Ls Investigation Lactose Tolerance Answer Key

Author qwiket
7 min read

Ls investigation lactose tolerance answer key provides a clear roadmap for students and educators who want to understand how genetic variation influences the ability to digest lactose in adulthood. This guide walks through the purpose of the investigation, the experimental steps, the interpretation of results, and the correct answers that align with the learning objectives. By following the detailed explanation below, learners can confidently connect genotype to phenotype, appreciate the evolutionary significance of lactase persistence, and apply scientific reasoning to real‑world data.


Introduction to Lactose Tolerance and the LS Investigation

Lactose tolerance, more accurately termed lactase persistence, is the continued expression of the lactase enzyme beyond infancy, allowing individuals to break down lactose—the sugar found in milk—into glucose and galactose. Populations with a long history of dairy farming show higher frequencies of lactase‑persistent alleles, illustrating a classic example of gene‑culture coevolution.

The ls investigation lactose tolerance answer key accompanies a hands‑on or virtual laboratory activity where students examine DNA samples, simulate polymerase chain reaction (PCR) amplification, and analyze gel electrophoresis patterns to determine whether specific genetic markers associated with lactase persistence are present. The answer key not only lists the expected outcomes but also explains the reasoning behind each result, reinforcing concepts of Mendelian inheritance, allele frequency, and phenotypic expression.


Overview of the Investigation

Component Description
Objective Determine the lactase persistence genotype of unknown DNA samples and correlate it with predicted lactose tolerance phenotypes.
Materials DNA extracts (from simulated individuals), primers targeting the -13910 C/T regulatory region, Taq polymerase, dNTPs, buffer, agarose gel, DNA ladder, loading dye, electrophoresis chamber, UV transilluminator.
Procedure Summary 1. Set up PCR reactions for each sample.<br>2. Run thermal cycler (denaturation 95 °C 30 s, annealing 58 °C 30 s, extension 72 °C 45 s for 35 cycles).<br>3. Load PCR products onto a 2 % agarose gel.<br>4. Electrophorese at 100 V for 30 min.<br>5. Visualize bands and record genotypes.
Expected Outcomes Two possible band patterns: <br>• Homozygous persistent (TT) – single band at ~200 bp.<br>• Heterozygous (CT) – two bands (~200 bp and ~180 bp).<br>• Homozygous non‑persistent (CC) – single band at ~180 bp.

The answer key aligns each observed band pattern with the correct genotype and then predicts the phenotype (lactose tolerant vs. intolerant) based on established dominance relationships.


Step‑by‑Step Walkthrough with Answer Key

Below is a detailed narrative of what students should observe at each stage, followed by the correct answers that the instructor can use for grading.

1. Preparing the PCR Master Mix

  • Correct action: Combine 10 µL 2× PCR buffer, 1 µL forward primer (10 µM), 1 µL reverse primer (10 µM), 0.5 µL Taq polymerase (5 U/µL), 1 µL dNTP mix (10 mM each), 62.5 µL nuclease‑free water, and 5 µL template DNA per reaction.
  • Answer key point: Master mix volume = 20 µL; final primer concentration = 0.5 µM each.

2. Thermal Cycling Conditions

  • Denaturation: 95 °C for 30 s (to separate strands).
  • Annealing: 58 °C for 30 s (primer binding to the -13910 region).
  • Extension: 72 °C for 45 s (polymerase synthesizes new strand).
  • Cycles: 35 repetitions, followed by a final extension of 72 °C for 5 min. - Answer key point: The annealing temperature is critical; too low yields nonspecific bands, too high reduces yield.

3. Gel Electrophoresis Setup

  • Agarose concentration: 2 % (suitable for separating 150‑250 bp fragments).
  • Running buffer: 1× TAE.
  • Voltage: 100 V (constant).
  • Run time: Approximately 30 min, until the dye front reaches ~¾ of the gel length.
  • Answer key point: Include a 100 bp DNA ladder to estimate fragment sizes accurately.

4. Interpreting the Gel

Sample Observed Bands (bp) Genotype (Answer Key) Predicted Phenotype
A 200 only TT (homozygous persistent) Lactose tolerant
B 200 & 180 CT (heterozygous) Lactose tolerant (dominant T allele)
C 180 only CC (homozygous non‑persistent) Lactose intolerant
D 200 only TT Lactose tolerant
E 180 only CC Lactose intolerant
F 200 & 180 CT Lactose tolerant

Explanation: The -13910 T allele creates a restriction site that, after PCR, yields a slightly longer amplicon (~200 bp). The C allele lacks this site, producing a shorter fragment (~180 bp). Heterozygotes display both fragments because each allele is amplified independently.

5. Calculating Allele Frequencies (Optional Extension)

If the class data includes 30 samples with the following counts: TT = 12, CT = 14, CC = 4, then:

  • Frequency of T allele = (2×TT + CT) / (2×total) = (2×12 + 14) / 60 = 38/60 ≈ 0.63.
  • Frequency of C allele = 1 – 0.63 = 0.37.

Answer key point: These frequencies can be compared to known population data (e.g., ~0.80 in Northern Europeans, ~0.10 in East Asian groups) to discuss evolutionary pressures.


Scientific Explanation Behind the Results

The persistence of lactase expression is primarily regulated by a single nucleotide polymorphism (SNP) located upstream of the LCT gene. The -13910 C/T variant influences a transcription factor binding site; the T allele enhances promoter activity in intestinal cells after weaning.

  • Molecular mechanism: In individuals carrying at least one T allele

-Molecular mechanism: In individuals carrying at least one T allele, the -13910 T substitution creates a novel binding site for the transcription factor Oct‑1 (and, in some contexts, for HNF4α). This site remains accessible in intestinal epithelial cells after the neonatal period, allowing Oct‑1 to recruit co‑activators and maintain an open chromatin configuration at the LCT promoter. Consequently, RNA polymerase II continues to transcribe lactase mRNA into adulthood, yielding the lactase‑persistent phenotype.

In contrast, the ancestral -13910 C allele lacks this Oct‑1 site. After weaning, repressive complexes (including Mi‑2/NuRD and histone deacetylases) bind the promoter, leading to chromatin condensation and transcriptional silencing of LCT. Homozygous CC individuals therefore exhibit the typical lactase‑non‑persistent pattern, with enzyme activity declining sharply after infancy. Heterozygotes (CT) possess one allele capable of sustaining transcription; the residual activity from the T‑containing allele is usually sufficient to confer lactose tolerance, reflecting the autosomal‑dominant nature of the trait.

  • Evolutionary context: The high frequency of the T allele in pastoralist populations (e.g., Northern Europeans, certain African groups) is interpreted as a classic case of gene‑culture coevolution. Domestication of dairy animals created a nutritional advantage for individuals able to digest lactose beyond infancy, driving positive selection on the -13910 T variant. Conversely, populations with limited historical dairy reliance retain low T allele frequencies, as seen in many East Asian and Indigenous American groups.

  • Assay considerations: The PCR‑RFLP approach described relies on the differential restriction pattern generated by the -13910 C/T SNP. Key points to ensure reliable interpretation include:

    1. Complete digestion – insufficient enzyme or suboptimal buffer can leave undigested product, mimicking a heterozygote pattern. Running a positive control (known TT DNA) alongside each batch helps verify complete cleavage.
    2. Allele‑specific amplification bias – preferential amplification of one allele can skew band intensities; using a high‑fidelity polymerase and limiting cycle number (as specified) minimizes this effect.
    3. Fragment size resolution – a 2 % agarose gel provides adequate separation for the 20 bp difference, but running a low‑range ladder (e.g., 50‑bp increments) improves accuracy, especially when samples are degraded.
    4. Alternative confirmation – for research or clinical settings, Sanger sequencing or allele‑specific qPCR can validate the genotype, particularly when ambiguous bands appear.
  • Educational take‑aways: This exercise links a molecular genotype to a readily observable phenotype, reinforcing concepts of transcriptional regulation, Mendelian inheritance, and natural selection. By calculating allele frequencies from class data, students also practice basic population‑genetics calculations and appreciate how genetic variation correlates with cultural practices such as dairying.


Conclusion
The -13910 C/T polymorphism upstream of the LCT gene serves as a molecular switch that determines whether lactase expression persists after weaning. The T allele creates a binding site for transcription factors like Oct‑1, sustaining an active promoter state and yielding lactose tolerance, whereas the C allele permits the repressive chromatin changes that lead to lactase decline and intolerance. Through PCR amplification, restriction digestion, and agarose‑gel electrophoresis, students can directly visualize the genotype‑phenotype relationship, calculate allele frequencies, and discuss the evolutionary forces that have shaped lactase persistence across human populations. This integrated approach not only solidifies core genetics and molecular‑biology principles but also highlights the interplay between genetics, diet, and cultural evolution.

More to Read

Latest Posts

You Might Like

Related Posts

Thank you for reading about Ls Investigation Lactose Tolerance Answer Key. We hope the information has been useful. Feel free to contact us if you have any questions. See you next time — don't forget to bookmark!
⌂ Back to Home