BIS 2B Lecture 12: Problems to HW Deviation

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beigetiger594

30 Oct 2018

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UC-Davis

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Biological Sciences

Course

BIS 2B

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Mark Schwartz; Sharon Strauss

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BIS 2B Lecture Notes - Lecture 11: Mendelian Inheritance, Allele Frequency, Tyrosinase

BIS 2B Lecture 12: Problems to HW Deviation

BIS 2B Lecture 13: Gene Flow to Natural Selection, Discussion 4

10/22 Problems to HW Deviation

Genetic problems:

1. In a dihybrid cross using one trait with co-dominance (Flower color where RR = red, Rr =

pink, rr = white) and one trait with complete dominance (Y_ = yellow peas, yy= green peas),

what percentage of offspring do you expect to exhibit pink flowers and green peas from crossing

an individual heterozygous at both traits (RrYy) to a white flowered, green peed individual?

RrYy

Rryy

rrYy

rryy

2. You have a self-compatible plant that exhibits three traits: flower color (yellow - Y); leaf shape

(entire -E); and seed shape (round - R). You know that the recessive characters for these three

traits are: a) flower color = white (yy); leaf shape = toothed (ee); and seed shape = flat (rr). You

conduct a test cross with a homozygous recessive to genotype your unknown plant that exhibits

the dominant phenology for each trait. Your plant makes lots of flowers so that you get lots of

offspring. The resulting generation has the following results: approximately 100 each of:

(1) yellow flowered, entire leaf, round seeded;

(2) white flowered, entire leaf, round seeded;

(3) yellow flowered, entire leaf, flat seeded; and

(4) white flowered, entire leaf, flat seeded.

What is the parent genotype?

- they all have entire leaves → homozygous dominant EE

- alternating color or flower and shape of seeds → heterozygous

3. Epistasis - The interaction between nonallelic genes at two or more loci resulting in one gene

masking the phenotypic expression of another gene

A yellow lab is bred with a brown lab

- yellow = _ _ e e

- brown = b b E _

4. Using probability

● Imagine that you have three loci, each has two alleles that sort independently: Yellow /

green (Y/y) peas; Smooth / wrinkled (Ss) peas; purple flowers / white flowers (Pp).

● Take a true breeding yellow, wrinkled, white plant and cross it to a true breeding green

smooth purple one to make the F1. Now cross the F1’s to get an F2

● What fraction will be…

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Rryy rryy rryy: you have a self-compatible plant that exhibits three traits: flower color (yellow - y); leaf shape (entire -e); and seed shape (round - r). You know that the recessive characters for these three traits are: a) flower color = white (yy); leaf shape = toothed (ee); and seed shape = flat (rr). You conduct a test cross with a homozygous recessive to genotype your unknown plant that exhibits the dominant phenology for each trait. Your plant makes lots of flowers so that you get lots of offspring. What is the parent genotype? they all have entire leaves homozygous dominant ee. Alternating color or flower and shape of seeds heterozygous: epistasis - the interaction between nonallelic genes at two or more loci resulting in one gene masking the phenotypic expression of another gene. A yellow lab is bred with a brown lab yellow = _ _ e e. Brown = b b e : using probability.

Related Questions

Problem A. You cross two homozygous parents: RRYY (round yellow) and rryy (wrinkled green) to produce an F1 generation. Round seeds are dominant, yellow seeds are dominant. You cross two F1s to produce the F2 generation. In the F2 generation, you find an individual with round yellow seeds. You do a testcross to find out the genotype of this âmysteryâ dominant/dominant phenotype. In the offspring of the cross, you find 1?2 have round yellow seeds and 1?2 have wrinkled yellow seeds.

1. Based on these data what was the genotype of the mystery F2?

Problem B. You carry out a dihybrid cross between a red-flowered rose with long pollen grains (A1A1 LL) and a white-flowered rose with round pollen (A2A2 l l). The flower color gene (A) shows incomplete dominance, and long pollen is completely dominant to round pollen. The genes for these traits are on different chromosomes so assort independently

2. What would be the phenotypes of the F1 generation?

3. What fraction of the F2 generation is expected to have red flowers and long pollen? (As always when traits are independent, what rule can you apply to save time?)

Problem D. In the example of epistasis presented in class, the agouti allele in mice (A) codes for striped hairs and is dominant to the allele for all black hairs (a). A loss-of-function allele for the pigment gene (b) produces no pigment and is recessive to the normal, dominant allele (B). The recessive allele at b masks the effect of the A gene. A dihybrid cross of AABB and aabb is carried out.

5. Name all of the genotypes in the F2 generation that can produce an albino (no pigment) phenotype.

Problem E. A cross is made between homozygous fruit flies that differ for eye color (the st gene) and

body color (the e gene). The F1 double heterozygote is shown below using chromosome notation.

st+ e

--------------------------------------

st e+

st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?

7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)

8. With what phenotype would you cross this F1 to measure recombination frequency?

cyansnail823

1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:

a.	by a single gene with two alleles that are codominant.

b.	by many genes with an additive effect.

c.	by epistatic interactions between two genes.

d.	mainly by the environment, with only a small genetic component.

2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.

a.	Her mother must be red-green colorblind.

b.	All of her brothers must be red-green colorblind.

c.	Her father must be red-green colorblind.

d.	All of the above statements must be true if a woman is red-green colorblind.

3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:

a.	structure of DNA is a double helix.

b.	two strands of the DNA molecule are joined by hydrogen bonds between the bases.

c.	four bases within DNA pair in a specific way.

d.	two strands of the DNA molecule are joined by covalent bonds between the bases.

4. In the genetic code, _________ one amino acid.

a.	one nucleotide specifies

b.	two nucleotides specify

c.	three nucleotides specify

d.	four nucleotides specify

5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:

a.	independent assortment.

b.	nondisjunction.

c.	segregation.

d.	crossing over.

6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:

a.	0.16 (16%)

b.	0.24 (24%)

c.	0.36 (36%)

d.	0.48 (48%)

7. Natural selection acts at the level of the:

a.	phenotype.

gene.

c.	population.

d.	nucleotide.

8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?

a.	This plant would produce only RY and ry gametes.

b.	This plant would produce only RrYy gametes.

c.	This plant would produce RY, Ry, rY, and ry gametes.

d.	You cannot determine which gametes this plant can produce without knowing the genotypes of its parents.

9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:

a.	percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides.

b.	four bases all occur in an equal frequency (25%) within each organism.

c.	percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides.

d.	genetic material is composed of proteins, not DNA.

10. During DNA replication:

a.	each strand of the double helix acts as a template for the synthesis of a new strand.

b.	the enzyme DNA polymerase adds nucleotides to the strand being synthesized.

c.	the bases A,C,G and T are required.

d.	All of the above are true of DNA replication.

11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:

12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.

a.	RR and rr.

RR and Rr

Rr and rr

d.	Each of the three possible genotypes will have a different phenotype.

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BIS 2B Lecture 12: Problems to HW Deviation

BIS 2B Lecture Notes - Lecture 11: Mendelian Inheritance, Allele Frequency, Tyrosinase

BIS 2B Lecture 12: Problems to HW Deviation

BIS 2B Lecture 13: Gene Flow to Natural Selection, Discussion 4

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