Question
Which processes are involved in the development of cancer?
Mutations occur in oncogenes.
Oncogenes prevent cancer.
Oncogenes affect cell cycle regulatory proteins.
I and II only
I and III only
II and III only
I, II and III
Answer/Explanation
Ans: B
The correct answer is:
**I and III only**
Explanation:
– **I. Mutations occur in oncogenes**: This is correct. Mutations in oncogenes can lead to their activation, promoting uncontrolled cell growth and contributing to cancer development.
– **II. Oncogenes prevent cancer**: This is incorrect. Oncogenes, when mutated or overexpressed, promote cancer rather than prevent it. Tumor suppressor genes, on the other hand, help prevent cancer by regulating cell growth.
– **III. Oncogenes affect cell cycle regulatory proteins**: This is correct. Oncogenes can alter the function of cell cycle regulatory proteins, leading to dysregulated cell proliferation, a hallmark of cancer.
Thus, the correct combination is **I and III only**.
In a person who is heterozygous for sickle-cell anemia, where is the mutation found?
A. In every gamete produced
B. Only in gametes carrying an X chromosome
C. In all brain cells
D. In blood plasma
Answer/Explanation
To determine where the mutation for sickle-cell anemia is found in a person who is heterozygous for the condition, we need to understand the genetic basis of sickle-cell anemia, what it means to be heterozygous, and how mutations are distributed in the body’s cells and gametes. Let’s break this down step by step.
### Step 1: Understand Sickle-Cell Anemia and Its Genetic Basis
Sickle-cell anemia is a genetic disorder caused by a mutation in the **HBB gene**, which encodes the beta-globin subunit of hemoglobin, the protein in red blood cells that carries oxygen. The mutation is a single nucleotide substitution (A to T) in the HBB gene, leading to the replacement of glutamic acid with valine at the sixth position of the beta-globin protein. This results in:
– **Normal hemoglobin (HbA)**: Produced by the wild-type allele.
– **Sickle hemoglobin (HbS)**: Produced by the mutant allele, causing red blood cells to sickle under low oxygen conditions, leading to symptoms like pain, anemia, and increased risk of infections.
Sickle-cell anemia is an **autosomal recessive disorder**, meaning:
– A person with two mutant alleles (HbS/HbS) has sickle-cell anemia.
– A person with one normal allele and one mutant allele (HbA/HbS) is heterozygous and is considered a **carrier** (also called having sickle-cell trait). They typically do not exhibit severe symptoms but can pass the mutant allele to their offspring.
The question specifies a person who is **heterozygous** for sickle-cell anemia, meaning they have the genotype HbA/HbS—one normal allele (HbA) and one mutant allele (HbS).
### Step 2: Where Is the Mutation Located in a Heterozygous Person?
The mutation is a change in the DNA sequence of the HBB gene. Since DNA is the genetic material in all nucleated cells of the body, we need to consider where this mutation exists in a heterozygous individual:
– **Genomic DNA**: The mutation is present in the DNA of the person’s cells. In a heterozygous individual, one copy of the HBB gene (on one chromosome 11, as HBB is on chromosome 11 in humans) carries the mutation, while the other copy is normal.
– **Somatic Cells**: These are the body’s non-reproductive cells (e.g., brain cells, liver cells, etc.). In a heterozygous person, **all somatic cells** that contain a nucleus will have one copy of the mutant allele because the mutation is part of the person’s genome, inherited from one parent.
– **Gametes**: These are the reproductive cells (sperm and eggs). During meiosis, the two alleles (HbA and HbS) segregate according to Mendel’s law of segregation, so each gamete will carry only one allele—either HbA or HbS. Therefore, **only half of the gametes** will carry the mutant allele (HbS).
– **Blood Plasma**: Blood plasma is the liquid component of blood and does not contain DNA (it’s acellular, meaning it lacks cells with nuclei). The mutation is in the DNA, so it cannot be found in blood plasma. However, the protein product (HbS) may be present in red blood cells, which are in the blood, but the question asks about the mutation itself, not the protein.
### Step 3: Evaluate the Options
Let’s analyze each option based on the distribution of the mutation in a heterozygous person:
– **A. In every gamete produced**:
– Gametes are produced through meiosis, where the chromosomes segregate. A heterozygous person (HbA/HbS) has two alleles for the HBB gene. During meiosis, these alleles segregate, so:
– 50% of the gametes will carry the HbA allele (normal).
– 50% of the gametes will carry the HbS allele (mutant).
– Therefore, the mutation is **not in every gamete**—only in half of them. This option is **incorrect**.
– **B. Only in gametes carrying an X chromosome**:
– The HBB gene is located on **chromosome 11**, which is an autosome (not a sex chromosome). The mutation for sickle-cell anemia is not linked to the X chromosome.
– In humans, females have two X chromosomes, and males have one X and one Y chromosome. Gametes carrying an X chromosome would be eggs (in females) or sperm carrying an X (in males). However, the HBB gene’s location on chromosome 11 means the mutation’s presence in a gamete is independent of whether the gamete carries an X chromosome.
– As established, 50% of gametes will carry the HbS allele, regardless of whether they carry an X or Y chromosome. This option is **incorrect** because the mutation is not specific to X-carrying gametes.
– **C. In all brain cells**:
– Brain cells are somatic cells, meaning they are non-reproductive cells of the body. In a heterozygous person, **all somatic cells** contain the same genomic DNA, which includes one copy of the HBB gene with the mutation (HbS) and one normal copy (HbA).
– Since brain cells are nucleated (they contain a nucleus with DNA), the mutation will be present in the DNA of all brain cells in a heterozygous individual.
– This option is **correct** because the mutation is indeed found in the DNA of all brain cells.
– **D. In blood plasma**:
– Blood plasma is the liquid portion of blood and does not contain DNA. It contains proteins, ions, and other molecules, but not genomic DNA where the mutation resides.
– The mutation is in the DNA of nucleated cells (like white blood cells in the blood), but the question specifies “blood plasma,” which is acellular. The protein product (HbS) may be present in red blood cells, which are in the blood, but red blood cells lack a nucleus in humans, and the question is about the mutation (in DNA), not the protein.
– This option is **incorrect** because the mutation cannot be found in blood plasma.
### Step 4: Double-Check and Consider Misconceptions
– **Gametes (Options A and B)**: The key here is understanding Mendelian segregation. A heterozygous person does not pass the mutation to every gamete, ruling out option A. Option B’s mention of the X chromosome is irrelevant because sickle-cell anemia is an autosomal condition, not sex-linked.
– **Brain Cells (Option C)**: Somatic cells, including brain cells, inherit the same genome as the zygote from which the person developed. In a heterozygous individual, this means all brain cells will have one copy of the mutant allele in their DNA.
– **Blood Plasma (Option D)**: It’s important to distinguish between the mutation (a change in DNA) and the protein product (HbS). The mutation itself is in the DNA, not in the plasma.
### Conclusion
In a person who is heterozygous for sickle-cell anemia (HbA/HbS), the mutation in the HBB gene is present in the DNA of all nucleated somatic cells, including brain cells. It is not in every gamete (only 50% of gametes carry the mutation), it is not specific to X-carrying gametes (since HBB is on chromosome 11), and it is not in blood plasma (which lacks DNA).
**Final Answer: C**
One type of gene mutation involves a base substitution.
What are the consequences of the base substitutions in the two new sequences of DNA?
A. Both are mutations that would result in different polypeptides.
B. Sequence 2 would result in a changed polypeptide but sequence 1 would not.
C. All three DNA sequences would translate into the same polypeptide.
D. Only the original DNA and sequence 2 would translate into the same polypeptide.
Markscheme
D
To determine the consequences of the base substitutions in the two new DNA sequences, we need to compare the original DNA sequence with the two mutated sequences, transcribe them into mRNA, translate the mRNA into amino acids, and assess whether the resulting polypeptides differ. The problem provides the original and mutated DNA sequences, along with mRNA codons for valine and glutamic acid, which suggests that these amino acids are relevant to the polypeptide being synthesized.
### Step 1: Understand the Given DNA Sequences
The sequences are:
– **Original DNA sequence**: GAC TGA GGA CTC TCT CAG A
– **Mutated sequence 1**: GAC TGA GGA **CAT** CTC TCA GA
– **Mutated sequence 2**: GAC TGA GGA **CTT** CTC TCA GA
The mutated bases are bolded:
– In sequence 1, the 10th base (C in the original) is changed to **A**.
– In sequence 2, the 10th base (C in the original) is changed to **T**.
The sequences are written without spaces, but since DNA is read in codons (groups of three nucleotides) during translation, we’ll break them into codons. The sequences appear to be the **coding (sense) strand** of DNA, because the problem provides mRNA codons directly (e.g., GUU for valine), which suggests that the DNA sequence given is the one that matches the mRNA sequence (with T replaced by U). In molecular biology, the sense strand has the same sequence as the mRNA (except T becomes U), while the template strand is complementary and used during transcription. Since the problem doesn’t specify which strand is the template, we’ll assume the given sequence is the **sense strand**, as this is common in such problems when mRNA codons are provided for comparison.
### Step 2: Break the DNA Sequences into Codons
The original DNA sequence is GAC TGA GGA CTC TCT CAG A. Let’s break it into codons (groups of three nucleotides):
– GAC TGA GGA CTC TCT CAG A
– Codon 1: GAC
– Codon 2: TGA
– Codon 3: GGA
– Codon 4: CTC
– Codon 5: TCT
– Codon 6: CAG
– Extra base: A (incomplete codon, likely not translated in this context)
The sequence has 19 bases, which gives 6 complete codons (18 bases) and 1 extra base (A). In a real gene, translation would continue until a stop codon, but for this problem, we’ll focus on the complete codons. The mutation occurs at the 10th base, which is the first base of the 4th codon (CTC).
– **Original codon 4**: CTC
– **Sequence 1 codon 4**: **CAT** (C to A substitution)
– **Sequence 2 codon 4**: **CTT** (C to T substitution)
The other codons remain unchanged across all sequences.
### Step 3: Transcribe the DNA to mRNA
Since the given DNA sequence is the sense strand, the mRNA sequence will be identical to the DNA sequence, except that **T** is replaced with **U**. Let’s transcribe each sequence:
– **Original DNA**: GAC TGA GGA CTC TCT CAG
– **mRNA**: GAC UGA GGA CUC UCU CAG
– Codons: GAC | UGA | GGA | CUC | UCU | CAG
– **Sequence 1 DNA**: GAC TGA GGA CAT TCT CAG
– **mRNA**: GAC UGA GGA CAU UCU CAG
– Codons: GAC | UGA | GGA | CAU | UCU | CAG
– **Sequence 2 DNA**: GAC TGA GGA CTT TCT CAG
– **mRNA**: GAC UGA GGA CUU UCU CAG
– Codons: GAC | UGA | GGA | CUU | UCU | CAG
### Step 4: Translate the mRNA Codons into Amino Acids
We need to translate each codon into its corresponding amino acid. The problem provides specific mRNA codons for valine and glutamic acid, which suggests these amino acids are relevant to the codon affected by the mutation (the 4th codon):
– **mRNA codons for valine**: GUU, GUC, GUA, GUG
– **mRNA codons for glutamic acid**: GAA, GAG
Let’s translate each codon for all sequences, focusing on the 4th codon where the mutation occurs, but we’ll also check the other codons to ensure we understand the full polypeptide.
#### Original Sequence mRNA: GAC UGA GGA CUC UCU CAG
– **GAC**: Aspartic acid (Asp)
– **UGA**: Stop codon (translation would normally stop here)
– **GGA**: Glycine (Gly)
– **CUC**: Leucine (Leu)
– **UCU**: Serine (Ser)
– **CAG**: Glutamine (Gln)
The presence of UGA (a stop codon) at the second position is unusual, as it would terminate translation early, resulting in a polypeptide of just one amino acid: **Asp**. However, in the context of this problem, the question implies that a polypeptide is formed, and the mutation in the 4th codon (CUC) is the focus. This suggests that UGA might be a typo or that the sequence is meant to be interpreted differently. Let’s reconsider the sequence interpretation.
### Step 5: Reassess the Sequence Interpretation
The UGA stop codon at the second position (codon 2) would halt translation, making the mutations in codon 4 irrelevant because translation wouldn’t proceed that far. This doesn’t align with the question’s focus on the consequences of mutations in codon 4. Let’s hypothesize that the sequence might be mislabeled or that the sense strand interpretation needs adjustment. If UGA were a coding codon (e.g., GGA, as in your previous question’s context), the sequence would make more sense. However, let’s try interpreting the given sequence as the **template strand** instead, since that would change the mRNA sequence and potentially avoid the stop codon issue.
If the given sequence is the **template strand**:
– **Original template**: GAC TGA GGA CTC TCT CAG
– Transcribe 3′ to 5′ (GAC TCT CTC GGA AGT CAG, reversed):
– G → C
– A → U
– C → G
– T → A
– C → G
– T → A
– C → G
– T → A
– C → G
– G → C
– G → C
– A → U
– A → U
– G → C
– T → A
– C → G
– A → U
– G → C
– **mRNA (5′ to 3′)**: CUG ACU UCC GAG AGA GUC
– **Codons**:
– CUG: Leucine (Leu)
– ACU: Threonine (Thr)
– UCC: Serine (Ser)
– GAG: Glutamic acid (Glu) (matches the provided codon for glutamic acid)
– AGA: Arginine (Arg)
– GUC: Valine (Val) (matches the provided codon for valine)
This interpretation avoids the stop codon issue and makes the sequence more consistent with a polypeptide being formed. The 4th codon (GAG, glutamic acid) is where the mutation occurs.
– **Sequence 1 template**: GAC TGA GGA CAT TCT CAG
– mRNA: CUG ACU UCC GUA AGA GUC
– 4th codon: GUA (Valine)
– **Sequence 2 template**: GAC TGA GGA CTT TCT CAG
– mRNA: CUG ACU UCC GAA AGA GUC
– 4th codon: GAA (Glutamic acid)
### Step 6: Compare the Polypeptides
– **Original**: Leu-Thr-Ser-**Glu**-Arg-Val
– **Sequence 1**: Leu-Thr-Ser-**Val**-Arg-Val
– The 4th amino acid changes from glutamic acid to valine due to the substitution (GAG to GUA).
– **Sequence 2**: Leu-Thr-Ser-**Glu**-Arg-Val
– The 4th amino acid remains glutamic acid (GAA, which also codes for glutamic acid, as per the provided codons).
### Step 7: Evaluate the Options
– **A. Both are mutations that would result in different polypeptides**:
– Sequence 1 results in a different polypeptide (Glu to Val).
– Sequence 2 does not (Glu to Glu, same amino acid).
– This option is **incorrect**.
– **B. Sequence 2 would result in a changed polypeptide but sequence 1 would not**:
– Sequence 1 changes the polypeptide (Glu to Val).
– Sequence 2 does not change the polypeptide (Glu to Glu).
– This option is **incorrect** (it’s the reverse of what we found).
– **C. All three DNA sequences would translate into the same polypeptide**:
– Sequence 1 results in a different polypeptide.
– This option is **incorrect**.
– **D. Only the original DNA and sequence 2 would translate into the same polypeptide**:
– Original and Sequence 2 both produce Leu-Thr-Ser-Glu-Arg-Val.
– Sequence 1 produces Leu-Thr-Ser-Val-Arg-Val, which is different.
– This option is **correct**.
### Conclusion
The base substitution in sequence 1 (C to A) changes the 4th codon from GAG (glutamic acid) to GUA (valine), resulting in a different polypeptide. The substitution in sequence 2 (C to T) changes the 4th codon from GAG to GAA, both of which code for glutamic acid, so the polypeptide remains the same as the original.
**Final Answer: D**