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I just need help with these two questions:

The Principle of Independent Assortment would suggest a 9:3:3:1 phenotype ratio (dom/dom; rec/dom; dom/rec; rec/rec) in a cross, involving parents heterozygous at each of two genes. When the same experiment is done using different traits including flower color and pollen shape, crimson flowers are dominant (red is recessive) and long pollen is dominant (round is recessive). A cross involving heterozygous parents shows the following results:

4831 (= 69.5%) with crimson flowers and long pollen,

390 (= 5.6%) with crimson flowers and round pollen,

393 (= 5.6%) with red flowers and long pollen,

1338 (= 19.3%) with red flowers and round pollen.

a. (5 points) Since there is no 9:3:3:1 ratio, would you conclude that these genes are on the same chromosome or on different chromosomes?

b. (5 points) Explain how the crimson/round and red/long flowers might have arisen at a low frequency?

Thank you!

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Trinidad Tremblay
Trinidad TremblayLv2
29 Sep 2019
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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 “mystery” 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?

In the tomato, red fruit (R) is dominant over yellow fruit (r), and yellow flowers (F) are dominant over white flowers (f). A cross was made between true-breeding plants with red fruit and yellow flowers, and plants with yellow fruit and white flowers. All F1 offspring are heterozygotes for both traits. The F1 female plants were then test crossed to males with yellow fruit and white flowers (homozygous recessive for both traits). The following results were obtained:

333 red fruit, yellow flowers

64 red fruit, white flowers

58 yellow fruit, yellow flowers

345 yellow fruit, white flowers

Total of 800

1. Conduct a Chi square analysis to see if the two genes are linked ( Check a solved problem S1, page 148 in the first edition text book, and page 150 in the second edition problem #1 for an idea)…follow the same steps)

a. Propose a hypothesis to be tested

b. Find the expected (predicted) number of offspring ( Hint what is the ratio of test cross results?)

c. Calculate and Interpret the Chi –square result

d. If the two genes (fruit color and flower color) are linked, calculate the map distance between the two genes.

2. Set a sample test cross, Explain the rationale behind a testcross, why is it necessary for one of the parents to be homozygous recessive for the genes of interest?

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

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