Simulation Stella.docx

ASSIGNMENT 1: SIMULATION (STELLA)

SBP 60204 TECHNOLOGY AND INNOVATION IN TEACHING AND LEARNING BIOLOGY

EN. AZMI BIN IBRAHIM

PREPARED BY:

NOR RASHIDAH BINTI AHMED FUAD

M20181001082

Genes on separate chromosomes assort independently because of the separation of homologous chromosomes during Anaphase I of Meiosis I. So an individual that is heterozygous for two genes (Aa Bb) on two separate chromosomes will produce four different possible gametes which are AB, Ab, aB, and ab. However, genes located on the same chromosome will only assort independently if the two genes recombine during crossing over during Prophase I of Meiosis I.

Gene 1 in the simulation is at the end of chromosome 2 at genetic map unit 0.0. This gene in fruit flies is the gene for the antennae length and is near the telomere (end) of chromosome 2. This gene has two alleles short antennae (a), and long antennae (A). Therefore phenotype "A" is long antennae, and phenotype "a" is short antennae. A heterozygous fly (Aa) would have long antennae (phenotype A). Gene 2 in this simulation is closer to the Gene 1 at genetic map unit 6.1. This gene has two alleles long wing (b) and curly wing (B). Therefore phenotype "B" is curly wing, and phenotype "b" is long wing. A heterozygous fly (Bb) would have curly wing (phenotype B).

Based on the graph above we can predict that the offspring from the testcross between Aa Bb (long antennae, curly wing) with aa bb (short antennae, long wing) will produce four different phenotypes which are A and B phenotypes (long antennae, curly wing), a and b phenotypes (short antennae, long wing), A and b (long antennae, long wing) and a and B (short antennae, curly wing). The number of offspring that have same phenotype as parent is higher compare to the recombinant phenotype. This is because the distance between these two gene that located on the same chromosome is very close to each other at 6.1 map unit.

Gene 1 in the simulation is at the end of chromosome 2 at genetic map unit 0.0. This gene in fruit flies is the gene for the antennae length and is near the telomere (end) of chromosome 2. This gene has two alleles short antennae (a), and long antennae (A). Therefore phenotype "A" is long antennae, and phenotype "a" is short antennae. A heterozygous fly (Aa) would have long antennae (phenotype A). Gene 2 in this simulation is very far from Gene 1 at genetic map unit 67.0. This gene has two alleles vestigial wing (b) and long wing (B). Therefore

phenotype "B" is long wing, and phenotype "b" is vestigial wing. A heterozygous fly (Bb) would have long wing (phenotype B). Based on the graph above we can predict that the offspring from the testcross between Aa Bb (long antennae, long wing) with aa bb (short antennae, vestigial wing) will produce four different phenotypes which are A B, A b, a B and a b. This is because the distance between these two gene that located on the same chromosome is very far to each other which is 67.0 map unit. The number of offspring that have same phenotype as parent is nearly the same with the recombinant phenotype.

Gene 1 in the simulation is at the end of chromosome 2 at genetic map unit 0.0. This gene in fruit flies is the gene for the antennae length and is near the telomere (end) of chromosome 2. This gene has two alleles short antennae (a), and long antennae (A). Therefore phenotype "A" is long antennae, and phenotype "a" is short antennae. A heterozygous fly (Aa) would have long antennae (phenotype A). Gene 2 in this simulation is closer to the Gene 1 at genetic map unit 13.0. This gene has two alleles dumpy wing (b) and long wing (B). Therefore

phenotype "B" is long wing, and phenotype "b" is dumpy wing. A heterozygous fly (Bb) would have long wing (phenotype B).

Based on the graph above we can predict that the cross between heterozygous parents Aa Bb (long antennae, long wing) will produce four different phenotypes which are dominant phenotype A and B (long antennae, long wing), recessive phenotype a and b (short antennae, dumpy wing), A and b (long antennae dumpy wing) and a and B (short antennae, long wing) . The number of offspring that have phenotype same as parent is higher compare to the recombinant phenotypes. From the result we can conclude that the distance between these two gene that located on the same chromosome is very close to each other at 13.0 map unit.

Gene 1 in the simulation is at the end of chromosome 2 at genetic map unit 0.0. This gene in fruit flies is the gene for the antennae length and is near the telomere (end) of chromosome 2. This gene has two alleles short antennae (a), and long antennae (A). Therefore phenotype "A" is long antennae, and phenotype "a" is short antennae. A heterozygous fly (Aa)

would have long antennae (phenotype A). Gene 2 in this simulation is very far from Gene 1 at genetic map unit 54.5. This gene has two alleles purple eyes (b) and red eyes (B). Therefore phenotype "B" is red eyes, and phenotype "b" is purple eyes. A heterozygous fly (Bb) would have red eyes (phenotype B). Based on the graph above we can predict that the cross between heterozygous parents Aa Bb (long antennae, long wing) will produce four different phenotypes which are A B, A b, a B and a b. This is because the distance between these two gene that located on the same chromosome is very far to each other at 54.5 map unit. From the result we can conclude that the genes which located on the same chromosome will assort independently and have higher chance of recombination between two genes. These genes will recombine during crossing over at Prophase I of Meiosis I.