Two Point Test Cross Problems (Forwards and Backwards) 1. In the imaginary mongs

Question

Two Point Test Cross Problems (Forwards and Backwards)
1. In the imaginary mongs, eyes can be brown or red (red is dominant to brown) and wings can

be short or long (long is dominant). A female mong heterozygous for both genes is mated with a double homozygous recessive male mong. From that mating the following offspring are obtained:

red, long 45 brown, long 92 red, short 87 brown, short 48

a) What is the linkage arrangement of the loci? (Determine if the genes are linked; if they are, give the map distance and indicate whether the mutant alleles are in cis or trans arrangement in the female heterozygote mong.)

2. In corn, the pg locus (pale green seedlings) is 5 map units from the gl locus (glossy leaves) on chromosome 9. The mutant phenotypes given are determined by recessive alleles. The dominant wild-type conditions are dark green seedlings and dull leaves, respectively.

a) Predict the types and frequencies of phenotypes that would occur from the test cross of a pg+ gl/pg gl+ dihybrid.

Combining Two Point Data

3. Use the two-point recombination data to map the genes.
a) Show the order and the length of the shortest interval.

Gene loci

a,b
a,c
a,d
a,e
b,c 50 d,e

Linkage and Combinatorial Probability

50 15 38

b,d b,e c,d

8 c,e

4. Nail-patella syndrome in humans is characterized by congenital abnormalities of the fingernails (and sometimes toenails) and the patellae (kneecaps). The gene for this disorder

BIOL 321 | Problem Set 3

is dominant and is located on chromosome 9 about 10 map units from the ABO locus. Suppose a man with nail-patella syndrome and type A blood marries a normal woman with type B blood. The mothers of both the husband and the wife are normal and have type O blood.

a) The husband and the wife have two children, one with type A blood and the other with type B. What is the probability that both children have normal fingernails and patellae?

b) The couple are now expecting another child. What is the chance that the child will have nail-patella syndrome and type O blood?

c) Amniocentesis reveals that the fetus has type AB blood. What is the chance that it has nail-patella syndrome?

Three Point Test Cross Problems

5. The following tomato plant traits were studied in a three-point test cross. Normal (round) fruit or ovate (long) fruit, flat leaves or curled leaves and solid color or mottled coloration. F1 plants (heterozygous for all three genes) have normal fruit, flat leaves and solid coloration. F1 plants are crossed with a strain that has ovate fruit, curled leaves and mottled coloration and the following offspring were obtained:

Normal, curled, mottled 3 Normal, curled, solid 26 Normal, flat solid 304 Normal, flat, mottled 68 Ovate, curled, mottled 310 Ovate, curled, solid 60 Ovate, flat, mottled 24 Ovate, flat, solid 5

a) Determine the map of the three genes (show gene order and distance).

b) Sketch the two chromosomes of the trihybrid used to generate these offspring.

c) Calculate the coefficient of coincidence (C) and interference (I).

Explanation / Answer

This has multiple question. As per Chegg’s policy., I am answering the first question.

Cross is between RrLl (red long) and rrll (brown short)

It is a test cross. Expected phenotypic ratios are 1:1:1:1

Progeny are

Red long 45    

Brown long 92     

Red short 87        

Brown short 48   

This offspring frequency is not in 1:1:1:1 ratio. So, there is linkage.

Here recombination in rrll parent has no consequence on progeny because same gamete genotype occurs even if cross over happens or not.

Cross over occurring in RrLl parent is important and it determines the offspring genotypes.

Based on off spring frequency, we know that brown long and red short are in high number. So, these are parental genotypes. So, rL are in one chromosome and Rl are in another chromosome of RrLl parent. So, the genes are in trans confirmation.

Red long 45     Recombinant

Brown long 92      Parental

Red short 87         Parental

Brown short 48    Recombinant

% cross over is equal to number of recombinantsx100/total number of offsprings

% cross over = (45+48) x 100/272 = 93 x 100/273 =34..07%

1% cross over is equal to one map unit distance between the two genes

So, the map distance is 34.07

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.