BIOL 2107 Study Guide - Quiz Guide: Microsoft Powerpoint, White Spot, Labradoodle

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scarletfly776

7 Feb 2016

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Short answer (9 marks) i have a labradoodle (labrador/poodle mix) named moxie and i want to understand her genetics but i am too cheap to have her genotype determined in a commercial lab. We know from lecture that coat and nose colour in labrador retrievers and related mixed breeds is controlled primarily by the brown (b) and extension (e) genes. B is completely dominant over b and e is completely dominant over e. in addition, the parti (sp) allele of the spotting gene (s) gives rise to spotted patterns. Sp is recessive to the solid allele (s). There is no need to consider any other loci. Use the following phenotypic descriptions to answer the questions below. Moxie"s father jack has a red coat and he has a black nose (ignore the single white spot). Prudence has a spotted white and brown coat with a brown nose. My dog moxie is golden in colour and her nose is brownish.

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Adapted from Question 22. Essentials of Genetics, 9th ed., Klug et al., Pearson.
Labrador retrievers may be black, brown, or golden in color. This coat color is controlled by an interaction
of two genes at independent loci. Gene B encodes dark color, where a dominant allele codes for black and
recessive codes for brown. Gene E programs the expression of color, where a dominant allele results in
expression (either black or brown, depending on the genotype at the B locus), and recessive prevents
expression (resulting in a yellow coat color, regardless of the genotype at the B locus).
Although each color may breed true, many different outcomes occur in different matings where the
parents are not true breeding. The following results show some of the possibilities.
Parent Mating Progeny Outcome
(a) Black x Brown all black
(b) Black x Golden all black
(c) Black x Brown 1/2 black, 1/2 brown
(d) Black x Brown 3/4 black, 1/4 golden
(e) Brown x Brown 3/4 brown, 1/4 golden
(f) Black x Golden 2/4 golden, 1/4 black, 1/4 brown
(g) Black x Golden 4/8 golden, 3/8 black, 1/8 brown
(h) Black x Black 9/16 black, 4/16 golden, 3/16 brown
Select two of these matings to discuss in your paper. One mating must be (a), (b), (c), or (d); the other
must be (e), (f), (g), or (h). For each mating, indicate the parental genotypes as completely as possible, and
how this produces the corresponding progeny outcome. Describe the basis for these results. (i.e. Explain
the reasoning you used to arrive at your conclusion). Although not required, it may be helpful to start with
keyouts to determine the genotypes that would breed true for each color.
Requirements
for Evaluation
Possible
Points
Content:
ï§ Discussion demonstrates mastery of related course concepts
o Genotypes, gamete types, and gene interactions are described with
sufficient detail to indicate their contribution to the corresponding
phenotypic ratio
o Appropriate terms are included and used correctly
ï§ Response is focused, well-developed, and includes specific detail
ï§ Overall structure is clear, logical, and well organized
ï§ Body of completed paper is 300-500 words
ï§ No references necessary
14
Style:
ï§ Clarity: clear, easy to understand
ï§ Structure: varied sentence pattern
ï§ Tone: professional and polished; personal style; maintains interest
3
Mechanics:
ï§ Correct grammar, syntax, spelling, and punctuation
3

aquamarinecamel440

2. In horses, the roan coat color pattern is caused by a completely dominant allele (Rn^Rn) at the Kit locus, such that heterozygotes (Rn^RnRn⁺) are roan. The Kit locus is linked to the Extension locus, which determines chestnut coat color. If a horse has two copies of the recessive e allele, then the horse will be chestnut (red). The E allele is completely dominant to e and animals with one copy of the E allele will have a base color of bay (brown) or black. The Kit and Extension loci recombine at a rate of 7%.

A heterozygous roan, homozygous bay/black stallion (Rn^RnRn⁺ EE) is mated to a mare with genotype Rn^RnRn⁺ Ee (roan, bay/black). The mareâs two haplotypes are Rn^Rn E and Rn⁺ e. What will be the expected genotype frequency of the progeny?

3. Hemophilia locus: Corneal dystrophy locus:

Two alleles Two alleles

H = wild-type/normal D = wild-type/normal

h = hemophilia d = corneal dystrophy

H is completely dominant to h D is completely dominant to d

Both loci are sex-linked

The recombination frequency between these two loci is 0.10 (or 10%)

A male dog with both hemophilia and corneal dystrophy has a normal female pup without either disorder. This female pup is now two years old and is expecting her first litter.

A. If her pup is male, what is the probability that he will have both hemophilia and corneal dystrophy?

B. What is the probability that her male pups would not have hemophilia and would have corneal dystrophy?

C. If a male pup has corneal dystrophy, what is the probability that he will not have hemophilia?

Please show work. I want to learn how to do these problems.

Short answers please

1. What is the ultimate effect in an XY human with its SRY genemutated?

2. In humans, the allele for brown eye (B) is dominant to blue(b) eye color. If two heterozygous parents produce children, whatis the probability that:

a. The first child has brown eyes, and the next two childrenhave blue eyes?

b. If the couple above eventually produces a total of sixchildren, what is the probability that three of them will be boyswith brown eyes, and three of them girls with blue eyes?

3. In a paternity dispute, a type AB woman claimed that one ofthe four men was the father of her type A child. Which of thefollowing men could be the father of the child on the basis of theevidence below? Give your reasoning.

Man #1 = type A

Man #2 = type B

Man #3 = type O

Man #4 = type AB

4. Several mating were made among all-white cats. The progentfrom these mating showed the following phenotypic frequency:

All-white cats = 12/16

Black cats = 3/16

Gray cats = 1/16

a. Diagram the cross showing the genotypes of the parents, andtheir progeny as given above,

b. Which gene is epistatic? Explain your answer.

5. Mitosis and meiosis differ in that

a. homologues are seperated in prophase of mitosis

b. homologues are seperated in prophase I of meiosis

c. homologues are seperated in prophase II of meiosis

d. sister chromatids are seperated in prophase I of meiosis

6. The phases of meiosis that most resemble those of mitosisare

a. the meiosis I phase

b. the meiosis II phase

c. meiosis I, except for the pairing of homologues

d. meiosis II, except for the pairing of homologues

7. Genetic variability in sexually reproducing organism s arisesfrom

a. crossing over events between sister chromatids

b. crossing over events between nonsister chromatids

c. egg and sperm genetic recombination from distinctindividuals

d. egg and sperm recombination from an individual

8. A female contains an extra X chromosome, resulting fromnondisjunction. When this gamete is fertilized with normal sperm,it would result in an individual of a genotype

a. XXX

b. XXY

c. XYY

d. all of the above

e. only a and b

9. A diploid organism has the genotype AaBbCcDdEe. How manydifferent types of gametes can this individual produce?

a. 5

b. 10

c. 25

d. 32

e. 625

10. Assuming complete dominance, what is the phenotype ratio ofthe offspring of the cross AaBb x aabb?

a. 1:1

b. 3:1

c. 1:2:1

d. 1:1:1:1

e. 9:3:3:1

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BIOL 2107 Study Guide - Quiz Guide: Microsoft Powerpoint, White Spot, Labradoodle

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