During facilitated diffusion - it is still passive but needs help of proteins to help pass.

slavikrds [6]

I believe its A because a bonds with t g bonds with c and c bonds with g, translated to rna it's UCG.

3 0

3 years ago

What is an essential part of the maculae involved in static equilibrium? what is an essential part of the maculae involved in st

lara [203]

Essential part of the maculae involved in static equilibrium is otoliths. The otolith organs sense gravity and linear acceleration such as from due to initiation of movement in a straight line. Persons or animals without otolith organs are imbalanced. The otolithic organs sense motion according to their orientation.

8 0

3 years ago

State three examples of human activities that could endanger botana curus

statuscvo [17]

Botana curus is a plant that produces curus which is a substance that is used to cure different kind of cancer.

This is an endangered species that cannot be grown in the laboratory. The ability to produce curus in the laboratory is not feasible.

The human activities that could have endangered botana curus is:

Pollution: It lead to slow down the rate of growth of the plant.

Overexploitation: The over use of plant for its medicinal purpose.

Habitat: The natural habitat of the plant has been changed due to various human activities.

8 0

3 years ago

Read 2 more answers

At the end of microbiology lab, you decide to practice your Gram-staining skills. The teaching assistant has no fresh cultures b

dangina [55]

Answer:

Loss of cell wall integrity or

Explanation:

Grams staining i a staining method used to classify bacteria into two main classes. Gram positive and Gram negative.

They are differentiated according to the ability of their cell wall to retain the primary stain color of the crystal violet stain or the counter stain pink.

Bacillus subtilis is a Gram positive bacteria hence it is supposed to stain purple.

Therefore, when it stained pink is an error. it shows loss of the cell wall integrity due to old age, overdecolorization or over fixing with heat.

4 0

3 years ago

Michelle was curious about her mother's blood type. Michelle knew that her own blood type was o,

babymother [125]

Blood type is determined by the I gene. IAIA and IAi determines A blood type. IBIB and IBi determines B blood type. ii determines 0 blood type. Michelle's mother must at least carry one recessive allele i.

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The blood type is determined by a gene, which is triallelic.

IA dominant over i. The person presents A type of antigens.

IB dominant over i. The person presents B type of antigens.

IA and IB are codominant. The person presents both types of antigens, A and B.

i is the recessive allele. The person does not present any antigen.

⇒ H0m0zyg0us dominant individuals IAIA, or heter0zyg0us Individuals IAi, will express the A blood type.

⇒ H0m0zyg0us dominant individuals IBIB, or heter0zyg0us Individuals IBi, will express the B blood type.

⇒ Individuals with h0m0zyg0us recessive genotype, ii will express the 0 blood type.

⇒ Individuals carrying IA and IB alleles will express the AB blood type.

To solve the problem, we will use this information.

We know that

Michelle is blood type 0

Father is blood type A

Brother and sister are blood type A

Mother??

⇒ If Michelle is blood type 0, then her genotype is ii. She must have received one recessive allele from her father and one from her mother.

⇒ This suggests that Michelle's father, who is blood type A, is heter0zyg0us IAi. He must have at least one dominant allele IA that determines his phenotype, and one i allele.

⇒ Michelle's brother and sisters are also type A. They might be either h0m0zyg0us dominant IAIA or heter0zyg0us IAi.

⇒ Michelle's mother should have at least one recessive allele i. She could be either IAi or IBi, or ii. There are three possible option for her genotype.

Let us make the Punnet squares to understand it better. We know the genotype of the father, but there are three options for the mother.

Option 1

Cross) father x mother

Parentals) IAi x IAi

Gametes) IA i IA i

Punnett square) IA i

IA IAIA IAi

i IAi ii

F1) 25% of the progeny is expected to be h0m0zyg0us dominant IAIA

50% of the progeny is expected to be heter0zygus IAi

25% of the progeny is expected to be h0m0zyg0us recessive ii

75% of the progeny is expected to be blood type A ⇒ IAIA and IAi ⇒ Sisters and brother

25% of the progeny is expected to be blood tye 0 ⇒ ii ⇒ Michelle

Option 2

Cross) father x mother

Parentals) IAi x IBi

Gametes) IA i IB i

Punnett square) IA i

IB IBIA IBi

i IAi ii

F1) 25% of the progeny is expected to be IAIB

25% of the progeny is expected to be heter0zygus IAi

25% of the progeny is expected to be heter0zygus IBi

25% of the progeny is expected to be h0m0zyg0us recessive ii

25% of the progeny is expected to be blood type AB ⇒ IAIB

25% of the progeny is expected to be blood type A ⇒ IAi ⇒ Sisters and brother

25% of the progeny is expected to be blood type B ⇒ IBi

25% of the progeny is expected to be blood tye 0 ⇒ ii ⇒ Michelle

Option 3

Cross) father x mother

Parentals) IAi x ii

Gametes) IA i i i

Punnett square) IA i

i IAi ii

F1) 50% of the progeny is expected to be heter0zygus IAi

25% of the progeny is expected to be h0m0zyg0us recessive ii

50% of the progeny is expected to be blood type A ⇒ IAi ⇒ Sisters and brother

50% of the progeny is expected to be blood tye 0 ⇒ ii ⇒ Michelle

Despite the fact we can not determine for sure the mother's genotype, we know that she carries at least one recessive allele, which is why Michelle's blood type is 0.

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8 0

2 years ago