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Doss [256]
2 years ago
5

The lineage of pea plants produced round seeds for four generations. The plants that fertilized these pea plants also produced r

ound seeds. However, in the fifth generation, the plant produced to two plants with wrinkled seeds. Can that be possible? Explain your answer using your pedigree chart.
Physics
1 answer:
frutty [35]2 years ago
3 0

Answer:

When two heter0zyg0us individuals are crossed -simple d0minant inheritance-, or a heter0zyg0us individual is crossed with a h0m0zyg0us recessive one, they can produce h0m0zyg0us recessive subjects among the progeny. Heter0zyg0us individuals among the fourth generation got crossed with another plant (h0m0zyg0us recessive or heter0zyg0us) and produced two h0m0zyg0us recessive plants expressing wrinkle seeds.

---------------------------    

The first step is to understand the given information. So,

The first four generations produced round seeds

Plants that fertilized these plants also produced round seeds (see generation II)  

Among the fifth generation there are two plants with wrinkled seeds

Let us assume that a diallelic gene determines the shape of seeds.

D0minant allele R codes for round seeds

Recessive allele r codes for wrinkle seeds

Now, let us analyze the pedigree. According to the information provided here, we can tell that

black figures represent h0m0zyg0us d0minant individuals,

grey figures represent heter0zyg0us individuals

empty figures represent h0m0zyg0us recessive individuals

Remember that Generations are represented with roman numbers.

We will name the plants with numbers from 1 to 26.

According to the information in the statement and the information in the pedigree, we can say that grey individuals from the fourth generation are heter0zyg0us expressing round seeds. They can produce wrinkle seeds, if they are fertilized by another heter0zyg0us plant or a h0m0zygous recessive one.

Option 1:

Two heter0zyg0us plants are crossed, and their offspring have wrinkled seeds.

Option 2:

One heter0zyg0us plant is crossed with a h0m0zyg0us recessive one, and their offspring have wrinkled seeds. Let us see the cross

Look at the attached files for a better understanding      

---------------------------------

Related link: brainly.com/question/2952835?referrer=searchResults

Explanation:

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2. (1) A piece of rubber is 50 cm long when a weight of
Harrizon [31]

Answer

By F = -kx {-ve just indicating the sign of the force}

=>35 = k x (85-50) x 10^-2

=>k = 100 N/m

Again by F = -kx

8 0
3 years ago
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
You have 4.9 g of an unknown substance. To identify the substance, you decide to measure its specific heat and find that it requ
kifflom [539]

The unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Explanation:

When heat energy is supplied to a certain substance, the temperature of the substance increases according to the equation:

Q=mC_s \Delta T

where

Q is the amount of energy supplied

m is the mass of the sample

C_s is the specific heat capacity of the substance

\Delta T is the change in temperature

In this problem, we have

m = 4.9 g is the mass

Q = 668.85 J is the specific heat capacity

\Delta T = 39 K is the change in temperature

Solving for C_s, we find the specific heat capacity of the substance:

C_s = \frac{Q}{m\Delta T}=\frac{668.85}{(4.9)(39)}=3.5 J/gK

Looking at tables of specific heat capacity, we can see that the unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
2 years ago
True or false this is for a test
OleMash [197]

Answer:

true

Explanation:

7 0
3 years ago
Read 2 more answers
how does the size of objects impact the pull of gravity between Earth and a baseball thrown into the air
Reil [10]
it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.
3 0
2 years ago
Read 2 more answers
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