HMB265H1 Lecture Notes - Lecture 8: Synaptonemal Complex, Gametogenesis, Synapsis

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sapphiremink668

16 Oct 2016

School

UTSG

Department

Human Biology

Course

HMB265H1

Professor

Maria Papaconstantinou

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HMB265H1 Full Course Notes

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Genes on the same chromosome may not independently assort. Syntenic: two genes located on the same chromosome. Genetic linkage: genes that travel together more often than expected. Recombination: separation of genes on the same chromosome. Male progeny received x genes from mother, thus they have the same phenotype. Contrarily, female progeny are dihybrids (dominant from both parents) Dihybrid females should thus make four different combinations of genes on x chromosome: w y+, w+ y, w+ y+, and w y; should occur with equal frequency, in a ratio of 1:1:1:1. Morgan predicted that, because the two genes are on the same chromosome, they are linked (genes on different chromosomes are not linked and assort independently). All female f1 progeny were w+ y+ / w y dihybrids. Similarly, parental combinations total more than 50%, while recombinant allelic types totaled less than 50%. Genes behave as if connected to each other.

Related Questions

1.) In tomato plants, spines (s) are recessive to nospines (S), and white fruit color (w) isrecessive to red fruit color (W). The genes for spines andfruit color are located on the same chromosome; findings frommapping experiments indicate that they are 13.2 m.u. apart. Atomato plant having spines and white fruit color is crossed with aplant that is homozygous for no spines and red fruit color. The F1are crossed with plants that have spines and white fruit color.Which of the following phenotypes and proportions are expected inthe progeny of this cross? Check all that apply.

a) Red, no spines = 43.4%; white, spines = 43.4%

b) White, no spines = 43.4%; red, spines = 43.4%

c) Red, no spines = 6.6%; red, spines = 6.6%

d) White, no spines = 6.6%; white, spines = 6.6%

e) White, no spines = 6.6%; red, spines = 6.6%

f) Red, no spines = 43.4%; red, spines = 43.4%

g) Red, no spines = 6.6%; white, spines = 6.6%

h) White, no spines = 43.4%; white, spines = 43.4%

2.) In Drosophila, the map positions of genes are givenin map units numbering from one end of a chromosome to the other.The X chromosome of Drosophila is 66 m.u. long.The X-linked gene for body color-with two alleles,y ⁺for gray body and y foryellow body-resides at one end of the chromosome at map position0.0. A nearby locus controlling bristle form, withf ⁺for normal bristles and ffor forked bristles is located at map position 1.5. A thirdX-linked gene for eye color, with alleles w⁺for red eye and w for white eye, islocated at map position 56.7. Each gene resides on the Xchromosome, and at each locus the wild-type allele is dominant overthe mutant allele. Do you expect any of these gene pair(s) toassort independently?

a) Yes, y and w are each expected to assortindependently of f

b) Yes, all the genes are expected to assort independently

c) Yes, y and f are each expected to assortindependently of w

d) No, there are no genes expected to assortindependently

rubywallaby524

1. The eye color of wild-type Drosophila flies is red. Differentmutations in a single or in multiple genes lead to flies with whiteeyes. Mutations in the w gene on the X chromosome lead to whiteeyes instead of the normal red. You have isolated both a white-eyedmutation (designated W-1) that gives a dominant phenotype, and awhite-eyed mutation (designated w-2) that gives a recessivephenotype.

(a) A white-eyed male from the W-1 line is crossed to awild-type female. What color eyes will the female progeny from thiscross have?

(b) What color eyes will the male progeny from the cross in part(a) have?

(c) One of the female progeny from the cross in part (a) ismated to a white-eyed male from the w- 2 line. What fraction of thewhite-eyed progeny from this cross will be female?

2. The body color of Drosophila adult flies is regulated bymultiple genes, including the autosomal ebony gene. Ebony body (e)is recessive to the wild type (yellowish) body (e+). A second gene,which is X-linked, controls bristle shape. Normal bristle shape(f+) is dominant to the mutant forked bristles (f). A female withforked bristles that is heterozygous for the body color gene ismated with an ebony body male (his bristles are normal).

(a) What are the genotypes of the parents?

(b) Using a Punnett square, determine the expected proportionsof each possible progeny phenotype. Since there is a sex-linkedgene here, be sure to include consider sex as part of thephenotype.

3. Describe in your own words what a balancer chromosome is andwhy it is important in fly genetics (you will have to do someresearch for this one!)

mauvepug893

True-breeding brown-eyed, short-winged female flies are crossed to true-breeding wild-type male flies with red eyes and long wings. The F1 progeny consists of wild-type females and short-winged males. An F2 is generated by interbreeding of the F1. F2 males and females have the following ratios: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the mode of inheritance for both the eye color and wing length genes. Using clear genetic symbols, state the genotypes and describe the phenotypes of the P, F1 and F2 progeny. For the F2, include a table of genotypes of gametes of both parents and progeny genotypes and corresponding phenotypes. Would you predict that these genes are on the same chromosome?

orchidcaribou84

HMB265H1 Lecture Notes - Lecture 8: Synaptonemal Complex, Gametogenesis, Synapsis

HMB265H1 Full Course Notes

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