BIOLOGY 1A03 Lecture Notes - Lecture 5: Meiosis, Punnett Square, Gamete

61 views4 pages

erincat447

25 Mar 2018

School

Department

Course

Professor

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

doinhumolalala and 39849 others unlocked

BIOLOGY 1A03 Full Course Notes

Verified Note

9 documents

Document Summary

This means that the probability of a new plant having the big a big a genotype is 25% or . Similarly, the probability of inheriting both little a alleles is times. The probability of the f2 plants having the heterozygous genotype (big a little a) is going to be twice that, or 50%, because there are two ways that this combination could occur, as seen in the punnett square. When the f1 offspring produce gametes, these are produced such that half the possible gametes would contain big a and half would contain little a. The expected ratio of f2 generation offspring with the dominant to recessive trait would be 3:1. Therefore, punnett square analysis accounts for the 3:1 phenotypic ratio of traits that is observed in the f2 generation. However, by looking solely at the phenotype, it is not possible to know of the underlying genotype. Mendel hypothesized that there was an underlying genotypic ratio of.

Related Questions

I am somewhat new to this site although I am a member, so I havethese questions I have answerd most of them my self yet I wouldlike to see if any experts, or people with some knowledge couldcould go ahead and do them so i can see if I am correct or not. Idont actually expect anybody to be able to make a punnet square, asI myself had to make my punnet squares on paper. But these are thequestions below, I will put my answers the ones I have gotten in () but if you are either able to tell me if im correct or if notshow me some sort of steps to becoming correct? Thanks ZackCross.

Solve all of these problems using a Punnett Square. ( I Havedone this mostly on paper)

In certain bacteria, an oval shape (A) is dominant over round(a) and thick cell walls (B) are dominant over thin (b). A bacteriaheterozygous for both oval shape and a thick cellwall was crossed with a round bacteria with a thin cell wall.Describe the phenotype of the offspring.

(My answer not sure if correct but I got : Oval Thick wall: Oval Thin wall : Round thick wall: Round thin wall

1:1:1:1)

2. The ability to roll the tongue sideways (A) is dominant overthe inability to roll one's tongue (a). In humans, non-attachedearlobes (B) are dominant over attached earlobes (b). Two parentswith attached earlobes can roll their tongues. They are bothheterozygous for the tongue rolling trait.

What is the genotype of the mother? ( I got Bb Aa)
What is the genotype of the father? ( I got Bb Aa)
What are the possible gametes from the mother? ( I gotBA Ba Ab ab)
What are the possible gametes from the father? ( I got BA Ba Abab)
Draw the Punnett square, and combine the gametes to produceoffspring( I did on paper)
What is the phenotypic ratio of these offspring? (Didn'tunderstand?)

A man heterozygous for non-attached earlobes can roll histongue. A woman with attached earlobes can roll her tongue. Assumeboth are heterozygous for the tongue-rolling trait.
1. What is the genotype of the mother?
2. What is the genotype of the father?
3. What are the possible gametes from the mother?
4. What are the possible gametes from the father?
5. Draw the Punnett square, and combine the gametes to produceoffspring
6. What is the phenotypic ratio of these offspring?
In summer squash, white fruit color (W) is dominant over yellowfruit color (w) and disk-shaped fruit (D) is dominant oversphere-shaped fruit (d). If a squash plant purebreeding for whiteand purebreeding for disk-shaped fruit is crossed with a plantpurebreeding for yellow, sphere-shaped fruit, what will thephenotypic and genotypic ratios be for:
1. the F₁ generation?
2. the F₂ generation?

Any Help would be greatly apretiated I don't understand how thepoint system works but I did purchase the membership package thingso I do believe I will be able to award someone with these karmapoints which I purchased.

1) The seed color in a species of flowering plant can be either green or white, with the trait determined by two independently assorting loci. At one locus, the G allele produces green color and is dominant to the recessive g allele for white. At the other locus the dominant, C, allele code for color production and its recessive counterpart, c, codes for colorless or white seeds. A pure-breeding green seeded plant is crossed with a white-seeded plant homozygous for both loci.

a) Identify all possible genotype(s) for the white-seeded parent and the F1 genotype(s) that each would produce when crossed with the true-breeding green seeded plant.

b) Assume that the genotype of the original white-seeded parent is gg, cc, F1 plants from the cross described in a. are crossed to produce the F2. Identify the F2 phenotypes and their expected ratio.

c) What would be the probability of producing a white-seeded plant in a cross between two F2 plants of the genotypes Ggcc and ggCc.

2) A snapdragon plant that bred true for white petals was crossed to a plant that bred true for purple petals with no spots, and all the F1 had white petals. The F1 was selfed and among the F2 three phenotypes were obtained in the following numbers: white 240 purple no spots 61 purple with white spots 19 total 320

a) Propose and justify an explanation for these results, showing genotypes for all generations. Make up your own symbols

b) Calculate the chi square value for the data and decide on your hypothesis. Use the chi-square table provided in probability module

In rats, several independently assorting autosomal genes affectcoat color. Gene A controls the distribution of yellow pigment inhair, and gene B causes black pigmentation. The two genes interactas follows: AâBâ (gray), Aâbb (yellow), aaBâ (black), and aabb(cream). These genotypes are only expressed in the presence of thedominant allele of a third gene, C; rats with genotype cc arealbino.

a. Deduce the genotype of each albino mice, to the extent thatis possible, in the following table. Explain your answers.

True-breeding parents	F₁	F₂ offspring
Gray x albino-1	All gray	3/4 gray : 1/4 albino
Gray x albino-2	All gray	9/16 gray : 3/16 yellow : 4/16 albino

b. Deduce the genotype and phenotype of each parent in thefollowing table. Explain your answers.

Parents	Numbers of offspring
Cross 1	135 gray, 83 albino, 47 yellow, 44 black, 16 cream
Cross 2	103 albino, 74 black, 25 cream

c. A gray-colored rat is mated with one that is yellow. Theoffspring include an albino rat and a cream-colored rat. Diagramthis cross. Be sure to include the Punnett square and thephenotypic ratio in the offspring.

ivorysquirrel791

BIOLOGY 1A03 Lecture Notes - Lecture 5: Meiosis, Punnett Square, Gamete

BIOLOGY 1A03 Full Course Notes

Document Summary

Get access

Related Documents

BIOLOGY 1A03 Lecture Notes - Lecture 5: Mendelian Inheritance, Meiosis, Ovule

BIOLOGY 1A03 Lecture Notes - Lecture 14: Heterogametic Sex, Y Chromosome, Somatic Cell

BIOLOGY 1A03 Lecture Notes - Lecture 6: Sister Chromatids, Hmga2, Centromere

Related Questions