BISC 102 Lecture 10: Lecture 10

73 views2 pages

raspberrywildebeest827

3 Mar 2017

School

Simon Fraser University

Department

Biological Sciences

Course

BISC 102

Professor

Erin Barley

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

Document Summary

Genotype = genetic information that codes for a trait (ex/ genes for flower colour) Phenotype = expression of the genotype (ex/ blue colour of flower) Gene = sequence of dna that codes for protein by transcription and translation. Variants of a gene = alleles (gene = flower colour, alleles = blue, red) Diploid genotypes: diploid organism pairs of alleles located on homologous chromosomes. Genotypes: aa (homozygous [2 of the same] for a), aa (heterozygous[different]), aa (homozygous for a) Many conditions coded by recessive (especially lethal ones) but some are coded by dominant alleles (ex/ achondroplasia = type of dwarfism: a (dominant) interferes with reception of growth factor; a (recessive) does not interfere. Heterozygotes have achondroplasia (only small amount of proteins needed to interfere reception of growth factor: homozygosity for achondroplasia = fatal. Abo blood typing (multiple alleles): 3 alleles a = a membrane protein (dominant), b = b membrane protein (dominant), o = no membrane proteins (recessive)

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

1. A plant with red flowers is crossed with a white-flowered plantof the same species. All the seeds, when grown, produce plants withred flowers. Assuming that the flower color is controlled by asingle pair of alleles, which allele is dominant and which isrecessive?

2. In cats, the allele (S) for short fur is dominant to the allele(s) for long fur.
What is the genotype of a true-breeding, long-furred cat?

3. In cats, the allele (S) for short fur is dominant to the allele(s) for long fur.
What is the phenotype of a cat with the genotype Ss?

4. In cats, the allele (S) for short fur is dominant to the allele(s) for long fur.
In an Ss genotype, which allele is expressed in thephenotype?

5. In cats, the allele (S) for short fur is dominant to the allele(s) for long fur.
Which of the fo1lowing genotypes is (i) heterozygous (ii)homozygous dominant?
SS, Ss, ss

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+