BIOA01H3 Lecture : lecture 12 for BGYA01

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chocolaterhinoceros334

19 Jul 2010

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Document Summary

Last class we started our consideration of transmission genetics the study of how genes are passed from one generation to the next. We looked at the law for segregation of alleles and independent assortment. We also looked at how you can use your knowledge of these two laws to create punnett squares that allow you to predict the outcomes of particular mating. If you have attempted the assigned problems you will see that truly understanding these two laws and their consequences (combine how to correctly use a punnett square or probability theory) is very powerful. Last class we saw how genotypes give rise to phenotypes when the alleles examined have only simple dominance /recessive relationships. Today we are going to see some things that complicate our predictions. for some genes both parents don"t have two versions of each gene some alleles are not recessive or dominant.

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

Complete the four-part question.

If you have an organism in which the A allele is dominant to the a allele, a heterozygote will produce gametes of either A or a. This is what leads to two phenotypes among the offspring in an F2 cross (ie. Aa x Aa). In genetic terms this is called

a. dominant

b. recessive

c. segregation

d. independent assortment

e. none of the above

If you have an organism of genotype AaBb (A dominant to a, and B dominant to b), there will be 4 phenotypic classes among F2 progeny in a 9:3:3:1 ratio. In genetic terms this is an example of

a. dominant

b. recessive

c. segregation

d. independent assortment

e. none of the above

Which type of nuclear division produces diploid cells from diploid cells? Haploid cells from diploid cells?

a. fission, fusion

b. meiosis, mitosis

c. mitosis, meiosis

d. linkage, independent assortment

e. none of the above

If a pea plant shows a recessive phenotype

a. it can be either TT or Tt

b. it can be either Tt or tt

c. it can be only TT

d. it can only be tt

e. it can be TT, Tt ot tt

brownwildebeest141

Which of these statements is incorrect?

Syntenic genes are located on the same chromosome.

Independent assortment results in recombinant chromosomes.

You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome.

Linked genes are always syntenic.

What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?

0.49 cM

4.9 cM

49 cM

490 cM

What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)

Human error: they should have been more careful about theirexperimental setup.

Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits.

Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes.

Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean.

When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?

9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio.

Genotypes of the offspring can be determined based on theirphenotype.

If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation.

B and C

Why do we map genes?

To understand how genes interact with each other

Comparative genomics analysis