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2 years ago

Which species is an example of an autotroph?

Biology

2 answers:

diamong [38]2 years ago

8 0

<span>Which species is an example of an autotroph?
</span>
d. hawthorne tree

xxTIMURxx [149]2 years ago

4 0

An autotroph is an example of a producer, something that builds its own nutrients from molecules such as carbon dioxide and water. In this case, the correct answer is d. A hawthorne tree uses the process of photosynthesis to produce nutrients such as simple sugars which can then be used to build organic molecules inside the plant to fuel respiration and other biochemical processes.

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When NAD+ becomes NADH, it is being

ipn [44]

NAD+ accepts a hydrogen ion (H+) and two electrons (2e−), as it becomes reduced to NADH + H+. The NADH moves to the electron transport chain and donates a pair of electrons (becomes oxidized) to the first compound in the chain.

Explanation:

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3 years ago

I would really appreciate it if someone can helpp!!!

skad [1K]

Answer: IM NOT SHORE but when you have to BB its 100% black and the parent of that one will pass on B always. the white mouse is bb this is a resistive trait.

Explanation:i hope i help you sorry if it didnt but i tired my best

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3 years ago

What term is used for a community of plants and animals that live in a particular region?

MAVERICK [17]

A community of plants and animals that live and interact together in a particular region is called an ecosystem.

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3 years ago

Give an example of mechanical and chemical digestion?

Karo-lina-s [1.5K]

Answer:

Mechanical digestion begins in the mouth as the food is chewed. Chemical digestion involves breaking down the food into simpler nutrients that can be used by the cells.

Explanation:

Chemical digestion begins in the mouth when food mixes with saliva. Saliva contains an enzyme (amylase) that begins the breakdown of carbohydrates.

7 0

2 years ago

Read 2 more answers

Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of eight offsp

Aleks [24]

Full question from alternative source:

Albinism in humans is inherited as a simple recessive trait. for the following families, determine the most likely genotypes of the parents

Part A

Two normal parents have five children, four normal and one albino. What are the most likely genotypes of the parents?

Part B

A normal male and an albino female have six children, all normal. What is the most likely genotypes of the parents?

Part C

A normal male and an albino female have six children, three normal and three albino. What are the most likely genotypes of the parents?

Part D

Suppose that one of the normal children from Part B and one of the albino children from Part C become the parents of 8 offspring. What would be the expected number of normal and albino offspring?

Answer:

Part A - both heterozygous for the trait (Aa)

Part B - female aa, male likely AA

Part C - female aa, male Aa

Part D - 4 normal, 4 albino (50:50)

Explanation:

Part A

We are told that albinism is a recessive trait. Lets call it a and the normal allele A. Since neither of the parents have it, that must mean they each possess at least one copy of the dominant allele (either AA or Aa). Most of their children are normal, but 1 is albino. The albino child must have inherited two copies of the a allele, one from each parent. This means each parent must possess an a allele and must both be Aa.

Part B

Since the trait is recessive, we already know that the albino female must have the genotype aa. If her husband in homozygous normal (AA), all their children will be heterozygous (Aa), and therefore normal, which is the case in this cross. Therefore aa and AA are the most likely genotypes. However, it is still possible that the male has the genotype Aa. This would mean they had a 50:50 chance of producing an albino offspring. We know they have six normal children, so this is unlikely.

Part C

Since the trait is recessive, we know the female has the genotype aa. Unlike the last family, this family has some albino children. This means that the husband must be a carrier of the trait, but he is unaffected, so must be Aa.

As described in part B, if he is Aa and she is aa, they will have 50:50 normal:albino children. This is the ratio in this family.

Part D

The normal children from Part B are all heterozygous for the trait (Aa), inheriting one normal allele from their father, and one albino allele from their mother. The albino children from part C are all aa. Therefore, the cross is Aa x aa. We can draw a punnett square:

A a

a Aa aa

aa a Aa aa

Half the children have the genotype Aa, and half have the genotype aa. That means half will be normal and half will be albino. Therefore, 4 normal, 4 albino.

4 0

3 years ago