BIOL 2107 Lecture Notes - Lecture 4: Heredity, Cystic Fibrosis, Consanguinity

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30 Jan 2017

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The punnett square is a simple way to visualize the segregation and random union of. Each f1 hybrid produces two kinds of gametes in a 1:1 ratio. F2 progeny: 3:1 ratio of phenotypes, 1/4 will breed true for the dominant trait, 1/2 will be hybrids, 1/4 will breed true for the recessive trait. F1 will result in f2 of 3:1/1:2:1. 3:1 is the phenotype ratio, 1:2:1 is the genotype ratio. Law of segregation: the gametes are going to be separated, having only one allele. Law of independent assortment: those gametes are going to independently match. Mendel"s dihybrid crosses revealed the law of independent assortment. Mendel tested whether two genes in dihybrids would segregate independently. First, he crossed true-breeding yellow round peas with true-breeding green wrinkled peas to obtain dihybrid f1 plants: yy rr x yy rr f1 yy rr. Then, the dihybrid f1 plants were selfed to obtain f2 plants: f1 yy rr x f1 yy rr f2.

Related Questions

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

To determine the genotype of an organism

All of the above

coralox210

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

BIOL 2107 Lecture Notes - Lecture 4: Heredity, Cystic Fibrosis, Consanguinity

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