Linkage worksheet, part 1 PDF

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Genetic maps and linkage problem-solving worksheet, part 1 SP2021 Use the genetic map of the genome of Drosophila melanogaster to answer the following two questions: 1. Which gene would likely to be inherited with the gene of yellow body: The gene for white eyes or the gene for sepia eyes? White eyes

2. Tan body and curly wings or clot eyes and curly wings?

Clot eyes Curly wings

3. Two cats that are heterozygous for the fur color gene (Bb) and for the gene for tail length (Ss).  For fur color, brown (B) is dominant to white (b)  For tail length, short tail (S) is dominant to long tail (s) Assume that the genes for fur color and tail length are:  Right next to each other on a chromosome (so that NO recombination occurs between them) and that  the alleles are in the coupling (cis) conformation (that is, the two dominant alleles are on the same chromosome and the two recessive genes are on the other) What would be the genotypes and phenotypes of the kittens? (Hint- make a Punnett square or use the multiplication rule to figure out the genotypes of the kittens)

Genotypes: 25% SSBB 50% SsBb 25% ssbb Phenotypes: 75% Brown fur, short tail 25% White fur, long tail

4. Now, assume again that genes for tail length and fur color are right next to each other on a chromosome (so that NO recombination occurs between them) but now:  the alleles are in the repulsion (trans) conformation (that is, each chromosome has one dominant allele and one recessive allele: chromosome as shown below

Genotypes: 25% SSbb

50% SsBb 25% ssBB Phenotypes: 25% White fur, short tail 50% Brown fur, short tail 25% White fur, long tai

5. Two genes, a and b, are on the same chromosome. During the pairing of the tetrad during prophase, crossing over occurs between a and b. In respect to these two genes, how many different types of gametes would this individual have at the end of meiosis? 4

6. After meiosis, which of the following pairs of genes are MOST likely to still be together (that is, to NOT have had recombination occur between them)? a. yellow body & bar eyes, which are 57 m.u. apart b. curved wings & star eyes, which are 74 m.u. apart c. sepia eyes & ebony body which are 50 m.u. apart d. spineless bristles & sepia eyes, which are 32 m.u. apart

7. (Calculating percentage recombination from map units) Using the map below, answer the following question: How frequently does recombination occur between the gene for body (Gray or black) and the gene for wings (long or dumpy) during one round of meiosis? 48.5 m.u. - 13.0 m.u. = 35.5 m.u. Since m.u. = percentage recombination = 35.5%

The following questions are about the testcross to determine the distance between the S gene and the B gene.

8. Are the wild type of the alleles SsBb cat on the same chromosome (in coupling (cis) conformation) or on different chromosomes (in repulsion (trans) conformation)? In repulsion trans conformation Circle the correct notation for the S and B genes in the SsBb cat.

9. If the two genes are completely linked (no crossing over), and you looked at 100 kittens, how many of each of the following would you expect to see?

___0___ # of brown cats with short tails (of the genotype SsBb) ___50___ # of brown cats with long tails (of the genotype ssBb) ___50___ # of white cats with short tails (of the genotype Ssbb) ___0___ # of white cats with long tails (of the genotype ssbb)

10. When you do the test cross, you observe the following numbers of kittens: 6 brown cats with short tails (of the genotype SsBb) 44 brown cats with long tails (of the genotype ssBb) 44 white cats with short tails (of the genotype Ssbb) 6 white cats with long tails (of the genotype ssbb) What is the recombination frequency between the color gene and the tail length gene?

6 +6 6 +44 + 44 + 6

x 100% = 12%

Recombination = number of recombinant progeny × 100% frequency total number of progeny...