Cancer-causing NF1 mutations in mice
- Dashboard
- Cancer-causing NF1 mutations in mice
Most Recent Score Report
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Previous Score Reports
| Questions Correct | Questions Answered | Time Spent | Status | Attempt Date | |
|---|---|---|---|---|---|
| -- | -- | -- | -- | -- |
Cancer-causing NF1 mutations in mice
Array
(
[passage] => WP_Post Object
(
[ID] => 553949
[post_author] => 12815
[post_date] => 2024-12-23 10:34:22
[post_date_gmt] => 2024-12-23 15:34:22
[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Neurofibromin is a protein encoded by the NF1 gene on chromosome 17. In humans, it contains multiple functional domains, including a Ras GTPase-activating protein-related domain (GRD), a cysteine-serine-rich domain (CSRD), and a central region with multiple binding sites for other proteins. Under normal circumstances, neurofibromin acts as a tumor suppressor in nerve cells, modulating Ras signaling by altering the ratio of active Ras-GTP to inactive Ras-GDP, which results in further alterations to the ERK signaling pathway.
Neurofibromatosis type 1 is caused by a mutation in the NF1 gene that results in skin hyperpigmentation and numerous neurofibromas, a type of benign cutaneous nerve sheath tumor. In rare cases a neurofibroma may turn into a malignant tumor, called a malignant peripheral nerve sheath tumor (MPNST), in a process known as malignant transformation. Researchers studied the phenotypic expression of neurofibromatosis in mice with NF1 mutations by performing the following experiments.
Experiment 1
Researchers created a large population of mice by initially mating a healthy female mouse with a homozygous diseased male mouse. The initial set of offspring is known as the first filial (F1) generation. The following generations mated at random until the population reached 10,000 mice. Table 1 shows data stratified between mice with and without the NF1 mutation. The variables analyzed include the number of mice with neurofibromas and the number of mice with MPNSTs.
Table 1 Number of cases of neurofibromas and MPNSTs in mice with or without an NF1 mutation
Experiment 2
Researchers created novel mouse models harboring a common nonsense mutation (exon 18 c.2041C>T; p.Arg681*) in addition to mouse models harboring a null allele due to exon 4 deletion (Δ4). To assess neurofibromin function in these models, researchers isolated and expanded mouse embryonic fibroblasts (MEFs) from embryos carrying different allelic mutations, then performed Western blot analyses and assayed neurofibromin and phosphorylated ERK (pERK) levels. The results are shown in Figures 1 and 2.
Figure 1 Western blot analyses for wildtype mice and mice harboring mutations
Figure 2 Expression of neurofibromin and pERK in wildtype mice and mice harboring mutations
Figures 1 and 2 from Kairong Li et al. June 2016. Disease Models and Mechanisms 9(7):dmm.025783
[post_title] => Cancer-causing NF1 mutations in mice
[post_excerpt] =>
[post_status] => publish
[comment_status] => closed
[ping_status] => closed
[post_password] =>
[post_name] => cancer-causing-nf1-mutations-in-mice
[to_ping] =>
[pinged] =>
[post_modified] => 2024-12-23 10:34:22
[post_modified_gmt] => 2024-12-23 15:34:22
[post_content_filtered] =>
[post_parent] => 0
[guid] => https://medlifemastery.com/?post_type=passage&p=553949
[menu_order] => 0
[post_type] => passage
[post_mime_type] =>
[comment_count] => 0
[filter] => raw
)
[questions] => Array
(
[0] => Array
(
[quiz_unique_key] => 602779517
[question] => If two mice from the F1 generation of Experiment 1 were crossed, what percentage of their offspring would have neurofibromatosis?
[value] => Array
(
[answer] => 4
[description] => Reason for Correct Answer:
Table 1 shows that all mice harboring an NF1 mutation have neurofibromatosis, which means that this mutation is dominant for neurofibromatosis.
The passage states that the F1 generation was made by crossing a healthy female mouse with a homozygous diseased male mouse (the P generation).
Therefore, you can express the genotypes of the P generation as nn x NN. The genotypes of the F1 generation would therefore all be nN:
Use another Punnett square to determine the results of a cross between two F1 generation mice:
As indicated, 75% would either be homozygous for the disease allele or heterozygous, so 75% would have the disease.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 3.75%
)
[1] => Array
(
[each_answer] => B. 25%
)
[2] => Array
(
[each_answer] => C. 50%
)
[3] => Array
(
[each_answer] => D. 75%
)
)
)
[1] => Array
(
[quiz_unique_key] => 1403770772
[question] => Within the final population of mice shown in Table 1, what is the heterozygous genotype frequency?
[value] => Array
(
[answer] => 4
[description] => Reason for Correct Answer:
Use the Hardy–Weinberg equations to answer this. Remember that:
p² + 2pq + q² = 1 and p + q = 1, where p and q are the allele frequencies in the population, p² and q² are the homozygous frequencies and 2pq is the frequency of heterozygous individuals.
The table shows that 3,600 mice have no NF1 mutation and 6,400 mice have at least one mutation (i.e. they are heterozygous OR homozygous for NF1 mutation). Therefore, you only know q² and can use that to determine q and p and then determine 2pq.
q² = 3,600/10,000 = 0.36
q = √0.36 = 0.6
If q = 0.6, then p = 1 − 0.6 = 0.4.
Therefore, the frequency of heterozygous individuals = 2pq = 2(0.6)(0.4) = 2(0.24) = 0.48, or 48%.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 12%
)
[1] => Array
(
[each_answer] => B. 16%
)
[2] => Array
(
[each_answer] => C. 36%
)
[3] => Array
(
[each_answer] => D. 48%
)
)
)
[2] => Array
(
[quiz_unique_key] => 1403770772
[question] => If a mouse with an NF1 mutation is chosen at random from the final population, what is the chance that the mouse will be homozygous dominant?
[value] => Array
(
[answer] => 3
[description] => Reason for Correct Answer:
Again, here is a reminder of the Hardy–Weinberg equations:
p + q = 1
P² + 2pq + q² = 1
The question stem states that it’s already known that the mouse has an NF1 mutation. This means that the mouse is from the population represented by p² + 2pq only.
We’ve already determined in the last question that q = √0.36 = 0.6 and p = 1 − 0.6 = 0.4. And we calculated the heterozygous genotype frequency as: 2pq = 2(0.6)(0.4) = 0.48.
The homozygous dominant genotype frequency is: p² = (0.4)² = 0.16
The total frequency of mice with a mutation is: p² + 2pq = 0.16 + 0.48 = 0.64
So, the chance that this mouse is homozygous is: 0.16/0.64 = 16/64 = ¼
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 1/12
)
[1] => Array
(
[each_answer] => B. ⅛
)
[2] => Array
(
[each_answer] => C. ¼
)
[3] => Array
(
[each_answer] => D. ⅓
)
)
)
[3] => Array
(
[quiz_unique_key] => 1403770772
[question] => Which statement would help explain the data shown in Table 1?
[value] => Array
(
[answer] => 2
[description] => Reason for Correct Answer:
The table shows that all mice with any NF1 mutation (which must include heterozygous mice) had neurofibromas, but not all have MPNSTs, which are the malignant tumor form, according to the passage.
The table shows that some mice without the NF1 mutation have MPNSTs, so you can eliminate Choice C.
In the previous calculations, we saw that the normal allele, which we called q, has a frequency of √0.36 = 0.6, which means that the disease allele has a frequency of 0.4. This came from the fact that 3,600 out of 10,000 mice did not have the mutation at all.
If the disease allele has a frequency of 0.4, you would expect the number of mice homozygous for the disease allele to be (0.4)² = 0.16. This would lead to 0.16(10,000) = 1,600 mice having MPNSTs if it were inherited in a purely autosomal recessive fashion. However, more than 1,600 have MPNST. So, you can eliminate Choice A.
The table also shows that 6,400 mice carry at least one copy of the allele. Not even close to all of these mice get MPNSTs, so the allele does not seem to exhibit complete dominance. So, you can eliminate Choice D.
The only answer that explains the fact that more than 1,600 but less than 6,400 mice get the disease is the fact that, in heterozygous individuals, the normal allele may become mutated, leading to malignant cancer generation.
This refers to the concept of a “two-hit hypothesis” in tumor development associated with the NF1 gene, in which the development of malignant tumors (cancers) typically requires additional “hits” or mutations in the other, normal copy of the NF1 gene. This could be either a spontaneous mutation occurring in the other copy or a different inherited mutation in the second allele. This “two-hit” model is a commonly observed phenomenon in tumor suppressor genes where both copies of the gene need to be mutated or inactivated to eliminate their tumor-suppressing effects and allow cancerous growth.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. MPNSTs develop only in individuals with two mutated NF1 alleles and thus exhibit an autosomal recessive inheritance pattern in reference to the NF1 mutation studied.
)
[1] => Array
(
[each_answer] => B. Malignant tumor development in heterozygous individuals requires additional mutation of the other NF1 allele.
)
[2] => Array
(
[each_answer] => C. Neurofibromas do not progress to MPNSTs in the absence of the NF1 mutation.
)
[3] => Array
(
[each_answer] => D. The mutant NF1 allele exhibits complete dominance in terms of malignant tumor development.
)
)
)
[4] => Array
(
[quiz_unique_key] => 1403770772
[question] => The results of Experiment 2, in conjunction with information provided by the passage, are most consistent with which conclusion?
[value] => Array
(
[answer] => 4
[description] => Reason for Correct Answer:
The passage indicates that neurofibromin has a GTPase-activating domain. Given that GTPases break down GTP, it is unlikely that neurofibromin would promote the conversion Ras-GDP to Ras-GTP and more likely that it promotes the reverse process.
The graphs in Figure 2 show that the NF1 mutations result in decreased neurofibromin and increased pERK expression compared with wildtype cells.
The NF1ᴬʳᵍ⁶⁸¹* mutation seems to have the same effect as the knockout (“null” gene) – in regards to neurofibromin expression.
Remember that the passage states that NF1 is a tumor suppressor; its mutation appears to promote the formation of both neurofibromas and their progression (malignant transformation) to MPNSTs per Table 1.
Because double mutations of NF1 lead to both MPNSTs and elevated levels of pERK, it is likely that pERK promotes malignant transformation to MPNSTs.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. NF1ᴬʳᵍ⁶⁸¹* retains more protein functionality than a complete knockout of the gene.
)
[1] => Array
(
[each_answer] => B. Neurofibromin activates a pathway that results in the phosphorylation of ERK.
)
[2] => Array
(
[each_answer] => C. Neurofibromin acts to convert Ras-GDP to Ras-GTP.
)
[3] => Array
(
[each_answer] => D. pERK promotes malignant transformation to MPNSTs.
)
)
)
[5] => Array
(
[quiz_unique_key] => 4010399236
[question] => Could the Western blots shown in Figure 1 be used to quantify the expression of neurofibromin and pERK proteins (e.g. to generate the data shown in Figure 2)?
[value] => Array
(
[answer] => 2
[description] => Reason for Correct Answer:
While Western blotting does visualize the presence or absence of proteins, quantification of band intensities is commonly performed to obtain quantitative or semi-quantitative data about protein expression levels.
UV-vis spectroscopy is not a method typically used to quantify Western blot results, especially for proteins like neurofibromin and pERK. It measures absorbance but is not suitable for quantifying Western blot band intensities.
Western blot results can usually be quantified by comparing the band intensities of the proteins of interest to a control band of known protein concentration.
In this case, because the results are quantified as % expression and the wildtype expression levels are normalized to 100%, the values can be determined by comparing the intensity of the neurofibromin and pERK bands in cells harboring mutations to the intensity of the corresponding bands of the wildtype mice.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. Yes, Western blot results can be quantified through methods like UV-vis spectroscopy, which measures the absorbance of specific wavelengths of light, to determine neurofibromin and pERK quantities accurately.
)
[1] => Array
(
[each_answer] => B. Yes, Western blot results could be quantified using image analysis techniques to compare the neurofibromin and pERK and band intensities in samples derived from mutated and wildtype mice.
)
[2] => Array
(
[each_answer] => C. No, Western blot results cannot be quantified as the technique only provides qualitative information about the presence or absence of neurofibromin and pERK proteins in the sample.
)
[3] => Array
(
[each_answer] => D. No, in the case of the Western blot shown in Figure 1, there is no control band to provide a basis for comparison, so another technique must have been used to generate the graphical data.
)
)
)
)
[total_question] => 6
[correct_answers] => Array
(
[553949|1] => D
[553949|2] => D
[553949|3] => C
[553949|4] => B
[553949|5] => D
[553949|6] => B
)
[hide_display_feedback_settings] =>
[hide_solutions] =>
)
Figure 1
Figure 2
