BIOL 121 Lecture Notes - Lecture 8: Phenylalanine, Allele, Zygosity

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lavenderwhale632

24 Jan 2017

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BIOL 121 Full Course Notes

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Double check the learning objectives with what we go over in lecture. Eye colour and body colour are linked (same chromosome). They are also on the sex chromosome (x linked) Parent 2: male with white eyes and tan body. Xr = red eyes xr = white eyes. Xt = tan body xt = yellow body. Observed: a novel phenotypes red/yellow male (xrt y) & white/tan make (xrt y) based on . Relative position of genes on a chromosome. First person (alfred henry sturtevant) to come up with the idea of how to map alleles on the same chromosome. He did so by looking at the rate of recombination. When the recessive and dominant alleles mix: e. g red (dominant) flowers are crossed with while (recessive) flowers. When heterozygotes have an intermediate phenotype compared to the previous generation. Heterozygote has phenotype that is characteristic of both the homozygous dominant and homozygous recessive.

Related Questions

Which of these statements is incorrect?

Syntenic genes are located on the same chromosome.

Independent assortment results in recombinant chromosomes.

You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome.

Linked genes are always syntenic.

What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?

0.49 cM

4.9 cM

49 cM

490 cM

What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)

Human error: they should have been more careful about theirexperimental setup.

Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits.

Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes.

Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean.

When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?

9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio.

Genotypes of the offspring can be determined based on theirphenotype.

If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation.

B and C

Why do we map genes?

To understand how genes interact with each other

Comparative genomics analysis

To determine the genotype of an organism

All of the above

coralox210

I just need help with these two questions:

The Principle of Independent Assortment would suggest a 9:3:3:1 phenotype ratio (dom/dom; rec/dom; dom/rec; rec/rec) in a cross, involving parents heterozygous at each of two genes. When the same experiment is done using different traits including flower color and pollen shape, crimson flowers are dominant (red is recessive) and long pollen is dominant (round is recessive). A cross involving heterozygous parents shows the following results:

4831 (= 69.5%) with crimson flowers and long pollen,

390 (= 5.6%) with crimson flowers and round pollen,

393 (= 5.6%) with red flowers and long pollen,

1338 (= 19.3%) with red flowers and round pollen.

a. (5 points) Since there is no 9:3:3:1 ratio, would you conclude that these genes are on the same chromosome or on different chromosomes?

b. (5 points) Explain how the crimson/round and red/long flowers might have arisen at a low frequency?

Thank you!

jadedog477

GENETICS : You perform a dihybrid cross between two dogs that are each BbFf. Where not specified otherwise, assume black fur B is completely dominant to white fur b and furry F is completely dominant to hairless f

1. What genotypes at one locus will be epistatic to what genotypes at the other locus, if hair color and hair presence were the only measured phenotypes? (you may find it helpful to write out the genotypes and the phenotypes each would produce, and remember that epistasis is involved in this question).

What phenotypic ratios would you expect in the following scenarios (assume, unless otherwise noted, that the epistasis in part 1 still applies)?

2. Black B is completely dominant to albino b, furry F is completely dominant to hairless f

3. Black displays incomplete dominance to albino , and furry is codominant with hairless

4. Animals that are heterozygous at either locus are black and furry, overriding the other locus' geneotype

5. The black allele B displays epistasis when present at even one copy that masks the effects of the hairless f allele

6. It takes two black B alleles to mask the effects of the hairless f allele

7. Both the black alleles B and furry alleles F are lethal in their homozygous dominant states

8. Both the albino alleles b and hairless alleles f are lethal in their homozygous recessive states

9. Female dogs die when they have homozygous recessive or homozygous dominant states for either locus (dogs use XY inheritance for sex like we do). Note sex of doss in your phenotype ratios.

10. The black allele B although completely dominant, is the only 70% penetrant in heterozygotes and is 90% penetrant in homozygotes for the B allele

11. The hairless allele f is dominant in males and recessive in females

BIOL 121 Lecture Notes - Lecture 8: Phenylalanine, Allele, Zygosity

BIOL 121 Full Course Notes

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