| Literature DB >> 25452487 |
H E Chrispeels1, M L Klosterman2, J B Martin1, S R Lundy1, J M Watkins1, C L Gibson1, G K Muday3.
Abstract
This study tests the hypothesis that undergraduates who peer teach genetics will have greater understanding of genetic and molecular biology concepts as a result of their teaching experiences. Undergraduates enrolled in a non-majors biology course participated in a service-learning program in which they led middle school (MS) or high school (HS) students through a case study curriculum to discover the cause of a green tomato variant. The curriculum explored plant reproduction and genetic principles, highlighting variation in heirloom tomato fruits to reinforce the concept of the genetic basis of phenotypic variation. HS students were taught additional activities related to mole-cular biology techniques not included in the MS curriculum. We measured undergraduates' learning outcomes using pre/postteaching content assessments and the course final exam. Undergraduates showed significant gains in understanding of topics related to the curriculum they taught, compared with other course content, on both types of assessments. Undergraduates who taught HS students scored higher on questions specific to the HS curriculum compared with undergraduates who taught MS students, despite identical lecture content, on both types of assessments. These results indicate the positive effect of service-learning peer-teaching experiences on undergraduates' content knowledge, even for non-science major students.Entities:
Mesh:
Year: 2014 PMID: 25452487 PMCID: PMC4255351 DOI: 10.1187/cbe.14-01-0007
Source DB: PubMed Journal: CBE Life Sci Educ ISSN: 1931-7913 Impact factor: 3.325
Student demographic data for 2012 and 2013 cohorts
| 2012 | 2013 | |||
|---|---|---|---|---|
| Year | MS undergrads | HS undergrads | MS undergrads | HS undergrads |
| Number of students | 23 | 41 | 23 | 31 |
| Gender | 43.5% M | 34.1% M | 20.8% M | 22.6% M |
| 56.5% F | 65.9% F | 79.2% F | 77.4% F | |
| Class | 34.8% Freshmen | 41.5% Freshmen | 56.5% Freshmen | 64.5% Freshmen |
| 56.5% Sophomores | 51.2% Sophomores | 43.5% Sophomores | 29.0% Sophomores | |
| 8.7% Juniors | 7.3% Juniors | 0% Juniors | 3.2% Juniors | |
| 0% Seniors | 0% Seniors | 0% Seniors | 3.2% Seniors | |
Figure 1.Comparison of mean (± SE) scores of MS undergraduates (n = 19) vs. HS undergraduates (n = 38) from pre- and posttests of SLP content knowledge from 2012. (A) Pretest. (B) Posttest. Asterisks (*) indicate significant difference between groups as determined by a Mann-Whitney U-test followed by false-discovery rate control. Hash marks (#) indicate significant difference between pre- and posttest score within the group, as determined by an exact Wilcoxon signed-rank test followed by false-discovery rate control. W, Z, and p values for all comparisons are reported in Table S2.
Examples of answers from one MS undergraduate and one HS undergraduate to pre- and posttest questions from 2013a
| MS undergraduate | HS undergraduate | |||
|---|---|---|---|---|
| Question | Pretest answer | Posttest answer | Pretest answer | Posttest answer |
| 1A. Below are listed the events in fertilization and fruit production. List the numbers of the events in the correct order. | 3-6-2-5-7-4-8-1 (4) | 3-6-2-5-7-8-4-1 (3) | 3-6-2-5-7-8-1-4 (3) | 3-6-2-7-5-4-8-1 (3) |
| 1B. If a plant cannot produce pollen, how would that affect its ability to produce fruit? Describe and explain. | “Plants are not asexual and a plant that cannot produce pollen cannot experience the genetic variation of a sexual organism. I suppose it could still produce fruit, but I’m not entirely sure.” (0) | “It would still be able to produce fruit, it would just have to receive pollen from another plant.” (1) | “I suppose it wouldn't be able to create fruit because fruit is the fertilized ‘egg’ ready to be planted after pollenation [ | “It would not be able to produce fruit because it lacks the pollen that fertilizes the plant ‘egg’ and produces fruit.” (2) |
| 2A. Define the terms “genotype” and “phenotype.” | “I know I have studied this but it has been a long time, I can't recall what they mean.” (0) | “A genotype describes the set of alleles found in an organism's genes. A phenotype describes the trait that shows.” (3) | “Genotype is a genetic trait while phenotypes are physical traits.” (1) | “Genotype—an organisms [ |
| 2B. In two or three sentences, explain why two different genotypes (representing the same gene) might give the same phenotype. | No answer (0) | “It is possible for one phenotype to show through several genotypes. If the organism is heterozygous, they have a recessive allele that does not affect the genotype, or they are homozygous and have two of the dominant allele.” (1) | “Because genotypes can sometime be one passive and one dominant so even if a certain genotype is present it may not show through a phenotype therefore simply producing the phenotype of the dominant genotype only.” (0) | “Sometimes genotypes can have heterozygous and homozygous traits. These traits appear different in the genotype because of the presence or lack of one recessive allele. These look the same because there is a dominant allele present which always shows in the phenotype.” (2) |
| 3A. Define the terms “homozygous” and “heterozygous.” | “Homozygous—has all one type of allele. Heterozygous—has different alleles.” (2) | “Homozygous = two of the same allele in the genotype. Heterozygous = two different alleles in the genotype.” (4) | “Homozygous—one gene. Heterozygous—two genes.” (0) | “Homozygous—a trait with either both alleles being dominant or both recessive. Heterozygous—trait that has both a recessive and a dominant trait.” (3) |
| 3B. Brown eye color is dominant over blue eye color. Imagine that a heterozygous brown-eyed woman and a blue-eyed man have children. Set up a Punnett square to demonstrate the fraction of brown-eyed children from this couple. Use the letters B and/or b to designate alleles. | No answer (0) | |||
| 4A. List three essential components that are required to carry out a PCR experiment. | No answer (0) | “Heat resistant DNA polymerase, heat, small pieces of DNA.” (2) | No answer (0) | “Primer, polymerase, DNA sample.” (2) |
| 4B. The polymerase chain reaction is dependent on multiple cycles of polymerization. Explain why multiple cycles are needed for the technique to work as it should. | No answer (0) | “The goal is to exponentially increase the rate at which the DNA is synthesized. If there aren't multiple cycles, this result will not be achieved.” (0) | No answer (0) | “Because with each cycle DNA grows exponentially, the point of PCR is to have a lot of DNA to work with therefore we need a few cycles to get adequate DNA.” (4) |
| 5A. What is the objective of the technique of DNA gel electrophoresis? In other words, what does the technique do? | No answer (0) | “It identifies various small pieces of DNA by how fast they move through the gel.” (1) | No answer (0) | “It spreads out pieces of DNA allowing one to compare sizes of DNA pieces by comparing how far down the pieces moved in the gel.” (3) |
| 5B. You are running a DNA electrophoresis gel and you discover that you accidentally reversed the positive and negative electrodes. What will happen to the DNA that you loaded at the top of your gel? Explain. | No answer (0) | “They won't move at all.” (2) | No answer (0) | “It will most likely stay at the top, DNA is negatively charged therefore having the positive charge at the top would make the DNA stay at the top.” (4) |
| 6A. Some diseases are caused by mutations. What in the cell is mutated? Briefly describe three types of mutations that might occur. | “The chromosomes are damaged.” (0) | “The DNA of the cell is mutated. A deletion mutation completely removes one base. An insertion mutation copies an extra base.” (1) | “The DNA, sometimes an extra chromosome can be created, the ‘genetic instructions’ can simply be copied wrong.” (1) | “The DNA is mutated, deletion—loss of a nucleotide, substitution—nucleotide being replaced by another, and addition—nucleotide is added.” (4) |
| 6B. On the gel electrophoresis image below are samples from a healthy person and someone with a mutation that causes cystic fibrosis (CF). What kind of mutation would explain this difference? Explain your answer. | No answer (0) | No answer (0) | No answer (0) | “A deletion mutation, because the DNA sample of the CF patient sits lower in the gel compared to a healthy person we can assume the DNA to be smaller and therefore contain less DNA, the product of a deletion mutation.” (4) |
aEach question is worth a total of four points. The score earned for each question is shown in parentheses after the answer.
Figure 2.(A) Comparison of mean (± SE) pre- and posttest scores for all undergraduates (n = 52) from 2013. Asterisks (*) indicate significant difference between pre- and posttest scores for each question as determined by an exact Wilcoxon signed-rank test followed by false-discovery rate control. (B) Comparison of mean scores (± SE) from posttests of MS undergraduates (n = 24) and HS undergraduates (n = 28). Asterisks (*) indicate significant difference between groups as determined by a Mann-Whitney U-test followed by false-discovery rate control. W, Z, and p values for all comparisons are reported in Table S2.
Figure 3.Comparison of mean (± SE) scores of MS undergraduates (n = 24) vs. HS undergraduates (n = 28) from pre- and posttests of SLP content knowledge from 2013. (A) Pretest. (B) Posttest. Asterisks (*) indicate significant difference between MS undergraduates and HS undergraduates as determined by a Mann-Whitney U-test followed by false-discovery rate control. Hash marks (#) indicate significant difference between pre- and posttest scores, within the group, as determined by an exact Wilcoxon signed-rank test followed by false-discovery rate control. W, Z, and p values for all comparisons are reported in Table S2.
Figure 4.Comparison of final exam scores from 2012. (A) Percentage of correct answers to individual questions pertaining to material taught by undergraduate students. Questions 1–4, genetics questions common to both MS and HS curriculum; questions 5–6, questions specific to HS curriculum. Asterisks (*) indicate significant difference between MS undergraduates and HS undergraduates as determined by a Mann-Whitney U-test followed by false-discovery rate control. (B) Mean scores (± SE) represented as percentages of the four common curriculum questions, the two HS-specific questions, and the other exam questions. Asterisks (*) indicate significant difference between MS and HS for that question set, as determined by a Mann-Whitney U-test followed by false-discovery rate control. Hash marks (#) indicate significant difference from other exam questions, within the group, as determined by an exact Wilcoxon signed-rank test followed by false-discovery rate control. (C) Mean scores (± SE) represented as percentages of the 70 lecture-only questions and the four lab and lecture questions. MS, n = 23; HS, n = 37. W, Z, and p values for all comparisons are reported in Table S2.