BIOL 1000 Lecture 4: nov 30 bio notes

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21 Feb 2019

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Pp p pp pp: pp= heterozygous, pp= homozygous, one parent must be homozygous, slightly different coding in one give rise to the same proteins with slightly diferent traits/functions/amino proteins. Due to a recessive allele is mutated (abnormal) which gives rise to non- functional protein = c. Dominant allele is normal which gives rise to functional proteins = c. Mutations can happen anywhere in the sequence. Result of the presence of abnormal and normal proteins. Dihybrid cross: rryy x rryy rryy x rryy (f1) Let"s say r and y came from mom. Let"s say r and y came from dad. Case one: can the alleles assort independently (separate) what are the possible gene combinations? (independent assortment, each gamete should get one copy of each, possible combinations: ry, ry, ry, ry. Case two: what if they don"t separate, then what are the possibilities? (if they are linked), possible combinations: ry and ry.

Related Questions

Question 1

What would be the dimensions of a Punnett square for the cross Hh x hh?

		4 x 4
		2 x 2
		2 x 1
		1 x 1

Question 2

In dogs, the gene for fur color has two alleles. The dominant allele (G) codes for grey and the recessive allele (g) codes for black fur. The female dog is heterozygous. The male dog is homozygous recessive. Figure out the phenotypes and genotypes possible in their puppies by using a Punnett Square. (provide answers on one line with commas between - xx, xx, xx, xx)

Genotypes: , ,

Phenotypes: ,

Question 3

The fundamental Mendelian process which involves the separation of alleles located at the same spot on the chromosome would be called ___.

		segregation
		independent assortment
		continuous variation
		discontinuous variation

Question 4

Using the information given on dominant and recessive traits, state the phenotype for each genotype. Purple flowers are dominant to white flowers in pea plants.

-A.B.C.	PP_____________
-A.B.C.	Pp____________
-A.B.C.	pp_______________

A.	purple flowers
B.	purple flowers
C.	white flowers

Question 5

In dogs, the gene for fur color has two alleles. The dominant allele (G) codes for grey and the recessive allele (g) codes for black fur. The female dog has black fur and the male dog is homozygous dominant. Make a Punnett square to determine the chance of getting each genotype and phenotype below.

-A.B.	Probability of GG:
-A.B.	Probability of Gg:
-A.B.	Probability of gg:
-A.B.	Probability of gray fur:
-A.B.	Probability of black fur:

A.	100
B.	0

Question 6

Using the information given on dominant and recessive traits, state the phenotype for each genotype. Round seeds are dominant to wrinkled seeds in pea plants.

-A.B.C.	RR______________
-A.B.C.	Rr _____________
-A.B.C.	rr ____________

A.	round seeds
B.	wrinkled seeds
C.	round seeds

Question 7

Using the information given on dominant and recessive traits, state the phenotype for each genotype. Hairy knuckles are dominant to non-hairy knuckles in humans.

-A.B.C.	HH_____________
-A.B.C.	Hh ___________
-A.B.C.	hh _________________

A.	non-hairy knuckles
B.	hairy knuckles
C.	hairy knuckles

Question 8

An allele is ___.

		one of the bases in DNA
		an alternate form of a gene
		another term for epistasis
		present only in males and is responsible for sex determination

Question 9

In a cross of a round hybrid pea with a true breeding round parent (Rr x RR), what genotypic proportions would be observed in the offspring?

		Half heterozygous, half homozygous dominant
		Half round, half wrinkled
		All heterozygous
		All round

Question 10

What would be the dimensions of a Punnett Square depicting a dihybrid cross?

		1 x 4
		2 x 4
		4 x 4
		2 x 2

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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BIOL 1000 Lecture 4: nov 30 bio notes

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