BIO 122 Lecture Notes - Lecture 10: Gregor Mendel, Dihybrid Cross, Punnett Square

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fuchsiacoyote139

30 May 2018

School

Drexel University

Department

Bioscience & Biotechnology

Course

BIO 122

Professor

Gurney

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Prior to Week 10 – Chapter 14 Mendel and the Gene

Know

• That Gregor Mendel worked on pea plants to understand heredity.

• That characters are features (like eye color) and traits are identifiable characteristics (like green eye color).

• That the pea plant normally will self-fertilize, but because of the size and anatomy of its flowers, it can be

manipulated to cross fertilize experimentally.

• What true breeding means

o Breed two purebreds

o Homozygous dominant or homozygous recessive

• That a hybrid refers to the offspring of two true-breeding parents differing in one trait (like height or flower

color).

• What dominant and recessive traits refer to

o Dominant masks recessive

• The distinction between a “gene” and an “allele”

o Gene is the genetic code for a characteristic

• Ie height

o Allele is the variation of the gene

• Ie tall of short

• What homozygous and heterozygous refer to

o Homozygous

• both dominant or both recessive allele

o Heterozygous

• one dominant and one recessive allele

• How to set up a Punnett Square for a single trait and a two-factor cross.

o Single trait

• Four boxes

o 2 factor

• 16 boxes

Understand

• That a monohybrid cross is the mating of two true breeding parents differing in one trait (refer back to

definition of “hybrid”).

o Ie just different height or just different flower color

• That a dihybrid cross is the mating of two true breeding parent plants differing in two traits

o Ie different height and flower colors

• The difference between a genotype and a phenotype.

o Genotype what is genetically given

o Phenotype

• What is expressed

• How the separation of homologous pairs during Meiosis I explains segregation of alleles.

o They separate and go to separate gametes

o Each is carried by a separate sperm or egg

• How the random way that homologous pairs line up during Meiosis I explains independent assortment for

genes on separate chromosomes.

o Any allele can go to any gamete

• That fertilization randomly combines gametes with each other.

• How the results of crossing two true-breeding organisms would result in different offspring from the

crossing of two heterozygotes.

o Two true breeding cross

• Have 4 identical offspring, all heterozygous

o 2 heterozygous cross

• 2 heterozygous

• 2 homozygous recessive

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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

In need of assistance. Please help answer the following.

1) Pure-breeding yellow guinea pigs crossedwith pure-breeding white guinea pigs produce only cream-coloredoffspring. This pattern indicates incomplete dominance. Rough hairis found to be dominant to smooth hair. Give the F1 and F2genotypic and phenotypic ratios of a cross of a smooth white guineapig with a homozygous rough yellow guinea pig.

2) A breeder of cattle has a herd of white cowsand a roan bull. Hair color in this breed is controlled by anincomplete dominant gene. The two homozygous forms are either redor white, and the heterozygous is roan. Answer the followingquestions please

What color calves are expected and in whatproportion?

Outline a procedure to develop an all-redherd.

3) In horses, black coat color is influenced bythe dominant allele (B), and chestnut coat color is influenced bythe recessive allele (b). Trotting gait is due to a dominant gene(T), pacing gait to the recessive allele (t). If a homozygous blacktrotter is crossed with a chestnut pacer:

What will be the appearance of the F1 and F2generations?

Which phenotype will be the most common?

Which genotype will be the most common?

Which of the potential offspring will be certain tobreed true?

4) In garden peas, one pair of alleles controlsthe height of the plant and a second pair of alleles controlsflower color. The allele for tall (D) is dominant to the allele fordwarf (d), and the allele for purple (P) is dominant for the allelefor white (p). A tall purple crossed with a tall purple produces Â¾tall purple and Â¼ tall white. What is the genotype of the parents?

5) Three alleles control the ABO blood types.I^A and I^B are codominant genes, so thecombination of I^AI^B produces the AB bloodtype. The third allele I^ois recessive to the other twoalleles. Indicate which of these parents could produce the givenchild and submit your response to the instructor.

A X AB produce B child.

A X O produce A child.

A X B produce O child.

A X AB produce O child.

A X AB produce B child.

B X B produce O child.

AB X AB produce A child.