BM1000 Lecture Notes - Lecture 6: Gamete, Abo Blood Group System, Mendelian Inheritance

47 views19 pages

coffeebear832

16 Jul 2018

School

James Cook University

Department

Course

BM1000

Professor

Lionel Hebbard

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

What happens when we cross a pea plant and look at two genes: Mendel crossed two plants which were homozygous for both traits rryy x rryy. The genes of seed shape and seed colour maintain their association and are linked. F1 plants produce two types of gametes: ry and ry. F2 generation would only show two possible phenotypes: round yellow and wrinkled gr (such as p generation) and the ratio would be 3:1 just as in monohybrid cross. The segregation of r from r is independent of segregation of y from y and the genes ar not linked. F1 plants produce four types of gametes in equal numbers: ry, ry, ry and ry. F2generation would show four possible phenotypes: round yellow, round green, wrink yellow and wrinkled green. Punnet squares: ed green es are rinkled pe: 9 round and yellow (parent in the p generation) 1 wrinkled and green (parent of p generation)

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.

mauvefox570

My prof says this is done wrong as the break point is between R and Y. Could you please help me out? With explanation? Thank you!

10. In a certain plant, a dominant allele for one gene, R, produces round fruit and the recessive allele, r, produces oblong fruit. A dominant allele of a second gene, Y, produces yellow flowers and the recessive allele, y, produces white flowers. The two genes are linked on chromosome IV. A plant (r y/r y) with normal chromosomes is crossed to a plant (R Y/R Y) homozygous for a reciprocal translocation between chromosomes IV and VIII. The presence of the translocation causes semi-sterility. The F1 progeny are phenotypically yellow flowered with round fruit and are semi-sterile. A backcross to the parent with normal chromosomes yields the 750 progeny shown in the table below.

	Number of Progeny
	Fertile	Semi-sterile
Yellow, round	3	283
White, oblong	295	2
Yellow, oblong	73	12
White, round	10	72

a. From these data, draw a map showing all relevant map distances (in centimorgans) for the genes and the translocation breakpoint. Show your work.

Answer:

Cross over gametes: Yellow, oblong & White, round

Total number of Progeny = 381

Frequency of cross over gametes=

83 / 381 = .22 Morgans

22 cM

85 / 369 = .23 Morgans

23 cM

The genes are located on the same side of the chromosome.

Break point R Y

. . 22cM .

No; the breakpoint is located between the two genes. Also you muat use both classes of dluble crossovers to calculate the two intervals. Please redo

b. Briefly explain the basis of the semi-sterility of translocation-bearing plants.

Answer: The semi-sterility in F1 plants are caused by reciprocal translocation. As the organism is heterozygous for reciprocal translocation meaning that offsprings are produced less than normal. This reciprocal translocation may have occurred at meiosis during chromosome segregation. No; the translocation already existed.

Please read your text and redo 3/10

lilacsea-lion754

BM1000 Lecture Notes - Lecture 6: Gamete, Abo Blood Group System, Mendelian Inheritance

Document Summary

Get access

Related Documents

BIS 2B Lecture 12: Problems to HW Deviation

BIOL 155 Lecture Notes - Lecture 38: Dihybrid Cross, Mendelian Inheritance, Zygosity

BIO152H5 Chapter Notes - Chapter 13: Dihybrid Cross, Punnett Square, Nettie Stevens

Related Questions