyruvate kinase and lactate dehydrogenase have to be sequenced to check for any changes. The activity of phosphofructo kinase in the presence of its regulators ATP, 6BP and F2 is studied.
b)
If there are mutations in the coding regions of the above genes that were sequenced, it means that the protein function is altered causing discrepancy in the glycolysis process. The changes in the PFK affecting metabolism, PK affecting the red blood cells and HK affecting the glucose uptake can be studied in patient A.
c)
If there are mutations in the non-coding regions, they might not affect the expression of the gene or might even affect the expression of the gene, if the mutation is present in the regulatory regions.
3.
The gene sequences of the specific genes of study in the disease categories can be compared with the wild type gene sequences of patient A and B present in their full genomes. The comparison will confirm the nature of gene sequences in their diseased genes.
4.
Human cancer database will give the types of genes that can lead to cancer. The database also gives us the clue about the LDH levels in the case of a cancer patient. This data can be used to conclude whether patient A has cancer or not.
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