BIOL 311 Chapter Notes - Chapter 3: Punnett Square, Dihybrid Cross, Mutual Exclusivity

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mauvewolf957

13 Jun 2018

School

University of Calgary

Department

Biology

Course

BIOL 311

Professor

Isabelle Barrette- Ng

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

BIOL 311 Chapter 3: Independent Assortment of genes

Dihybrid: a double heterozygous ex. A/a B/b

• Mendel used a dihybrid cross to determine his second law of heredity

• Focused on two specific traits seed shape and colour

o Crossed true breeding parental lines of wrinkled yellow and round green

o Produced round and yellow F1 offspring indicating the dominance

o Then selfed the F1 to produce the F2 to obtain a 9:3:3:1 ratio of round yellow,

round green, wrinkled yellow and wrinkled green

Medel’s Secod Law of Idepedet Assortet: different gene pairs assort independently in

gamete formation

• Genes on the same chromosome generally do not assort independently

o Gene pairs on different chromosomes assort independently in meiosis

• The actual numbers obtained explain the 9:3:3:1 ratio as two randomly combined 3:1

ratios. The ratio between yellow and green is 3:1 and the ratio between wrinkled to

smooth seed is 3:1

• Tree diagram: ¼ round, ¾ are yellow and the other ¼ is green (same for ¼ wrinkled)

• The ratio can also bee explained by gamete frequency

• ½ gametes are R then ½ will be Y and ½ will be y

• ½ gamete are r then ½ will be Y and ½ will be y

o ¼ R;Y ¼ R;y ¼ r;Y ¼ r;y

• Both males and females will produce the same gametes so once the four gametes are

crossed together in a punnet square, he obtained a ratio

o The frequency of a game being produced in ¼ so the frequency of the fusion of

two gametes is (1/4)^2

o Phenotypes and genotypes of the gametes and progeny that underlie the ratio

• In order to test the hypothesis of the 1:1:1:1 ratio, Mendel testcrossed the dihybrid F2

generation and obtained equal proportions of gametes

o 1:1:1:1 is a diagnostic of independent assortment in one dihybrid

o 9:3:3:1 is a diagnostic of independent assortment is two dihybrids

Product rule: the probability of independent events both occurring together is the product of

the individual probabilities (A and B)

o Example of rolling a die. The pobability of olling to ’s is independent; olling

one four is p=1/6 pto ’s= 1/ *1/ = 1/

Sum rule: probability of either of two mutually exclusive events occurring is the sum of the

indiidual pobabilities A’ or A’’

o Pobability of to ’s is eual to the pobability of to ’s so the pobability of

olling to ’s o to ’s is 1/ + 1/= 1/18

• In a genetic example, if the gene pairs do assort independently we can use the product

rule. Use a simple punnet square between the same genes and obtain the proportion of

the desired genotype for all the genes; then multiply

Example: finding how many progeny need to be grown? Estimate how many progeny plants

need to be grown to aquire a specific genotype

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Document Summary

Biol 311 chapter 3: independent assortment of genes. Product rule: the probability of independent events both occurring together is the product of the individual probabilities (a and b: example of rolling a die. The p(cid:396)obability of (cid:396)olling t(cid:449)o (cid:1008)"s is independent; (cid:396)olling one four is p=1/6 p(cid:894)t(cid:449)o (cid:1008)"s(cid:895)= 1/(cid:1010) *1/(cid:1010) = 1/(cid:1007)(cid:1010) In a genetic example, if the gene pairs do assort independently we can use the product rule. Use a simple punnet square between the same genes and obtain the proportion of the desired genotype for all the genes; then multiply. Example: how many distinct genotypes will a cross produce: for a tetrahybrid a/a; b/b; c/c; d/d there will be 3 genotypes for each gene pair a/a, A/a, a/a: 3 genotypes for 4 gene pairs results in 3^4= 81. Synthesizing pure lines: pure lines are made by repeated generations of selfing to generate fully homozygous individuals, repeated selfing leads to an increased proportion of homozygotes.

Related Questions

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+

st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?

7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)

8. With what phenotype would you cross this F1 to measure recombination frequency?

cyansnail823

Question 1 Select one answer. 10 points Crossing over occurs:_______________.

during both meiosis I and meiosis II

during mitosis but not meiosis

during meiosis II only

during meiosis I only

Question 2 Select one answer. 10 points Homologous pairs ofchromosomes line up at the center of the cell in during which stageof meiosis?

Metaphase I,

Metaphase II,

Telophase I,

Telophase II

Question 3 Select one answer. 10 points A haploid organism hashow many copies of each chromosome?

1, 2, 3, 4

Question 4 Select one answer. 10 points The sister chromatidsseparate during which stage of meiosis?

Anaphase I, Anaphase II, Prophase I, Prophase II

Question 5 Select one answer. 10 points Describe the importanceof independent assortment: ________________________.

will assort independently of each other,

will sort to the same gamete,

are called alleles

All of these are true

Question 6 Select one answer. 10 points How many differentgametes are produced by a cell with the genotype AABb?

1, 2, 3,4

Question 7 Select one answer. 10 points The next two questionsare based on the following information. Part 1 of 2: In beansyellow (Y) is dominant to green (y) and smooth (S) is dominant towrinkled (s). What are the possible genotypes for the offspring ofthe following cross: YySS and YYSs.

Yy, SS, YY, Ss

YYSS, YySS, YYSs, YySs

YYSS, YYss, yySS, yyss

YS, yS, YS, yS

Question 8 Select one answer. 10 points Part 2 of 2: In beanswith yellow (Y) dominant to green (y) and smooth (S) dominant towrinkled (s): What are the phenotypes for the offspring in theexample in the previous question?

100% green wrinkled,

50% yellow smooth and 50% greensmooth,

100% yellow smooth,

50% yellow smooth and 50% yellowwrinkled

Question 9 Select one answer. 10 points Alleles are differentversions of a _______________.

chromosome,

genotype,

phenotype,

gene

pucesnake469

Canaries: Green (wild-type) Pied phenotype Yellow type

Canaries are popular pets and easy to breed in captivity. Additionally, they display great phenotypic diversity, which allows you to do some interesting genetic investigations with them.

When you take a canary from a true-breeding green strain and cross it with a canary from a true-breeding yellow strain, all offspring display the pied phenotype. Based on this you suspect that this color variation is inherited by a gene with a co-dominant inheritance pattern (showing both phenotypic traits in hybrid phenotypes).

You decide to test this hypothesis. To do so, you take pairs of your hybrid F1 individuals (of pied phenotype), and cross them to produce an F2 generation.

After many dozens of crosses and hundreds of offspring, you arrive at the peculiar ratio of:

3/16 green : 6/16 pied : 7/16 yellow

Which of the following hypotheses are consistent with these data? (Note that not all of these hypotheses are mutually exclusive. Assume that the two genes are independently assorting)

Hint: This problem is very hard to work from the given ratios to what is happening genetically. It is easier to "work backwards" and start with each choice below, and see if you can obtain all or important aspects of the results of your cross when making the assumption that the answer choice is correct.

Select all that apply:

Select one or more:

a. At least one of the genes involved is sex linked.

b. One gene controls production of darker pigment (melanin which appears green when mixed with yellow). A second gene acts as a patterning gene with one allele causing expression and the other non-expression of melanin and when in the heterozygous condition each allele controls melanin expression on different parts of the body so long as the bird's genotype of the first gene allows for the production of melanin.

c. this is a single-gene trait with co-dominance

d. one of the two genes involved displays complete dominance, while the other displays co-dominance.

e. the genes involved are epistatically interacting

f. more than one gene is involved

BIOL 311 Chapter Notes - Chapter 3: Punnett Square, Dihybrid Cross, Mutual Exclusivity

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