Question

In your large and randomly mating Angus herd, the locus B controls coat color and the locus H controls horn presence or absence. Black individuals are BB, or Bb and red individuals are bb. Horned individuals are HH or Hh and polled individuals are hh. In our herd, 70% of individuals are black and horned, 21% of individuals are black and polled, 5% of individuals are red and horned, and 4% of individuals are red and polled.

A) If you mate an individual with genotype Bbhh to another individual with genotype BbHh. Assuming locus B and locus H are independent, what is the probability of getting a black horned individual?

                     A. 3/8 B. 3/16 C. 9/16 D. 1/4

B) What do you expect the allele frequencies are in your population at both loci?

C) Do these traits in your population support or violate Mendel’s law of independent assortment? Justify your answer with the appropriate calculation.

D) What arrangement of the horned and coat color loci is more likely? Draw it on this homologous chromosome below.

E) Andrew is your favorite Angus bull, and you want to know if Andrew is a homozygote or a heterozygote in locus B. You mate it with one of his daughters having a black coat color. How many offspring would you need to produce to have a 99% probability of detection for the recessive allele b?

F) [TRUE/FALSE] You can be 100% sure that Andrew’s genotype is Bb if one of its offspring has red coat color.

QUESTION HAS ALL INFORMATION PROVIDED.

Answer 1

Solution (A)

Parental Cross- Bbhh x BbHh

Punnet square

	Bh	Bh	bh	bh
BH	BBHh	BBHh	BbHh	BbHh
Bh	BBhh	BBhh	Bbhh	Bbhh
bH	BbHh	BbHh	bbHh	bbHh
bh	Bbhh	Bbhh	bbhh	bbhh

Black Horned Genotypes are - BBHh, BbHh

BBHh-2

BbHh-4

• This means out of 16 progenies 6 are black horned. So, the probability is 6/16 i.e.. 3/8

Solution (B)

• Black polled and red polled individuals are 21 % and 4 % respectively.

That means, BBhh and Bbhh = 21%, bbhh = 4%

• So, the total number of polled genotypes (hh)  in the population is 21+4 i.e.. 25%.
• It means out of 100 individuals 25 individuals having polled genotype ( hh )
• So, genotype frequency of polled individual (hh) is  0.25

In Hardy Weinberg equation

p2+2pq+q2 =1

(p2=HH, 2pq=Hh, q2=hh)

here we got q2=0.25

Hence,

q=square root of 0.25

q= 0.5

"q" is the frequency of "h" allele.

From the equation p+q=1

We, can derive the value of "p" i.e. frequency of "H" allele

p=1-q

p=1-0.5

p= 0.5

So, the allelic frequency of "H" and "h" are 0.5 and 0.5 respectively.

The same way we can get the allelic frequency of B and b

• The percentage of red Horned and red polled individuals are 5% and 4% respectively
• That means, bbHH and bbHh is 5%, bbhh is 4%

So, the total number of red genotypes in the population is 5+4 i.e.. 9%.

It means out of 100 individuals 9 individuals having red genotype ( bb )

• So, genotype frequency of red individual (hh) is  0.09

i.e.. q2 (hh) = 0.09

q= square root of 0.09 = 0.3

"q" is the frequency of "b" allele

Then, p=1-q

p=1-0.3

p=0.7

"p" is the frequency of "B" allele.

• So, the allelic frequency of B and b are 0.7 and 0.3 respectively.

Solution (c)

Chi square test

Observed values ( for a population of 100 individuals)

Observed phenotypes	Number of individuals
Black Horned	70
Black Polled	21
Red Horned	5
Red Polled	4

Expected phenotype

• The Mendelian ratio for dihybrid cross , phenotypic ratio 9:3:3:1 (black horned : black polled : red horned : red polled)

So for a population of 100, expected phenotypes are

Expected	Number of individuals
Black phenotype	(100x9)/16 = 56
Black polled	(100x3)/16= 19
red horned	(100x 3)/16 = 19
red polled	(100x1) /16 = 6

Chi square Test

phenotype	observed (o)	expected (e)	o-e	(o-e)2	(o-e)2 /e
Black horned	71	56	15	225	4.017
black polled	21	19	2	4	0.210
red horned	5	19	-14	196	10.315
red polled	4	6	-2	4	0.666
					chi square value= 15.208

Chi square value is 15.208

1. Degree of freedom for dihybrid crosses is ( number of phenotypes -1), here, (4-1) i.e.. 3.
2. At 3 degrees of freedom and alpha level of 0.05 , the chi square value 15.208 is greater than the table value of 7.815.
3. Hence the null hypothesis is rejected. The difference between expected and observed frequencies are statistically significant.
4. The traits in Populations violate Mendel law of independent assortment.

Solution (D)

Solution (E)

Allelic frequency of "b" is 0.3.

expected "b" individual = 0.3 x Number of Offsprings

99/100 = 3/100 x number of Offsprings

number of Offsprings = 33

Solution (F)

If one of its Offsprings have red coat color, then Andrew must be Bb

Bb x Bb  only gives "bb" individual

So, the answer is true.