1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
butalik [34]
3 years ago
5

Three loci are found to be located (linked) on Chromosome 2 in fruit flies. At locus 1, wild is dominant over tall (a). For locu

s 2, wild is dominant over red (b) and for locus 3, wild is dominant over bent (c). Assume that a standard trihybrid backcross is made and the following progeny are produced: tall red bent 322 wild wild wild 306 wild red bent 6 tall wild wild 10 tall wild bent 32 tall red wild 153 wild red wild 41 wild wild bent 130 The distance between the locus for red (r) and the locus for bent (c) is ________ map units
Biology
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

38.8 MU  

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

• At locus 1, wild (a+) is dominant over tall (a).  

• For locus 2, wild (b+) is dominant over red (b)  

• For locus 3, wild (c+) is dominant over bent (c).  

A standard trihybrid backcross is made and the following progeny are produced:  

• tall red bent, a b c   322

• wild wild wild, a+ b+ c+  306  

• wild red bent, a+ b c   6  

• tall wild wild, a b+ c+   10  

• tall wild bent, a b+ c    32  

• tall red wild, a b c+  153  

• wild red wild, a+ b c+   41  

• wild wild bent, a+ b+ c  130

The total number of individuals, N, is 1000.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

<u>Parental) </u>

• tall red bent, a b c   322

• wild wild wild, a+ b+ c+  306  

<u>Double recombinant) </u>

• wild red bent, a+ b c   6  

• tall wild wild, a b+ c+   10  

<u>Simple recombinant)</u>

• tall wild bent, a b+ c    32  

• tall red wild, a b c+  153  

• wild red wild, a+ b c+   41  

• wild wild bent, a+ b+ c  130

By comparing them we will realize that between  

a b c   (parental)

a+ b c   (double recombinant)

and  between

a+ b+ c+  (Parental)

a b+ c+   (double recombinant)

They only change in the position of the alleles a/a+. This suggests that the position of the gene “a” is in the middle of the other two genes, “b” and “c”, because in a double recombinant only the central gene changes position in the chromatid.  

So, the order of the genes is:

---- b ---- a -----c ----

Now we will call Region I to the area between b and a, and Region II to the area between a and c.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between b and a genes, and P2 to the recombination frequency between a and c.

  • P1 = (R + DR) / N
  • P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  So:  

Parental)

•  b a c  

•  b+ a+ c+    

Double recombinant)

• b a+ c    

• b+ a c+    

Simple recombinant)

•  b+ a c      

•  b a c+  

• b a+ c+  

•  b+ a+ c

P1 = (R + DR) / N

P1 = (32+41+6+10)/1000    

P1 = 89/1000

P1 = 0.089

P2= = (R + DR) / N

P2 = (153+130+6+10)/1000

P2 = 299/1000

P2 = 0.299

Now, to calculate the recombination frequency between the two extreme genes, b and c, we can just perform addition or a sum:

P1 + P2= Pt

0.089 + 0.299 = Pt

0.388=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes for which every 100 meiotic products, only one results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.089 x 100 = 8.9 MU

GD2= P2 x 100 = 0.299 x 100 = 29.9 MU

GD3=Pt x 100 = 0.388 x 100 = 38.8 MU

You might be interested in
How is the protein created from the mRNA sequence which reads AUGAUAGUGUAC different from the protein created from one that read
adoni [48]
To transcribe the mRNA to make the DNA, we use genetic codes called codons. These are composed of three amino acids clumped together. It could be Adenine(A), Guanine(G), Uracil(U) and Cytosine (C). The difference is that first thread contains GUG, while the second one contains AGU. GUG is for Histidine while AGU is for Serine.
8 0
3 years ago
How many different species of shrimp are currently found in the waters surrounding the isthmus if panama
hichkok12 [17]
There are approximately 12 transisthmian shrimps species, called sister species because they cannot interbreed effectively, in the Isthmus of Panama. It is believed that the closure of the Panama resulted to allopatric speciation of the shrimp population within the region.
4 0
3 years ago
I need help asap please!!
fenix001 [56]

Answer:

commensalism

an association between organisms in which one benefits and the other derives nor harm

8 0
3 years ago
Three reasons why do you think women in today's society could be more obese than men in South Africa​
bezimeni [28]

Answer:

Three Reasons

Explanation:

1, due to the genome. By birth, there is a biologically higher percentage of body fat in women.

2, unemployment. As normally in a South African household, it is the man that goes out to work while the women stay at home. loss of mobility means the accumulation of fats.

3, Males' major sex hormone is testosterone, and females have high levels of estrogen and progesterone in their blood. Females build up fat more easily than males.

8 0
2 years ago
Which property do metalloids share with nonmetals?
vitfil [10]

Answer:

both can react to form acidic compounds

6 0
3 years ago
Read 2 more answers
Other questions:
  • Why do high-energy electrons need carrier molecules?
    6·1 answer
  • Why did Europeans call the bubonic plague the Black Death? because that is what the Chinese called it because of the black boils
    8·2 answers
  • Which of the following factors does not vary between a fast flowing river and a pond that receives no water input?. . oxygen lev
    6·1 answer
  • Explain the significance of the threat that an invasive species brings to an ecosystem.
    15·2 answers
  • What is a limitation of using pedigrees for positional cloning?
    8·1 answer
  • An atom is the smallest complete part of a(n) __________ that still has all the original properties of this substance.
    9·2 answers
  • Which of the following hormones increases the heart rate in response to stress?
    6·1 answer
  • What important gas did stromatolites (cyanobacteria) release into the atmosphere?
    8·1 answer
  • Parasitism is when
    15·2 answers
  • During one year, the birthrate in a country is 28 births per 1000 people, and the death rate is six deaths per 1000 people. What
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!