BI111 Lecture Notes - Lecture 1: Mendelian Inheritance, Panmixia, Genotype Frequency

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31 Mar 2017

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Higher in rna viruses than in dna viruses. Higher in mitochondrial dna than in genomic dna. Genetic code has both specificity and redundancy. Genetic code established very early in evolution of life and has remained virtually unchanged. L (dominant, short hair) l (recessive, long hair) (lower case l) **in order for a cat to be long hair, it must have 2 copies of the recessive gene homozygous)** **a cat can have short whether it has two copies of the dominant gene (homozygous) or one copy of the dominant and one copy of the recessive gene and still come out as short hair** Epistatic: where an allele of one gene hides or masks the visible output, or phenotype, of another gene (pattern) Pleiotropic: a single gene that controls more than one trait. The dense pigment gene, d codes for melanophilin, a protein involved in the transportation and deposition of pigment into a growing hair.

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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

Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.

	genotype	phenotype
1	AA BB CC DD (wild type)	normal vulva
2	aa BB CC DD	No vulva
3	AA bb CC DD	no vulva
4	AA BB cc DD	no vulva
5	AA BB CC dd	no vulva

You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:

a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.

b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.

c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.

d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.

To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.

	Genotype	Phenotype
6	A*A BB CC DD	multiple vulvas
7	AA B*B CC DD	multiple vulvas

To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).

	Genotype	Phenotype
8	A*A bb CC DD	multiple vulvas
9	A*A BB cc DD	multiple vulvas
10	A*A BB CC dd	no vulva
11	AA B*B cc DD	multiple vulvas

Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?

a. A----> B----->C----->D ---> Vulva

b. D----> B----->C----->A----> Vulva

c. B----> C----->A----->D----> Vulva

d. C----> B----->A----->D----> Vulva

e. C----> B----->D----->A----> Vulva

f. don't have enough data to make any conclusions

Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:

a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B

b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele

c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele

d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D

e. The A* allele enhances the homozygous recessive loss of function mutation in gene D

Q 1 â Manx cats are produced when a dominant allele is inherited for the Manx gene (M). Inheritance of two dominant alleles, however, results in death of the fetus. Coat color in cats is determined by two alleles with incomplete dominance. BB produces black colored cats, Bb produces grey cats, and bb produces white cats. A grey Manx cat is mated with a white Manx. What proportion of the living offspring will be white Manx kittens? (3 points)

Q 2 â In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. Weâll use the symbol B^b for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their proportions expected for the following cross: B⁺B⁺Ww male Ã B^bB^bWw female (2 points)

Q 3 â E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescent-white seedlings, and 260 yellow seedlings.

Give the genotypes for the green, virescent-white, and yellow seedlings

Explain how color is determined in these seedlings (which genotype gives which phenotype). (3 points)

Q 4 â You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However the L allele is only 60% penetrant in both homozygotes and heterozygotes. In a population of 212 giraffes in an African nature preserve, 80 giraffes are either homozygous or heterozygous for the long ossicone allele (L). What proportion of offspring would you expect to exhibit the long ossicone phenotype? (2 points)

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BI111 Lecture Notes - Lecture 1: Mendelian Inheritance, Panmixia, Genotype Frequency

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