Lesson 5: How to make a linkage map based on phenotype of offspring

Introduction:

The approximate distance of two genes that are closely located on the same chromosome can be determined by observing the phenotype of the offspring and calculating how the results differ from expected Mendelian cross. This lesson walks students through those calculations and shows how to make a linkage map of three traits on the same chromosome. It uses actual traits found in American Chestnut trees to teach this concept.

Learning outcomes:

Students will be able to calculate map unit distance between two linked genes on a chromosome. They will also be able to determine if the outcome of a genetic cross follows the expected outcome of a Punnett square.

Curriculum alignment:

Biology Goal 3.03 – Interpret and predict patterns of inheritance including independent assortment and Punnett squares.

Classroom time required:

45 minutes

Materials needed:

“How to make a linkage map” handout for each student, 1 calculator and 1 ruler for each student, overhead projector. If completing the optional activity you will also need 3 beads, 1 pipe cleaner per student. Pony beads work well with pipe cleaners.

Technology resources:

1 calculator per student, overhead projector

Pre-activities:

This lesson is intended to be completed after students have learned how to complete and analyze dihybrid crosses. Copy 1 handout for each student, gather other materials (calculators, pipe cleaners and beads) prior to lesson.

Activity:

1. Review a simple dihybrid cross problem with students to remind them that they are observing 2 traits at a time. Also, discuss the outcome of phenotypes of the offspring represented by the cross. Ask students what it would mean if the actual offspring ratios came out differently than the Punnett square said they should. Allow students time to give various responses- even if incorrect.
2. Give each student a copy of the “How to make a linkage map” handout. Have 1 student read the information at the top of the first page. Explain to students that sometimes the ratio of phenotypes in the offspring does not come out as expected, meaning that the traits did not assort independently. This means that the genes for the trait are most likely located on the same chromosome and by doing a few simple calculations, we can figure out where on the chromosome the gene is located.
3. Give students time to complete the dihybrid cross on the first page. Have a student complete the cross on the overhead . Check this for correctness and make sure all students have this correct answer. Have other students from the class help to determine the ratio of phenotypes seen in the offspring. (This first example is a 1:1:1:1 ratio.) 25% of 2000 equals 500 so that each phenotypic combination would be represented by 500 offspring.
4. Have students turn to the second page to see that the actual mating produced very different numbers of offspring than expected. The traits must be fairly close together on the chromosome and crossing over did not separate these two genes from each other. Have a student define “recombination” from their previous notes and/or textbook.
5. Explain to students that the offspring that have had their chromosomes recombined are always represented by the groups with the smaller numbers. In this first example that would be the Hairy/ Fertile with 150 offspring and the Non-hairy/ Fertile with 150 offspring.
6. Work through the calculation with the students as described under “How do you calculate linkage map distance.”
7. The line representing the chromosome has the gene for “hairiness” already located. To represent the gene for fertility, use a ruler to measure a distance 15 mm away from the hairiness gene. The direction, left or right, does not matter at this point. Have students complete this step with you.
8. Direct students to complete the next calculation in Example 2 on their own. Help individual students as needed. Once everyone has had time to complete this (about 3 minutes) make sure all have the correct answer. Then, have students continue on to Example 3. Again, make sure students have the correct answer after they have had adequate time to solve.
9. At this point students have 3 distances and 3 traits. On the second representation of the chromosome the order and direction matter. Two of the distances should add up to the third distance (15+10=25). Therefore, the traits have to be put in order to make this numerical sentence true. Give students time (2-3 minutes) to try to figure out the order. If no one has discovered the order at the end of this time, show them on the overhead. Discuss the remaining 2 questions in this final example as a class. Xbudding 10 units Xhairiness 15 units Xfertility
10. Students are now ready to complete the final problem on their own. Give them about 10 minutes for the calculations and drawing/ pipe cleaner representation.

Assessment:

Check the final problem and drawing/ pipe cleaner representation for correctness.
Xwidth 7 units Xcolor 13 units Xlength

Modifications:

Since most of this lesson is done with the teacher’s direct instruction, the only suggested modification is additional time for those with computational difficulties.

Critical vocabulary:

1. Phenotype: the observable properties of an organism
2. Crossing over: an interchange of genes or segments between homologous chromosomes
3. Punnett square: a square used in genetics to calculate the frequencies of the different genotypes and phenotypes among the offspring of a cross
4. Dihybrid cross: a Punnett square which represents the mating of two parents and observing 2 genes for each parent
5. Recombination: the processes of crossing-over and independent assortment of new combinations of genes in offspring that did not occur in the parents
6. Recombinant: An organism or cell in which genetic recombination has taken place
7. Linkage map: a representation of the relative positions of genes in a chromosome
8. Stipule: either of a pair of appendages at the base of the petiole in many plants