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Contact [7]
3 years ago
6

Punnett squares are used to show possible combinations of alleles or to predict the probability of a trait occurring in offsprin

g. Cats that have coats that are a patchwork of black and orange are called tortoiseshell cats, are typically female, and have genes that code for both black fur (Xo) and orange fur (XO) located on the X chromosomes. Y chromosomes do not code for color in tortoiseshell cats. A female cat that has only orange fur (XOXO) is crossed with a male cat that has only black fur (XoY). How many of the offspring will exhibit tortoiseshell coloring? 1 in 4. 2 in 4. 3 in 4. 4 in 4.
Biology
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

2 in 4

Explanation:

If a female cat with orange fur (XOXO) is crossed with a male cat with black fur (XoY) then:

P: XOXO  x  XoY

F1: XOXo  XOY  XOXo  XOY

This means that half of the offspring will be females with tortoiseshell color  (XOXo) and other half  of the offspring will be males with orange fur (XOY).

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Humans can take up available space, disrupt systems, or change water supplies in an ecosystem. What conclusion can you make abou
alexandr402 [8]

Answer:

Human activities affect the flow of energy and matter in an ecosystem and alter the energy balance in ecosystems through the unsustainable nature of what they do.  The energy flow is affected in several ways as a result of pollution, overpopulation, deforestation, burning fossil fuels, etc. Such changes have stimulated soil erosion, climate change, causing water unfit to consume, poor quality of air and so on. When humans cause a change in the energy balance, they impact the ability of the ecosystem to respond and adapt to changes in the environment. It is like getting a cut, but it never heals and grows bigger instead.

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2 years ago
Which scientific discipline is correctly matched to a contribution to society? biology; theory of relativity geology; theory of
ExtremeBDS [4]

Answer:

B) geology; theory of continental drift

Explanation:

6 0
3 years ago
2. Dominant trait: cleft chin (C) Mother’s gametes: Cc
andre [41]

.2. Offspring Genotypes will be Cc or cc.

     Offspring phenotypes : Cleft chin or no cleft chin.

    % chance child will have cleft chin: 50%

3.  % chance child will have arched feet: 25%

4.  % chance child will have blonde hair:  50%

5.  % chance child will have normal vision: 25%

 

Explanation:

CASE 1 :

 Dominant trait: cleft chin (C)

    Recessive trait: lacks cleft chin (c)

    Father’s gametes: cc

    Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and  c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents=  \frac{1}{2}C + \frac{1}{2} c........(i)

Gametes of cc parents =<u> </u>\frac{1}{2}c + \frac{1}{2}c .........(ii)

Combining (i) and (ii) we get,

\frac{1}{2}  Cc + \frac{1}{2} cc                              

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin  =\frac{0.5}{1} ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous   (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

     A from mother, a from father =Aa

     a from mother, A from Father = Aa

     a from mother, a from father = aa

Gametes of Aa parent =\frac{1}{2} A + \frac{1}{2} a

Gametes of other Aa parent = \frac{1}{2} A + \frac{1}{2} a

                                       <u>..................................................................................</u>

                                              \frac{1}{4} AA + \frac{1}{4} Aa

                                                                           +  \frac{1}{4} Aa +\frac{1}{4} aa

                                   <u>..........................................................................................</u>

                                <u>\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa</u>

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = \frac{0.25}{1} × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive  (bb)

Father’s gametes: Heterozygous  (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in  half  Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous  (Ff)

Father’s gametes: Heterozygous  (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF  then phenotype farsightedness

Genotype Ff then phenotype  farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

 

3 0
3 years ago
Some plants have stems called
inna [77]

Answer:

Runners

Explanation:

stems are usually upright and prop up the plant to gather sunlight, however, some stems grow along the ground. These stems are called runners and the plant itself is called a stolon. Runners are involved in asexual reproduction of plants. They have nodes along them which can grow into another plant.

8 0
3 years ago
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What are we doing to for remedy for a F.O.G?
MAVERICK [17]

Answer:

i need more details to answer.

Explanation:

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