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Naddik [55]
4 years ago
7

How genetic conditions are transmitted from one person to another

Chemistry
1 answer:
Zinaida [17]4 years ago
6 0

Answer:

there are 5 ways this could happen

Autosomal dominant inheritance:  a child recieves a normal gene from one parent and a defective gene from the other parent.

can occur on any of the 22 non-sex chromosomes and have a 50% inheritence rate, gender is not a factor, and disorder differs with inheritance.

examples:  Huntington's disease, neurofibromatosis, achondroplasia, familial hypercholesterolemia

Autosomal recessive inheritance:  both parents carry the defective gene but they are not affected by the disorder.

there is a 25% chance of defective gene from both parents, a 50% chance of inheriting one gene to become a carrier, gender is not a factor in the pattern of the defective gene.

examples:  Tay-Sachs disease, sickle cell anemia, cystic fibrosis, phenylketonuria (PKU)

X-linked (sex-linked) recessive inheritance:  mother carries the affective gene on one of the two X chromosomes.

males inherite X chromosomes from their mothers and Y from their father; which gives the son a 50% chance of inheriting the disorder.

daughters have a 50% chance, but they are not affected by the disorder.

examples:  Hemophilia A, Duchenne muscular dystrophy

X-linked Dominant:  females are affected more so than males; more common for males if they are in the same generation if the mom is affected (because females have two X-chromosomes)

example:  Hypophatemic rickets (Vitiamin Dresistant rickets, ornithine transcarbamylase deficiency.

Mitochondrial:  can affect both males and femlaes, can only be passed by females due to all mitochondria of all children is from the mother, and can appear in every generation.

examples:  Lebrer's hereditary optic neuropathy and Kearns-Sayre syndrome

Explanation:

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beks73 [17]

ANSWER IS CONDUCTION. HOPE THIS HELPED!

8 0
3 years ago
True or False<br>Are coefficients used to balance chemical reactions? ​
Gennadij [26K]

Answer: True

Explanation:

8 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
3) How many micrograms are found in 9.12 x 10-5 L of Water?
Ann [662]

Answer:

86.2 or 431/5

Explanation:

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6 0
3 years ago
For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
4 years ago
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