Question

In: Biology

An alien species was found with the following characteristics Flat feet dominant over round feet A...

An alien species was found with the following characteristics Flat feet dominant over round feet A vs a 10 toes dominant to 15 toes B vs b Curved ears dominant to straight ears C vs c A flat footed, curved eared alien with 10 toes was crossed with a round footed, 15 toed, straight eared individual and produced the following offspring

Flat, curved, 10 toes 588

Round, straight, 15 toes 599

Flat, 15 toes, straight 165

Flat 15 toes, curved 159

Round, 10 toes, curved 145

Round, 10 toes, straight 163

Flat, straight, 10 toes 15

Round, curved, 10 toes 23

What gene is in the middle?

How far apart are the genes?

Solutions

Expert Solution

Dominant allele A = Flat feet

Recessive allele a = Round feet

Dominant Allele B = 10 toes

Recessive allele b = 15 toes

Dominant allele C = Curved ears

Recessive allele c = Straight ears

a)

Two frequencies of offspring phenotypes in highest number are parental types and two frequencies of offspring phenotypes in the lowest number are double cross overs

ABC (Flat, 10 toes, curved) = 588 (Parental type - No cross over)

abc (Round, 15 toes, straight) = 599 (Parental type - NCO)

ABc (Flat, 10 toes, straight) = 15 (double cross overs)

abC (Round, 15 toes, curved ) = 23 (Double cross over)

When parental types compared with double cross over,

ABC (parental type) when compared to ABc (DCO), the uncommon gene is C/c

abc (parental) when compared to abC (DCO), the uncommon gene is C/c

The uncommon gen comes in the middle, So the gene C comes in the middle

The order of genes on the chromosome is

A.....C....B

b).

Parents: ACB/acb * acb/acb

Offspring :

ACB (Parental type (NCO)) = 588

acb (parental type (NCO)) = 599

aCB (Single cross over between genes A and C) = 145

Acb (Single cross over between genes A and C) = 165

ACb (Single cross over between genes C and B) = 159

acB (Single cross over between genes C and B) = 163

AcB (Double cross over) =15

aCb (Double cross over) = 23

Total progeny = 1857

Map distance between two genes = (SCO between two loci + DCO)/ Total progeny ) * 100

Map distance between A and C = ((145 + 165 + 15 +23 )/1857 ) * 100

= (348/1857) * 100 = 0.1874 * 100 = 18.74m.u

Map distance between C and B = ((159 + 163 + 15 +23)/ 1857) * 100

= (360/1857) * 100 = 0.1939 *100 = 19.39 m.u

Therefore, the distance between genes are

A.......18.74 m.u.....C................19.39..............B

..............................38.13 m.u...........................


Related Solutions

In the garden pea Pisum sativum, round pea texture (R) is dominant over wrinkled pea texture...
In the garden pea Pisum sativum, round pea texture (R) is dominant over wrinkled pea texture (r). A female that is homozygous recessive for pea texture is crossed to a male that is heterozygous for pea texture. Please answer the following questions (show your calculations): 1. A) genotype of the female (1 point) B) list the types of gametes the female can produce (2 points) C) do the cross in a Punnett square (4 points).   D) What is the phenotypic...
1- The changes in characteristics of a species over time. Group of answer choices fitness evolution...
1- The changes in characteristics of a species over time. Group of answer choices fitness evolution reproduction natural selection 2- A group of animals that are all the same species is referred to as what? Group of answer choices population kingdom community biosphere 3- Inherited characteristics that enhance an organisms ability to survive and reproduce are called what? Group of answer choices evolution fitness natural selection adaptations 4- How can an adaptation lose its value? Group of answer choices If...
You've discovered a new plant species in which yellow fruit color (Y) is dominant to green fruit color (y), and round fruit shape (R) is dominant to long fruit shape (r).
You've discovered a new plant species in which yellow fruit color (Y) is dominant to green fruit color (y), and round fruit shape (R) is dominant to long fruit shape (r). The genes for fruit color and shape are located on the same chromosome and are 12 m.u. apart. Determine the proportions of progeny of each phenotype produced for this cross. Yr/yR x yr/yr Reset Help 0% 6% 12% 25% 44% 50% 75% 88% 100% 1. yellow, round:25% 2. yellow,...
. Briefly explain the following characteristics for the four groups of plants. Moss: Dominant part of...
. Briefly explain the following characteristics for the four groups of plants. Moss: Dominant part of life cycle: reproduction method and sex organ: adaptation to live on dry land: vascular tissue: (present or absent) seeds: (present or absent) Fern: Dominant part of life cycle: reproduction method and sex organ: adaptation to live on dry land: vascular tissue: (present or absent) seeds: (present or absent) Conifers: Dominant part of life cycle: reproduction method and sex organ: vascular tissue: (explain) seeds: (what...
Which of the following are characteristics of a population? All members belong to the same species....
Which of the following are characteristics of a population? All members belong to the same species. There is interaction between members. There is interbreeding among members. All members live in the same geographic area. All of the above
For each of the following, identify what characteristics of a species male and female karyotype would...
For each of the following, identify what characteristics of a species male and female karyotype would indicate that mechanism of sex-determination. In each case, list a characteristic that would be specific to only that sex-determination system if possible. If no such characteristic is possible, specify which other sex-determination system would show the same karyotype characteristic. (A) Genic (B) XY sex chromosomes: (C) ZW sex chromosomes (D) Haplo-diploid genomes (E) XO sex chromosomes (F) Environmental
Phase Diagram: A sample of moist soil was found to have the following characteristics: Volume: 0.60...
Phase Diagram: A sample of moist soil was found to have the following characteristics: Volume: 0.60 Ft3 (as sampled) Mass: 60 Lb (as sampled) Mass: 50 Lb (after oven drying) Specific Gravity: 2.65 First Question: Find a-h? (show your work) Ratio Porosity Degree of Saturation Wet density (total density) Dry density Volume of Solid Volume of Void Weight of air
Find the characteristics and the minimal polynomial of the following matrices over R, then deduce the their corresponding Jordan Canonical Form J.
Find the characteristics and the minimal polynomial of the following matrices over \( \mathbb{R} \), then deduce the their corresponding Jordan Canonical Form J. A=\( \begin{pmatrix}1&-1&-1\\ 0&0&-1\\ 0&1&2\end{pmatrix}\: \)
Find the characteristics and the minimal polynomial of the following matrices over R , then deduce the their corresponding Jordan Canonical Form J.
Find the characteristics and the minimal polynomial of the following matrices over \( \mathbb{R} \), then deduce the their corresponding Jordan Canonical Form J. B=\( \begin{pmatrix}2&-1&-1&2\\ 0&1&-1&2\\ 2&-5&-1&6\\ 1&-3&-2&6\end{pmatrix} \)
Find the characteristics and the minimal polynomial of the following matrices over R, then deduce their corresponding Jordan Canonical Form J.
Find the characteristics and the minimal polynomial of the following matrices over R, then deduce their corresponding Jordan Canonical Form J. C=\( \begin{pmatrix}1&1&0&0\\ -1&-1&1&0\\ 0&1&1&0\\ -1&-1&1&1\end{pmatrix} \)  
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT