BIS 2B Study Guide - Midterm Guide: Meiosis, Outcrossing, Gene Flow

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13 Nov 2017

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BIS 2B Full Course Notes

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Lecture 15: mendelian genetics: laws, punnett squares, probabilities. Bis 2b midterm 2 study guide: law of segregation: two alleles of gene separate into different gametes. Lecture 23: predation (antagonistic relationships) dn/dt = rn(1-n/k) (logistic growth of prey) - anp (prey death by predator) dp/dt = canp (predator births) - dp (predator deaths) Prey equilibrium density n = d/ca a = searching capacity (potential to capture prey) c = potential to multiply (conversion efficiency for viruses, female egg load for insects: predator-prey dynamics. Increase in prey population allows increase in predator population. Increasing prey carrying capacity k does not change prey abundance but causes predator abundance to increase. R0 = pcd (average number of people individual infects during infectious period in mostly susceptible population) p = probability of transmission (infectiousness mx) D = duration of infectious period (survivorship of infection lx)

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Fill in the blank, each word can be used more than once.

word bank- crossing over, avoid predation, reproduce,independent assortment, phenotypes, frequencies, nucleotides,meiosis, protein, prophase, segregation, gene, allele, amino acids,fertilization, arrangement, chromosome.

A(n) _________ is one of multiple forms of the same gene.

These different forms of genes are in the same locus of a(n)_________, but they have slightly different sequences of________.

During ________ I of meiosis, genes of homologous chromosomesalign side-by-side.

Then in a process called _________, homologous chromosomesrandomly line up double-file at the center of the cell and thenseperate. Since alleles are randomly distributed, this processcreates variation among the gametes. After ________ with a gametefrom the other sex, variation among gametes will create variationamong the offspring.

Some offspring will have a more favorable combination of allelesthan others, and will therefore survive and ________ moresuccessfully, leading to more favorable alleles in the nextgeneration. However, sometimes conditions change, andonce-favorable phenotypes may becomes less favorable.

Scientists hope that they will soon be able to use biotechnologyto track the ________ of alleles over multiple generations, therebywatching evolution as it occurs.

taupewolf882

Ants are haplodiploid organisms. This means females are diploid (2n) and males are haploid (n). The gametes produced by both males and females are haploid (n). There is no homologous recombination in males during gamete formation. There is no sex chromosome.

In an ant species, it is known that wings formation (F) and setae density (D) are each controlled by a different, simple-Mendelian locus/gene. Individuals which are homozygous recessive for both genes have no wings (f) and no setae (d). Previous experiments also indicated that both loci/genes are on the same chromosome.

a. Design a test cross to map the genetic distance of the genes, by writing the genotype of the tester and its mate. (2 pts)

Genotype of tester:

Genotype of testerâs mate:

b.In order for the test cross to produce offspring with non-parental phenotypes, the tester has to be of a specific sex. What sex must the tester be? Briefly explain your choice in your own words. (Hint:what is a test cross?) (2 pts)

Sex of tester:

emeraldweasel173

1) Ants are haplodiploid organisms. This means females are diploid (2n) and males are haploid (n). The gametes produced by both males and females are haploid (n). There is no homologous recombination in males during gamete formation. There is no sex chromosome. In an ant species, it is known that wings formation (F) and setae density (D) are each controlled by a different, simple-Mendelian locus/gene. Individuals which are homozygous recessive for both genes have no wings (f) and no setae (d). Previous experiments also indicated that both loci/genes are on the same chromosome.

a. Design a test cross to map the genetic distance of the genes, by writing the genotype of the tester and its mate. (2 pts) Genotype of tester: Genotype of testerâs mate:

b. In order for the test cross to produce offspring with non-parental phenotypes, the tester has to be of a specific sex. What sex must the tester be? Briefly explain your choice in your own words. (Hint: what is a test cross?) (2 pts) Sex of tester:

c. 100 offspring were produced from a particular test cross. The phenotypic ratio of these offspring is observed to be the following:

Normal wings and setae density40

Wings and setae both absent 47

Normal wings but no setae 4

Normal setae density but no wings 9

Calculate and describe in words the genetic distance between the two loci/genes of interest. Remember to include the unit of measurement (6 pts).

carminecaribou726

BIS 2B Study Guide - Midterm Guide: Meiosis, Outcrossing, Gene Flow

BIS 2B Full Course Notes

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