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bazaltina [42]
3 years ago
12

A yellow female Labrador retriever was mated with a brown male. Half of the puppies were brown, and half were yellow. Explain ho

w the same female, when mated with a different brown male, could produce only brown offspring.
O The firsf male was bb Ee, and the second male was bb ee.
O The first male was bb Ee, and the second male was Bb Ee.
O The first male was bb Ee, and the second male was bb EE.
O The first male was bb EE, and the second male was bb Ee.
O The first male was Bb EE, and the second male was Bb Ee.
Biology
1 answer:
Over [174]3 years ago
6 0

Answer:

The first male was bb Ee, and the second male was bb EE.

Explanation:

In Labradors coat colour is controlled by two genes. Suppose the two genes are B and E. B produces black colour and recessive form bb gives brown colour. Gene E is epistatic over gene B in its recessive form which means that ee will produce yellow colour regardless of the genotype present of B gene.

The first case is possible if the female lab is bbee (yellow) and the male lab is bbEe (brown):

bbee X bbEe

        bE      be

be  bbEe    bbee

So half of the offspring will be brown (bbEe) and half of them will be yellow (bbee)

The second case is possible if the same female bbee mates with a brown male of different genotype which can be bbEE:

bbee X bbEE

         bE

be   bbEe

So all offspring will be brown (bbEe)

Hence, the first male was bbEe and the second male was bbEE.

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1. A brown mink crossed with a silver-blue mink produced all brown offspring. When these F1 were crossed among themselves, they
emmasim [6.3K]

Answer:

The correct answer is -

Parental - BB and bb

F1

Genotypes: 100% Bb  

F2

Genotypes: ¼ BB, ½ Bb, ¼ bb

Explanation:

It is given that a cross between a brown and a silver-blue mink produces all brown offspring and it is only possible if brown is dominant over silver blue in true breed cross, assume brown trait is represented by B and silver-blue is b then parents will be

BB and bb

   B    B

b  Bb   Bb

b  Bb   Bb

Cross between F1 individuals: Bb cross Bb

gametes : B, b and B, b

   B     b

B  BB   Bb

b   Bb    bb

the genotype of F2 : 1:2:1 = ¼ BB, ½ Bb, ¼ bb

3 0
3 years ago
Consider a population that is in Hardy–Weinberg equilibrium at a locus with two alleles, A and a, at frequencies of p and q, res
nasty-shy [4]

Answer:

B) 2pq

Explanation:

Hardy-Weinberg equilibrium refers to a model which explains the effect of evolution on the gene pool.

The model is based on the assumptions that if no evolutionary force like genetic drift, natural selection and many other will act on the population and therefore the gene pool  (gene frequency and the genotypic frequency) of a populations remains in equilibrium or constant throughout the generations.

The genotypic frequency in the model is calculated by  

P² =  genotype of a homozygous dominant trait

q² = genotype of a homozygous recessive trait

2pq = genotype of heterozygous trait.

Thus, P²+2pq+q²=1

In the given question, since the population is in equilibrium that is no evolutionary force is acting, therefore, the genotype frequency remains the same that is the frequency of Aa will remain same that is 2pq even after 100 generations.

Thus, Option-B is the correct answer.

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Answer:

B. Changing muscle length.

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Isometric exercises are stationary/static forms of exercise. They include contractions of the muscle without the movement of the affected joints. There is no change in length of the muscles since the joints are static.

This exercise is best fitted for strength maintenance. Examples include squatting,prayer and yoga pose etc.

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